Fluent Essays

1 Statistics II Midterm Ch 10-14 Name______________________ Dr. C. Monticelli Show all work as done on the Practice Midterm Assume all populations normal. 1) You must use the TI83/84 (or plus calculator); failure to do so will result in grade of 0. *Conclusions must have ‘support’ or ‘reject’ and the word ‘claim’ as done on my materials. *Interpret confidence intervals must be as done on my materials. PROBLEMS MUST BE DONE AS SHOWN ON MY MATERIALS FOR CREDIT! 2) Please handwrite the solutions in blue or black pen on the Midterm. Scan in the Midterm Solutions with a scanner from home, an office store like Staples, or with a free phone app such as CamScanner, Genius Scan etc.; these apps allow you to save pages into a single, multi-page pdf, which is preferred. OR Type the answers in MS Word copying and pasting from the Symbols link. Save your Midterm as a .pdf file. Please do NOT attach photos as they are hard to see and grade. 3) Click on SUBMIT ASSIGNMENT, choose file, type phone number in comments box, SUBMIT. 4) Please do NOT email me your Midterm. 5) NO late Midterms will be accepted. 6) I will confirm receipt of all Midterms. If you did not receive an email confirmation or grade posting within 24 hours, then I did not receive your Midterm and you must contact me asap. 7) All work on the Midterm must be your own; no joint efforts allowed. ---------------------------------------------------------------------------------------------------------------------------- 1. a) Two different schools create their own versions of the same aptitude test, and a Department Chair administers both versions to the same randomly selected subjects with the results given below. At the .01 level of significance, test the claim that both versions produce the same mean. Did you subtract before – after or after – before? _______________________ claim ………………………………................ ________________________ null hypothesis…………………………………. ________________________ alternative hypothesis………………………….. ________________________ Calculator Screen Name……………………… ________________________ test statistic ………………………… ________________________ pvalue/alpha comparison………………………. ________________________ decision …………………………. ________________________ Conclusion …………………………. ________________________ b) Construct a 99\% confidence interval for, . Interpret interval in a sentence. Confidence Interval Name__________________________________ Interval___________________________________________ Interpret_____________________________________________ TestB(before) 109 118 104 127 126 99 104 108 113 TestC(after) 102 115 107 116 104 91 113 112 112 µd 2 2. Test the claim that the variances are the same. Use a .05 level of significance. claim ………………………………................ ________________________ null hypothesis…………………………………. ________________________ alternative hypothesis………………………….. ________________________ Calculator Screen Name……………………… ________________________ test statistic ………………………… ________________________ pvalue/alpha comparison………………………. ________________________ decision …………………………. ________________________ Conclusion …………………………. ________________________ 3. a) Two types of flares are tested for their burning times (in min) and sample results are given below. a) Test the claim that Brand Z has a mean greater than Brand W. Use a .03 significance level. Assume populations normal. claim ………………………………................ ________________________ null hypothesis…………………………………. ________________________ alternative hypothesis………………………….. ________________________ Calculator Screen Name……………………… ________________________ test statistic ………………………… ________________________ pvalue/alpha comparison………………………. ________________________ decision …………………………. ________________________ Conclusion …………………………. ________________________ BrandZ BrandW n1 = 30 n2 = 20 x1 = 61.8 x2 == 67.3 s1 =11.9 s2 = 6.4 BrandZ BrandW n1 = 25 n2 = 30 x1 = 20.4 x2 ==16.1 σ1 =1.5 σ2 = .9 3 b) Construct a 97\% confidence interval for . Interpret the interval. Confidence Interval Name__________________________________ Interval___________________________________________ Interpret ________________________________________ 4. a) Test the claim that the mean for Brand Z is greater than Brand W at the .04 significance level. Assume both populations are normal and the variances are equal. claim ………………………………................ ________________________ null hypothesis…………………………………. ________________________ alternative hypothesis………………………….. ________________________ Calculator Screen Name……………………… ________________________ test statistic ………………………… ________________________ pvalue/alpha comparison………………………. ________________________ decision …………………………. ________________________ Conclusion …………………………. ________________________ b) Construct a 96\% confidence interval for based on the sample data above. Interpret the interval in a complete sentence. Confidence Interval Name__________________________________ Interval___________________________________________ Interpret_____________________________________________ µ1 − µ2 BrandZ BrandW n1 =15 n2 = 25 x1 = 67.3 x2 = 61.8 s1 = 4.4 s2 =11.9 µ1 − µ2 4 5 a) A study is made of the defect rates of two machines used in manufacturing. Of 300 randomly selected items produced by the first machine, 7 are defective. Of 350 randomly selected items produced by the second machine, 20 are defective. At the .01 level of significance, test the claim that the two machines have the different rate of defects. claim ………………………………................ ________________________ null hypothesis…………………………………. ________________________ alternative hypothesis………………………….. ________________________ Calculator Screen Name……………………… ________________________ test statistic ………………………… ________________________ pvalue/alpha comparison………………………. ________________________ decision …………………………. ________________________ Conclusion …………………………. ________________________ b) Construct a 99\% confidence interval for . Interpret the interval in a complete sentence. Confidence Interval Name__________________________________ Interval___________________________________________ Interpret_____________________________________________ p1 − p2 5 6 a) Test the claim that Brand Z and Brand W have the different means. Use the .04 level. Assume the variances are different and the populations normal. claim ………………………………................ ________________________ null hypothesis…………………………………. ________________________ alternative hypothesis………………………….. ________________________ Calculator Screen Name……………………… ________________________ test statistic ………………………… ________________________ pvalue/alpha comparison………………………. ________________________ decision …………………………. ________________________ Conclusion …………………………. ________________________ b) Construct a 96\% confidence interval for . Interpret the interval in a complete sentence. Confidence Interval Name__________________________________ Interval___________________________________________ Interpret_____________________________________________ BrandZ BrandW n1 = 25 n2 = 50 x1 = 87.3 x2 = 81.8 s1 = 7.4 s2 =11.9 µ1 − µ2 6 7. Listed below are results from two different tests designed to measure achievement. (x)TestB 64 48 51 59 60 43 41 42 35 50 45 (y) testC 91 68 80 92 91 67 65 67 56 78 71 a. Plot the scatter diagram below. Label x and y axes. Do a rough sketch. b. Find the value of the linear correlation coefficient r by the TI83 shortcut- state calculator screen name c) Test the claim of no linear relation by the TI83 p-value method. = .01 claim ………………………………................ ________________________ null hypothesis…………………………………. ________________________ alternative hypothesis………………………….. ________________________ Calculator Screen Name……………………… ________________________ test statistic ………………………… ________________________ pvalue/alpha comparison………………………. ________________________ decision …………………………. ________________________ Conclusion …………………………. ________________________ d) Find the estimated equation of the regression line by TI83 shortcut e) Plot the regression line on the scatter diagram in part a). f) Assuming a significant linear correlation, predict the score a student would get on Test C, given he got a 37 on test B. g) What percentage of the total variation can be explained by the regression line? α 7 8. Responses to a survey question are broken down according to employment and the sample results are given below. At the .05 significance level, test the claim that the response and employment status are independent. Yes No Undecided Employed 40 25 5 Unemployed 30 15 7 claim ………………………………................ ________________________ null hypothesis…………………………………. ________________________ alternative hypothesis………………………….. ________________________ Calculator Screen Name……………………… ________________________ test statistic ………………………… ________________________ pvalue/alpha comparison………………………. ________________________ decision …………………………. ________________________ Conclusion …………………………. ________________________ 9. In studying the occurrence f genetic characteristics, the following sample data were obtained. At the .04 significance level, test the claim that the characteristics occur with the same frequency claim ………………………………................ ________________________ null hypothesis…………………………………. ________________________ alternative hypothesis………………………….. ________________________ Calculator Screen Name……………………… ________________________ test statistic ………………………… ________________________ pvalue/alpha comparison………………………. ________________________ decision …………………………. ________________________ Conclusion …………………………. ________________________ Characteristic B C D E F G frequency 38 40 55 45 35 49 8 10. At the .02 significance level, test the claim that the three brands have the same mean level if the following sample results have been obtained. claim ………………………………................ ________________________ null hypothesis…………………………………. ________________________ alternative hypothesis………………………….. ________________________ Calculator Screen Name……………………… ________________________ test statistic ………………………… ________________________ pvalue/alpha comparison………………………. ________________________ decision …………………………. ________________________ Conclusion …………………………. ________________________ BrandA BrandB BrandC 32 27 22 34 24 25 37 33 32 33 30 22 36 21 39 1 Statistics II Sample Project/Practice Midterm Ch 10-14 Name__________________ Dr. C. Monticelli Show all work as done on my materials. Assume all populations normal. 1. a) The table below shows the weights of seven subjects before and after following a particular diet for two months. Using a .01 level of significance, test the claim that the diet is effective in reducing weight. Did you subtract before – after or after – before? ________________________ claim ………………………………................ ________________________ null hypothesis…………………………………. ________________________ alternative hypothesis………………………….. ________________________ Calculator Screen Name……………………… ________________________ test statistic ………………………… ________________________ pvalue/alpha comparison………………………. ________________________ decision …………………………. ________________________ Conclusion …………………………. ________________________ b) Construct a 98\% confidence interval for, . Interpret the interval in a complete sentence. Confidence Interval Name__________________________________ Interval___________________________________________ Interpret_____________________________________________ Subject A B C D E F G Before 187 156 153 194 195 179 157 After 180 147 151 199 181 181 145 µd 2 2. A random sample of 16 women resulted in blood pressure levels with a standard deviation of 21.9 mm Hg. A random sample of 17 men resulted in blood pressure levels with a standard deviation of 20 mm Hg. Use a .025 significance level to test the claim that the blood pressure level for women has a larger variance than those for men. Use women as group one. claim ………………………………................ ________________________ null hypothesis…………………………………. ________________________ alternative hypothesis………………………….. ________________________ Calculator Screen Name……………………… ________________________ test statistic ………………………… ________________________ pvalue/alpha comparison………………………. ________________________ decision …………………………. ________________________ Conclusion …………………………. ________________________ 3. a) The Better Cookie Company claims its chocolate chip cookies have more chips than another chocolate chip cookie. 120 Better Cookies and 100 of the other type of cookie were randomly selected and the number of chips in each cookie was recorded. At the .02 level of significance test the claim that the population of Better Cookies has a higher mean number of chips. Group one is Better Cookies. claim ………………………………................ ________________________ null hypothesis…………………………………. ________________________ alternative hypothesis………………………….. ________________________ Calculator Screen Name……………………… ________________________ test statistic ………………………… ________________________ pvalue/alpha comparison………………………. ________________________ decision …………………………. ________________________ Conclusion …………………………. ________________________ Better Another SampleMean#ChocChips 7.6 6.9 populationSD 1.4 1.7 3 b) Construct a 96\% confidence interval for . Interpret the interval. Confidence Interval Name__________________________________ Interval___________________________________________ Interpret ________________________________________ 4. a) Test the claim that the mean for Sample A is less than Sample B at the .01 significance level. Assume both populations normal and the variances are equal. claim ………………………………................ ________________________ null hypothesis…………………………………. ________________________ alternative hypothesis………………………….. ________________________ Calculator Screen Name……………………… ________________________ test statistic ………………………… ________________________ pvalue/alpha comparison………………………. ________________________ decision …………………………. ________________________ Conclusion …………………………. ________________________ b) Construct a 98\% confidence interval for based on the sample data above. Interpret the interval in a complete sentence. Confidence Interval Name__________________________________ Interval___________________________________________ Interpret_____________________________________________ µ1 − µ2 SampleA SampleB x1 = 24.7 x2 = 24.6 s1 = 5.7 s2 = 4.2 n1 =10 n2 =8 µ1 − µ2 4 5 a) A sample of 50 randomly selected men with high triglyceride levels consumed 2 tablespoons of oat bran daily for 6 weeks. After 6 weeks, 60\% of the men had lowered their triglyceride level. A sample of 80 men consumed 2 tablespoons of wheat bran for six weeks. After six weeks, 25\% had lower triglyceride levels. Test the claim that there is a significant difference in the two proportions at the .01 level. Let Oat Bran be Group 1. claim ………………………………................ ________________________ null hypothesis…………………………………. ________________________ alternative hypothesis………………………….. ________________________ Calculator Screen Name……………………… ________________________ test statistic ………………………… ________________________ pvalue/alpha comparison………………………. ________________________ decision …………………………. ________________________ Conclusion …………………………. ________________________ b) Construct a 99\% confidence interval for . Interpret the interval in a complete sentence. Confidence Interval Name__________________________________ Interval___________________________________________ Interpret_____________________________________________ p1 − p2 5 6 a) Test the claim that populations A and B have equal means. Use the .05 level. Assume the variances are not equal and populations normal. claim ………………………………................ ________________________ null hypothesis…………………………………. ________________________ alternative hypothesis………………………….. ________________________ Calculator Screen Name……………………… ________________________ test statistic ………………………… ________________________ pvalue/alpha comparison………………………. ________________________ decision …………………………. ________________________ Conclusion …………………………. ________________________ b) Construct a 95\% confidence interval for . Interpret the interval in a complete sentence. Confidence Interval Name__________________________________ Interval___________________________________________ Interpret_____________________________________________ SampleA SampleB n1 = 32 n2 = 37 x1 =130 x2 =160 s1 = 65 s2 = 30 µ1 − µ2 6 7.Listed below are results from two different tests designed to measure productivity and dexterity for randomly selected employees. a. Plot the scatter diagram below. Label x and y axes. Do a rough sketch. b. Find the value of the linear correlation coefficient r by the TI83 shortcut- state calculator screen name c) Test the claim of no linear relation by the TI83 p-value method. = .05 claim ………………………………................ ________________________ null hypothesis…………………………………. ________________________ alternative hypothesis………………………….. ________________________ Calculator Screen Name……………………… ________________________ test statistic ………………………… ________________________ pvalue/alpha comparison………………………. ________________________ decision …………………………. ________________________ Conclusion …………………………. ________________________ d) Find the estimated equation of the regression line by TI83 shortcut e) Plot the regression line on the scatter diagram in part a). f) Assuming a significant linear correlation, predict the score a student would get on dexterity, given he got an 80 on productivity. g) What percentage of the total variation can be explained by the regression line? Pr oductivity( x) 59 63 65 69 58 77 76 69 70 64 Dexterity( y) 72 67 78 82 75 87 92 83 87 78 α 7 8. Responses to a survey question are broken down according to employment and the sample results are given below. At the .10 level of significance, test the claim that the response and employment are independent. claim ………………………………................ ________________________ null hypothesis…………………………………. ________________________ alternative hypothesis………………………….. ________________________ Calculator Screen Name……………………… ________________________ test statistic ………………………… ________________________ pvalue/alpha comparison………………………. ________________________ decision …………………………. ________________________ Conclusion …………………………. ________________________ 9. In analyzing the random number generator of a certain computer, the following results were obtained. At the .05 significance level, test the claim that the outcomes occur with the percentages 20\%, 10\%, 15\%, 15\%, 20\%, 20\%. claim ………………………………................ ________________________ null hypothesis…………………………………. ________________________ alternative hypothesis………………………….. ________________________ Calculator Screen Name……………………… ________________________ test statistic ………………………… ________________________ pvalue/alpha comparison………………………. ________________________ decision …………………………. ________________________ Conclusion …………………………. ________________________ Yes No Undecided employed 15 35 20 unemployed 25 25 10 outcome 1 2 3 4 5 6 frequency 18 12 14 16 21 15 8 10. At the .01 significance level, test the claim that the three brands have the same mean level if the following sample results have been obtained. claim ………………………………................ ________________________ null hypothesis…………………………………. ________________________ alternative hypothesis………………………….. ________________________ Calculator Screen Name……………………… ________________________ test statistic ………………………… ________________________ pvalue/alpha comparison………………………. ________________________ decision …………………………. ________________________ Conclusion …………………………. ________________________ BrandA BrandB BrandC 44 30 28 47 32 27 44 34 31 40 36 32 39 38 36 40 42 ud or ud Chi-Squared or x 2 L1 or L1 L2 or L2 L3 or L3 = L4 or L4 Chi-Squared CDF or x2 CDF < > squared or 2 squared or 2 u1 or u1 u2 or u2 u3 or u3 u4 or u4 p1 or p1 p2 or p2 r squared or r2 y α ≤ ≥ ≠ ÷ 1σ 1σ 2σ 2σ ρ 1 Chapter 9 HYPOTHESIS TESTS ABOUT ONE PARAMETER 9.2 Z TEST for a MEAN: testing claim about one mean and is known and the population is normal OR for any population when is known and n 30. Sample Problem: GIVEN SAMPLE STATISTICS *A survey claims that the average cost of a hotel room in Atlanta is $69.21. To test the claim, a researcher selects a sample of 30 hotel rooms and finds that the average cost is $68.43. The population standard deviation is $3.72. Test the above claim using =.05. Assume population normal. Test Statistic (TS) Z = -1.15 Test Statistic Z = -1.15 p=.2508 > =.05 do not reject H There is not enough evidence to reject the claim. Ho might be true. σ σ ≥ a 0 1 : 69.21 : 69.21 : 69.21 Claim H H µ µ µ = = ¹ a 0 Entry Display [STAT] [>][>]TESTS [1] for 1:Z-Test Calculate [ENTER] P=.2508 z= -1.15 (TS) 2 Sample Problem: GIVEN DATA *The average 1 –year old (both sexes) is 29 inches tall. A random sample of 30 one-year olds in a large day care franchise resulted in the following heights. At =.05 , test the claim that the average height differs from 29 inches. The population standard deviation is 2.6108. Assume the population is normal. 25 32 35 25 30 26.5 26 25.5 29.5 32 30 28.5 30 32 28 31.5 29 29.5 30 34 29 32 27 28 33 28 27 32 29 29.5 Entry Display Enter data into list one. [STAT] [>][>] TESTS [1] for 1:Z-Test Z-Test Inpt: Data [Stats] :29 : 2.6108 List: Freq:1 Calculate [ENTER] z= .944 (TS) p=.3451 Test Statistic Z = .944 p=.3451 > =.05 do not reject H There is not enough evidence to support the claim. Ho might be true. a σ 0 1 : 29 : 29 : 29 Claim H H µ µ µ ¹ = ¹ µο σ 1L :µ µo¹ a 0 3 9.3 T TEST for a MEAN: testing a claim about one mean and is not known and the population is normal OR for any population when is not known and n 30. Sample Problem: GIVEN STATISTICS *The average amount of rainfall during the summer months for the northeast part of the United States is less than 11.52 inches. A researcher selects a random sample of 10 cities in the northeast and finds that the average amount of rainfall for 1995 was 7.42 inches and the standard deviation was 1.3 inches. At =.05 test the above claim. Assume the population is normal. Test Statistic (TS) t = -9.97 Entry Display [STAT] [>][>] TESTS [2] for 2:T -Test Calculate [ENTER] t= -9.97 (TS) p= 1.8E-6 Test Statistic t =-9.97 p=1.8E-6 < =.05 reject H There is enough evidence to support the claim. H1 is true. σ σ ≥ a Claim:µ <11.52 H0 :µ =11.52 H1 :µ <11.52 a 0 4 Sample Problem: GIVEN DATA In one part of a test developed by a psychiatrist, the test subject is asked to form a word by unscrambling letters. Given are the times (in seconds) required by 15 randomly selected persons to unscramble letters (50, 16, 17, 50, 25, 52, 60, 40, 33, 30, 76, 16, 30, 74, 59) At the = .05 level of significance, test the claim that the mean time is equal to 50 seconds. Assume the population is normal. Input the times into a list, say L1. Press [STAT] [>] [TESTS] [T-Test] Highlight DATA and press [ENTER] Input the following: Highlight calculate and press enter. Test Statistic t= -1.57 p=.1395 > do not reject Ho There is not enough evidence to reject the claim. Ho might be true. a Claim:µ = 50 H0 :µ = 50 H1 :µ ≠ 50 a 5 9.4 Z Test for Proportion: 1Prop Z Test: To test a claim about one proportion. Sample Problem: GIVEN X AND N *A telephone company wants to advertise that more than 30\% of all its customers have more than two telephones. To support this ad, the company selects a sample of 200 customers and finds that 72 have more than two telephones. Test the claim at =.05 . Assume the population is normal. Test Statistic (TS) Entry Display [STAT] [>][>] TESTS [5] for 5:1-PropZTest 1-PropZTest po: .3 x: 72 n: 200 prop: > po Calculate [ENTER] z=1.85 (TS) p=.0320 Test Statistic Z = 1.85 p=.0320< =.05 reject H There is enough evidence to support the claim. H1 is true. Note: In some problems (such as the next problem) instead of being given x and n, you are given n and the sample proportion. To find x, simply multiply n and the sample proportion. a Claim:p > .30 H0 :p = .30 H1 :p > .30 a 0 6 Sample Problem: GIVEN SAMPLE PROPORTION AND N *The American Automobile Association (AAA) claims that 54\% of fatal car/truck accidents are caused by driver error. A researcher studies 30 randomly selected accidents and finds 47\% were caused by driver error. Test the above claim at =.05 . Assume the population is normal. n = 30 =.47 (round to the nearest whole number: .5-.9 ROUND UP, and .1 - .4 ROUND DOWN) Test Statistic (TS) Entry Display [STAT] [>][>] TESTS [5] for 5:1-PropZTest 1-PropZTest po: .54 x: 14 n: 30 prop: po Calculate [ENTER] z=-.806 (TS) p=.420 Test Statistic Z = -.806 p=.4203 > =.05 do not reject H There is not enough evidence to reject the claim. Ho might be true. a p̂ x = p̂ ⋅n = (.47)(30) =14.1→14 Claim:p = .54 H0 :p = .54 H1 :p ≠ .54 ¹ a 0 7 Ch 10 &11 HYPOTHESIS TESTS/CONFIDENCE INTERVALS ABOUT TWO PARAMETERS 11.1 and 10.1 2 SAMPLE Z TEST: Testing a Claim about Two Means: Independent Samples, known, Both populations normal OR both samples are at least 30. Sample Problem: GIVEN STATISTICS *In a study of women science majors, the following data were obtained on two groups, those who left their profession within a few months after graduation (leavers) and those who remained in their profession after they graduated (stayers). Test the claim that those who stayed had a higher science grade-point average than those who left. Use =.05. Assume populations are normal. [STAT][>][>]TESTS [3] for 3: 2-SampZTest 2-SampZTest Inpt: Stats TS Z = -2.01 p = .0222 < =.05 Reject the null hypothesis. There is enough evidence to support the claim. H1 is true. σ1 and σ 2 a Leavers Stayers n1 =103 n2 = 225 x1 = 3.16 x2 = 3.28 σ1 = .52 σ2 = .46 claim:µ1 < µ2 H0 :µ1 = µ2 H1 :µ1 < µ2 1 2 1 2 :.52 :.46 1:3.16 1:103 2:3.28 2:225 : X n X n Calculate s s µ µ< a 8 Note: To do a confidence interval, simply choose 2SampZInt. Sample Problem: GIVEN DATA *A researcher claims that the average length of the major rivers in the United States is the same as the average length of the major rivers in Europe. The data (in miles) of a sample of rivers are shown. At =.01, test the above claim. Given and . Assume populations are normal. United States Europe 729 560 434 481 724 820 329 332 360 532 357 505 450 2315 865 1776 1122 496 330 410 1036 1224 634 230 329 800 447 1420 326 626 600 1310 652 877 580 210 1243 605 360 447 567 252 525 926 722 824 932 600 850 310 430 634 1124 1575 532 375 1979 565 405 2290 710 545 259 675 454 300 470 425 Enter US data in L1 and Europe data in L2 [STAT][>][>]TESTS [3] for 3: 2-SampZTest 2-SampZTest Inpt: Data TS Z = -.856 p=.3920 > =.01 Do not reject the null hypothesis. a σ1 = 449.8703 σ 2 = 474.1258 1 2 0 1 2 1 1 2 : : : claim H H µ µ µ µ µ µ = = ¹ 1 2 1 2 :449.8703 :474.1258 1: 1 2: 2 1:1 2:1 : List L List L Freq Freq Calculate s s µ µ¹ a 9 There is not enough evidence to reject the claim. Ho might be true. _________________________________________________________________________________________________ 2 SAMPLE T TEST: Testing the Claim about Two Means with Independent Samples and not known, both populations normal OR both sample sizes are at least 30. Sample Problem: GIVEN STATISTICS *A real estate agent wishes to determine whether tax assessors and real estate appraisers agree on the values of homes. Summary statistics for a random sample of the two groups is below. Test the claim that there is a significant difference in the values of homes for each group. Use = .05. Assume both populations are normal and the variances are the same. Assume populations are normal. [STAT][>][>] TESTS [4] for 4: 2 SampTTest 2-SampTTest Inpt: Stats __________________________ t = -4.02 (TS) p = 8.03 E-4 < = .05 Reject the null hypothesis. There is enough evidence to support the claim. H1 is true. σ1 and σ 2 a REstate Tax Assessors n1 =10 n2 =10 x1 = $83,256 x2 = $88,354 S1 = $3256 s2 = $2341 1 2 0 1 2 1 1 2 : : : claim H H µ µ µ µ µ µ ¹ = ¹ 1 2 1:83256 1:3256 1:10 2:88354 2:2341 2:10 : : X sx n X sx n Pooled Yes Calculate µ µ¹ a 10 NOTE: Since the variances are given to be equal, we say YES to POOLED. NOTE: If the variances are given to be unequal, we say NO to POOLED. Example) To construct a 95\% Confidence Interval using the above problem: [STAT] [>][>] TESTS [O] for O: 2-Samp TInt 2-SampTInt Inpt: Stats Enter data same as above. C-level: .95 Pooled: Yes Calculate (-7762, -2434) Sample Problem: GIVEN DATA *A health care worker wishes to see if the average number of family day care homes per county is greater than the average number of day care centers per county. The number of centers for a selected sample of counties is shown. At = .01 test the above claim. Assume both populations are normal and the variances are equal. Number of Family day care homes Number of day care centers 25 57 34 5 28 37 42 21 44 16 16 48 Enter data for homes in L1 and centers in L2 [STAT][>][>] TESTS [4] for 4: 2 SampTTest 2-SampTTest Inpt: Data t = 1.44 (TS) p = .0895 > = .01 Do not reject the null hypothesis. There is not enough evidence to support the claim. Ho might be true. a claim:µ1 > µ2 H0 :µ1 = µ2 H1 :µ1 > µ2 1 2 1: 1 2: 2 1:1 2:1 : : List L List L Freq Freq Pooled Yes Calculate µ µ> a 11 11.2 and 10.2 Test Difference Between Two Means Dependent Samples BEFORE/AFTER means Dependent samples. Both populations normal OR both sample sizes are at least 30. In a before/after test, this two-sample test reduces to a one-sample t test on the differences. Sample Problem: claim after scores less *A new composition teacher wishes to see whether a new grammar program will reduce the number of grammatical errors her students make when writing a two- page essay. The data is shown below. Can it be concluded that the number of errors has been reduced? Assume both populations are normal. Use = .025 Student 1 2 3 4 5 6 Errors before 12 9 0 5 4 3 Errors After 9 6 1 3 2 3 Since claiming that the errors has reduced, assuming that is true, such as 10 before and 5 after. Then after – before = 5 – 10 = - 5, negative five which is less than zero. That is why the claim is stated as the mean of the differences less than zero. Enter Before data in L1 and After data in L2 Let L3 = L2 –L1 (AFTER – BEFORE) In list mode, using arrows on keypad, highlight L3, press enter, type L2-L1 ENTER. a claim:µd < 0 H0 :µd = 0 H1 :µd < 0 12 [STAT] [>][>]TESTS [2] for 2:T-Test T-Test Inpt: Data TS t = -2.24 p = .0378 > = .025 Do not reject the null hypothesis There is not enough evidence to support the claim. Ho might be true. _________________________________________________________________ Alternative way to do the previous problem Note: If I Let L3 = L1 – L2 (BEFORE – AFTER) then Since claiming that the errors has reduced, assuming that is true, such as 10 before and 5 after. Then before – after = 10 – 5 = 5, positive five which is greater than zero. That is why the claim is stated as the mean of the differences greater than zero. Enter Before data in L1 and After data in L2 Let L3 = L1 –L2 In list mode, using arrows on keypad, highlight L3, press enter, type L1-L2 ENTER. [STAT] [>][>]TESTS [2] for 2:T-Test T-Test Inpt: Data TS t = 2.24 p = .0378 > = .025 Do not reject the null hypothesis There is not enough evidence to support the claim. Ho might be true. µ0 :0 List :L3 Freq :1 µ :< µ0 Calculate a claim:µd > 0 H0 :µd = 0 H1 :µd > 0 0 3 0 :0 : :1 : List L Freq Calculate µ µ µ> a 13 IMPORTANT SUMMARY OF THE 3 CASES in Before/After Hypothesis Tests In a before/after test, this two-sample test reduces to a one-sample t test on the differences. There are 3 cases explained below. L1 = before data L2 = after data 1) Test the claim that the number of errors has been increased. For ex: before 7 and after 9. L3= L2 – L1 =after – before = 9 - 7 = +2 positive or > 0 Claim greater than zero Claim: >0. (The mean of the differences greater than zero) OR----------------------- L3 = L1 – L2= before – after = 7 - 9 = -2 negative or < 0 Claim less than zero Claim: <0 (The mean of the differences less than zero.) 2) Test the claim that the number of errors has been reduced. For ex: before 9 and after 7. L3= L2 – L1 =after – before = 7 - 9 = -2 negative or < 0 Claim less than zero Claim: <0 (The mean of the differences less than zero.) OR----------------------- L3 = L1 – L2= before – after=9 – 7 = + 2 positive or > 0 Claim greater than zero Claim: >0. (The mean of the differences greater than zero) 3) Test the claim that the number of errors is the same before vs. after. ex: before 9 and after 9. L3= L2 – L1 =after – before = L3 = L1 – L2= before – after= 9 – 9 = 0 Claim equals zero Claim: = 0. (The mean of the differences equals zero.) NOTE: To construct a confidence interval for the above, simply use TInterval with the above data. ------------------------------------------- µd µd µd µd µd 14 11.3 and 10.3 Testing the Claim about Two Proportions: 2-Prop Z Test Sample Problem: GIVEN THE SAMPLE PROPORTION AND N *In a sample of 80 Americans, 55\% wished that they were rich. In a sample of 90 Europeans, 45\% wished that they were rich. At = .01 test the claim that there is a difference between the two groups. Assume populations are normal. Note: X1 = .55 *80 = 44 X2 = .45*90= 40.5 or 41 _________________________________ [STAT][>][>]TESTS [6] for 6:2-PropZTest 2-PropZTest TS Z = 1.23 P = .2190 > = .01 Do not reject the null hypothesis There is not enough evidence to support the claim. Ho might be true. To calculate the 99\% Confidence Interval: [STAT][>][>]TESTS [ALPHA][B] for B: 2-PropZInt 2-PropZInt Enter data same as above. C-level: .99 Calculate (-.1026, .29145) a 1 2 0 1 2 1 1 2 : : : claim p p H p p H p p ¹ = ¹ ˆx p n= × 1 2 1:44 1:80 2:41 2:90 : X n X n p p Calculate ¹ a 15 Sample Problem: GIVEN X AND N *A survey of 80 homes in a Washington, D.C., suburb showed that 45 were air- conditioned. A sample of 120 homes in a Pittsburgh suburb showed that 63 had air- conditioning. At = .05 test the claim that there is a difference in the two proportions. Assume populations are normal. [STAT][>][>]TESTS [6] for 6:2-PropZTest 2-PropZTest Test Statistic Z = .521 P = .6022 > = .05 Do not reject the null hypothesis There is not enough evidence to support the claim. Ho might be true. a 1 2 0 1 2 1 1 2 : : : claim p p H p p H p p ¹ = ¹ 1 2 1:45 1:80 2:63 2:120 : X n X n p p Calculate ¹ a 16 11.4 Testing the Claim about Two Variances: 2 Sample F Test Sample Problem: GIVEN DATA * A tax collector wishes to see if the variances of the values of the tax-exempt properties are different for two large cities. The values of the tax-exempt properties are shown below. The data are given in millions of dollars. At = .05, is there enough evidence to support the tax collector’s claim that the variances are different? Assume populations are normal. City A City B 113 22 14 8 82 11 5 15 25 23 23 30 295 50 12 9 44 11 19 7 12 68 81 2 31 19 5 2 20 16 4 5 Enter City A data in L1 and City B data in L2 TI83 [STAT][>]CALC [1] for 1:1-Var Stats [2nd] [L1] [ENTER] see TI83 [STAT][>]CALC [1] for 1:1-Var Stats [2nd] [L2] [ENTER] see (It is customary to let larger standard deviation be s ) TI84 [STAT][>][>]TESTS [APLHA] [D] or [E] for 2-SampFTest 2-Samp F Test Inpt: Stats a 2 2 1 2 2 2 0 1 2 2 2 1 1 2 : : : claim H H s s s s s s ¹ = ¹ 225.97xs s= = 172.74xs s= = 1 17 TS F = 7.85 P=2.67E -4 < = .05 Reject the null hypothesis. There is enough evidence to support the claim. H1 is true. 1 2 1:72.74 1:16 2:25.97 2:16 sx n sx n Calculate s s¹ a 18 4.1 & 4.2 & 13.1 LINEAR CORRELATION (linear relationship between 2 variables) and REGRESSION(finding the line of best fit) Table of Total Points and Personal Fouls in basketball for players of a team Question: Can we predict Total Points (Y) by knowing Personal Fouls (X)? Note: The variable that you are predicting is Y. Assume populations are normal. Let’s enter the data into lists. [2ND] [MEM] [4] for 4:ClrAllLists [ENTER] clears all lists from the home screen [STAT] [1] for 1:Edit enter fouls (x) data into L1 and total points (y) data into L2 To do a scatter plot (a graph of the ordered pairs): Press [Y=] [CLEAR] To clear the y= editor. Press [2ND] [STATPLOT] [4] for 4:Plots Off [ENTER] (to turn plots off) Press [2ND] [STATPLOT] [1] for 1:Plot 1 [ENTER] Player x( L1) PersonalFouls Y ( L2)TotalPo int s 1 2 3 2 25 76 3 2 1 4 19 60 5 10 10 6 4 8 7 6 36 8 21 47 9 2 1 10 4 3 11 23 58 19 [ZOOM] [9] for 9: ZoomStat Note: There appears to be a positive correlation between x and y. _____________________________________________________________________________________________ *r linear correlation coefficient gives strength and direction of a linear relationship. Notice when r is positive, slope of the line is positive, and when r is negative, slope of the line is negative. _____________________________________________________________________________________________ To compute r the linear correlation coefficient: need to turn diagnostics on (You only need to do this one time.) [2ND] [CATALOG] arrow down to DIAGNOSTICS ON [ENTER] [ENTER] 20 TI83 [STAT] [>] [CALC] [4] for 4:LinReg(ax +b) [2ND] [ ] [,] [2ND] [, ] [ ] [ENTER] Note: To get above, [VARS] [>] [YVARS] [1] for 1:Function] [1] for 1: TI84 [STAT] [>] [CALC] [4] for 4:LinReg(ax +b) Note: To get above, [VARS] [>] [YVARS] [1] for 1:Function] [1] for 1: r = .935 (looks significant, however we need to do a formal hypothesis test) = .875 is the coefficient of determination 87.5 \% of the total variation in total points can be explained by the variation in the number of personal fouls. The other 12.5\% is attributable to other factors. Note: r and are displayed because the diagnostics are on. Note: -1 r 1, when r = -1 perfect negative linear correlation, when r = +1 perfect positive linear correlation, when r = 0 no correlation. To see the equation of the regression line: Simply press [Y=] y= 2.85x – 3.03 To graph the regression line on the scatter plot: Simply press GRAPH 1L [ ]2L 1Y 1Y 1Y 1Y 1Y r2 r2 r2 ≤ ≤ 21 y = mx + b Sample Linear Model; m is Sample Slope Population Linear Model ; is the Population Slope. To graph the regression line on the scatter plot: Simply press GRAPH IMPORTANT: The population correlation coefficient ”rho” and the population slope ”Beta sub one” always have the same sign, both will be positive or both will be negative. Additionally, whenever one of them is equal to zero, the other is equal to zero as well. When the slope of a line is zero, we have a horizontal line and changes in x do not result in change in y; this is NOT a linear relationship, and hence the correlation is zero as well. Test the claim of no linear correlation. Use a level of significance of .05. Claim: = 0 (r is the sample statistic and the population parameter is NOTE: This problem could also be worded as: Test the claim of zero slope. ( is the population slope and m is the sample slope.) [STAT][>][>][TESTS][ALPHA] [E] or [F] for LinRegT-Test Enter in the following: Note: To get above, [VARS] [>] [YVARS] [1] for 1:Function] [1] for 1: Highlight Calculate and press [ENTER] TS t = 7.93 p= 2.37E-5 < reject There is enough evidence to reject the claim. H1 is true. Do a 95\% confidence interval for or , we do a LinRegTInterval. (2.0379, 3.6635) y = β1x + β0 β1 ρ β1 ρ H H 0 1 0 0 : : ρ ρ = ≠ ρ Claim :β1 = 0 H0 : β1 = 0 H1 : β1 ≠ 0 β1 1Y 1Y α =.05 H0 β1 ρ 22 Interpreting this interval: We are 95\% confident that the true value of or lies in the interval. Note: LinReg T interval is located on my TI84 Plus CE under TESTS, but my TI83 Plus does not have it. This is NOT a topic that I will be testing you on. Predictions Since there is significant linear correlation between number of personal fouls and the number of total points, we can use the regression equation for predictions. Example: Find the best predicted total points given 10 personal fouls. [VARS] [>] [Y-VARS] [1] for 1:Function [1] for 1: (10) = 25.47 total points ----------------------------------------------------------- 13. 3 Multiple Regression In many cases, a better prediction model can be found for a dependent (response) Y variable by using more than one independent (explanatory) X variables. Just substitute the x values into the regression equation to compute the predicted value of y. β1 ρ 1Y Y1 23 12.1 GOODNESS of FIT TEST 1) Under TESTS, check to see if your calculator has the Chi-Square GOF Test already. My TI84 Plus C has this test under TESTS, option D: Chi-Square GOF TEST (best option) 2) If you do not have this test, you can either: - follow my handout on the GOF PROGRAM from Canvas. (recommended) - follow the directions below for the “long way” ------------------------------------------------------------------------------------------------------------ Ex) Mars Inc. claims that it’s M&M candies are distributed with the color percentages of 30\% brown, 20\% yellow, 20\% red, 10\% orange, 10\% green, and 10\% blue. A classroom exercise resulted in the observed frequencies listed below. At the .05 level, test the claim that the color distribution is as claimed by Mars Inc. Assume populations are normal. Claim= Ho: The distribution is as stated above. H1: The distribution differs from the claimed distribution. L1 L2 L3 Enter the observed frequencies in List 1 and the experimental probabilities in List 2 to figure out the exp f (expected frequency) column Recall E = np To generate List 3: = sum( [ENTER] Note: the SUM command is from 2ND STAT à à MATH and choose 5)SUM Using the Chi-Square GOF Test, enter data in L1 and L2, and generate L3 as explained above. You do not need L4. color obsf P f brown yellow red blue green orange exp exp . . . . . . 17 30 39 20 8 20 26 10 12 10 19 10 L3 L L1 2)* 24 STAT---->-----> TESTS Choose: D: Chi-Square GOF Test Observed: L1 (observed f) Expected: L3 (expected f) Df: 5 (#categories – 1) CALCULATE You will see……… Chi-Square TS = 50.06 P value = 1.34 E-9 < alpha = .05 Reject the null hypothesis. There is enough evidence to reject the claim. H1 is true. ____________________________________________________________________________________________ OR “GOF Program” from Canvas Module Follow my directions to type program in your calculator (easy to do!), then Enter the data in L1 and L2 and generate L3 as explained above. Then press PRGM, 1: GOF, ENTER __________________________________________________________________________________________ OR “long way” To generate List 4: [2ND] [QUIT] To find the Test Statistic =50.06 Now, find the p-value: [2nd][DISTR] [7] or [8] for : cdf( TS, E99, df) [ENTER] df= #categories - 1 (Note: to get ‘E’ press [2nd] [EE] ) (50.06, E99, 5) [ENTER] p = 1.35 E-9 < = .05 Reject the null hypothesis. There is enough evidence to reject the claim. H1 is true. L L L L4 1 3 2 3= −( ) / 2 4( )sum Lχ = 2χ 2χ α 25 Ex) Mars Inc. claims that it’s M&M candies are distributed with the same frequency. The colors are: Brown, yellow, red, orange, green, and blue. A classroom exercise resulted in the observed frequencies listed below. At the .05 significance level, test the claim that the color distribution is as claimed by Mars Inc. Assume populations are normal. WE ARE ASKED TO TEST THE CLAIM THAT THE COLORS OCCUR WITH THE SAME FREQUENCY. SINCE WE HAVE 6 COLORS, EACH MUST OCCUR 1 DIVIDED BY 6 or .1667. Claim= Ho: The distribution is as stated above. H1: The distribution differs from the claimed distribution. L1 L2 L3 Enter the observed frequencies in List 1 and the experimental probabilities in List 2 To figure out the exp f (expected frequency) column Recall E = np To generate List 3: = sum( [ENTER] Note: the SUM command is from 2ND STAT à à MATH and choose 5)SUM Continue the problem either using the Chi-Square GOF Test or GOF Program or the “long way” as explained in the last example. Chi-Square TS = 30.48 P value = 1.18E-5 < alpha = .05 Reject Ho There is enough evidence to reject the claim. H1 is true. ----------------------------------------------------------------------------------------------------------- NOTE: For some of the problems at ALEKS, you are given the Observed Frequencies (L1) and the Expected Frequencies (L3). In my notes, we are given the Observed Frequencies (L1) and the Expected Probabilities (L2), and then we calculate the Expected Frequencies (L3). For problems like these at ALEKS, put the Observed Frequencies in L1 and the Expected Frequencies in L3 and nothing in L2. color obsf expP exp f brown 17 .1667 yellow 39 .1667 red 8 .1667 blue 26 .1667 green 12 .1667 orange 19 .1667 L3 L L1 2)* 26 12.2 TESTS OF INDEPENDENCE Ex) Responses to a survey question are broken down according to employment and the sample results are given below. At the.10 significance level, test the claim that the response and employment status are independent. Assume populations are normal. Claim=Ho: Response and employment status are independent. H1: Response and employment status are dependent. enter the data in Matrix [A]: [2ND] [MATRIX] [>] [>] [EDIT] press [1] for matrix A Note: We have a 2 (rows) x 3 (columns) matrix. [STAT] [>][>] [TESTS] [ALPHA][C] for C: Test Highlight Calculate and press [ENTER] TS p=.0512 < =.10 reject the null hypothesis There is enough evidence to reject the claim. H1 is true. Yes no undecided Employed Unemployed 30 15 5 20 25 10 2χ 2 5.94χ = α 27 14.1 One Way ANOVA: Testing claims about 3 or more population means. Ex) At the .05 significance level, test the claim that the mean distance traveled by each type of golf ball is the same. Assume populations are normal. ____________________________________________________________________________________ Note other ways of stating the alternative hypothesis: H1: At least one mean is different from the others. H1: Means are not all equal. _________________________________________________________________ Enter data into lists 1,2 and 3 [STAT][>] [>][TESTS] [ALPHA][F] or [H] for ANOVA [2ND] [ ][,] [2ND] [ ][,] [2ND][ ] [ENTER] TS F = .653 p=.531 > = .05 Do not reject the null hypothesis There is not enough evidence to reject the claim. Ho might be true. … 1 HOMEWORK WORKSHEETS section by section Assume all populations normal except for the last chp on non-parametric statistics. 9.2 Z Test for Mean 1. A random sample of 50 medical school applicants at a university has a mean raw score of 31 on the multiple-choice portions of the Medical College Admission Test (MCAT). A student claims that the mean raw score for the school’s applicants is more than 30. Assume the population standard deviation is 2.5. At alpha = .01, test the claim. Assume population normally distributed. 2. A consumer group claims that the mean annual consumption of cheddar cheese by a person in the United States is at most 10.3 pounds. A random sample of 100 people in the United States has a mean annual cheddar cheese consumption of 9.9 pounds. Assume the population standard deviation is 2.1 pounds. At alpha = .05, test the claim. Assume population normally distributed. 3. The lengths of time (in years) it took a random sample of 32 former smokers to quit smoking permanently are listed. Assume the population standard deviation is 6.2 years. At alpha = .05, test the claim that the mean time it takes for smokers to quit smoking permanently is 15 years. Assume population normally distributed. 15.7 13.2 22.6 13.0 10.7 18.1 14.7 7.0 17.3 7.5 21.8 12.3 19.8 13.8 16.0 15.5 13.1 20.7 15.5 9.8 11.9 16.9 7.0 19.3 13.2 14.6 20.9 15.4 13.3 11.6 10.9 21.6 9.3 T Test for Mean 1. A county is considering raising the speed limit on a road because they claim that the mean speed of vehicles is greater than 45 miles per hour. A random sample of 25 vehicles has a mean speed of 48 miles per hour and a standard deviation of 5.4 miles per hour. At alpha = .10, test the claim. Assume population normally distributed. 2. An oceanographer claims that the mean dive depth of a North Atlantic right whale is 115 meters. A random of 34 dive depths has a mean of 121.2 meters and a standard deviation of 24.2 meters. At alpha = .10 test the claim. Assume population normally distributed. 3. You receive a brochure from a large university. The brochure claims that the mean class size for full-time faculty is fewer than 32 students. You randomly select 18 classes taught by full-time faculty and determine the class size of each. The results are shown in the table below. At alpha = .05 test the claim. Assume population normally distributed. 2 Class Sizes 35 28 29 33 32 40 26 25 29 28 30 36 33 29 27 30 28 25 9.4 1-Prop Z Test 1. A medical researcher claims that less than 25\% of U.S. adults are smokers. In a random sample of 200 U.S. adults, 19.3\% say that they are smokers. At alpha = .05, test the claim. 2. A research center claims that at most 75\% of U.S. adults think that drivers are safer using hands-free cell phones instead of using hand-held cell phones. In a random sample of 150 U.S. adults, 77\% think that drivers are safer using hands-free cell phones instead of hand-held cell phones. At alpha = .01, test the claim. 3. A research center claims that more than 80\% of females, ages 20-29 are taller than 62 inches. In a random sample of 150 females ages 20-29, 79\% are taller than 62 inches. At alpha = .10 test the claim. 4. A humane society claims that less than 35\% of U.S. households own a dog. In a random sample of 400 U.S. households, 156 say they own a dog. At alpha = .10, test the claim. 11.1 and 10.1 2 Sample Z Test and 2 Sample Z Interval 1. To compare the braking distance for two types of tires, a safety engineer conducts 35 braking tests for each type. The mean braking distance for Type A is 42 feet. Assume the population standard deviation is 4.7 feet. The mean braking distance for Type B is 45 feet. Assume the population standard deviation is 4.3 feet. At alpha = .10 test the claim that the mean braking distances are different for the two types of tires. Assume populations are normally distributed. 2. An energy company wants to choose between two regions in a state to install energy- producing wind turbines. A researcher claims that the wind speed in Region A is less than the wind speed in Region B. To test the regions, the average wind speed is calculated for 60 days in each region. The mean wind speed in Region A is 14.0 miles per hour. Assume the population standard deviation is 2.9 miles per hour. The mean wind speed in Region B is 15.1 miles per hour. Assume the population standard deviation is 3.3 miles per hour. At alpha = .05, test the claim. Assume populations are normally distributed. 3. The mean ACT score for 43 male high school students is 21.1. Assume the population standard deviation is 5.0. The mean Act score for 56 female high school students is 20.9. Assume the population standard deviation is 4.7. At alpha = .01 test the claim that male and female high school students have equal ACT scores. Assume populations are normally distributed. 3 4. A sociologist claims that children ages 6-17 spent more time watching television in 1981 than children 6-17 do today. A study was conducted in 1981 to find the time that children ages 6-17 spent watching television on weekdays. The results (in hours per weekday) are shown below. Assume the population standard deviation is .6 hour. Assume populations are normally distributed. 2.0 2.5 2.1 2.3 2.1 1.6 2.6 2.1 2.1 2.4 2.1 2.1 1.5 1.7 2.1 2.3 2.5 3.3 2.2 2.9 1.5 1.9 2.4 2.2 1.2 3.0 1.0 2.1 1.9 2.2 Recently, a similar study was conducted. The results are shown below. Assume the population standard deviation is .5 hour. 2.9 1.8 0.9 1.6 2.0 1.7 2.5 1.1 1.6 2.0 1.4 1.7 1.7 1.9 1.6 1.7 1.2 2.0 2.6 1.6 1.5 2.5 1.6 2.1 1.7 1.8 1.1 1.4 1.2 2.3 At alpha = .05, test the claim. 5. Construct a 95\% confidence interval for the difference between the mean annual salaries of microbiologists in Maryland and California using the following data: Assume populations are normally distributed. Microbiologists in Maryland: Microbiologists in California: 2 Sample T Test and 2 Sample T Interval 1. A marine biologist claims that the mean length of mature female pink seaperch is different in fall and winter. A sample of 26 mature female pink seaperch collected in fall has a mean length of 127 millimeters and a standard deviation of 14 millimeters. A sample of 31 mature female pink seaperch collected in winter has a mean length of 117 millimeters and a standard deviation of 9 millimeters. At alpha = .01, test the claim. Assume the population variances are equal. Assume populations are normally distributed. X1 = $102,650 n1 = 42 σ1 = $8795 X2 = $85,430 n2 = 38 σ 2 = $9250 4 2. A personnel director from Pennsylvania claims that the mean household income is greater in Allegheny County than it is in Erie County. In Allegheny County, a sample of 19 residents has a mean household income of $49,700 and a standard deviation of $8800. In Erie County, a sample of 15 residents has a mean household income of $42,000 and a standard deviation of $5100. At alpha = .05 test the claim. Assume the population variances are not equal. Assume populations are normally distributed. 3. The tensile strength of a metal is a measure of its ability to resist tearing when it is pulled lengthwise. A new experimental type of treatment produced steel bars with the tensile strengths (in newtons per square millimeter) listed below. Assume populations are normally distributed. Experimental Method: 391 383 333 378 368 401 339 376 366 348 The old method produced steel bars with the tensile strengths (in newtons per square millimeter) listed below. Old Method: 362 382 368 398 381 391 400 410 396 411 385 385 395 At alpha = .01, test the claim that the new treatment makes a difference in the tensile strength of steel bars. Assume the population variances are equal. Assume populations are normally distributed. 4. To compare the mean times spent waiting for a kidney transplant for two age groups, you randomly select several people in each age group who have had a kidney transplant. The results are shown below. Construct a 95\% confidence interval for the difference in mean times spent waiting for a kidney transplant for the two age groups. Assume populations are normally distributed. Sample Statistics for Kidney Transplants 35-49 50-64 X1 =1805days X2 =1629days s1 =166days s2 = 204days n1 = 21 n2 =11 5 11.2 and 10.2 Dependent Samples (Before/After) T Test and T Interval 1. A researcher claims that a post-lunch nap decreases the amount of time it takes males to sprint 20 meters after a night with only 4 hours of sleep. The table shows the amounts of time (in seconds) it took for 10 males to sprint 20 meters after a night with only 4 hours of sleep, when they did not take a post-lunch nap and when they did take a post-lunch nap. At alpha = .01, is there enough evidence to support the researcher’s claim? Male 1 2 3 4 5 Sprint Time (w/o nap) 4.07 3.94 3.92 3.97 3.92 Sprint Time With nap 3.93 3.87 3.85 3.92 3.90 Male 6 7 8 9 10 Sprint Time (w/o nap) 3.96 4.07 3.93 3.99 4.02 Sprint Time With nap 3.85 3.92 3.80 3.89 3.89 2. A physical therapist claims that one 600-milligram dose of Vitamin C will increase muscular endurance. The table below shows the numbers of repetitions 15 males made on a hand dynamometer (measures grip strength) until the grip strengths in three consecutive trials were 50\% of their maximum grip strength. At alpha = .05 test the claim. Assume populations are normally distributed. Participant 1 2 3 4 5 6 7 8 Repetitions (using placebo) 417 279 678 636 170 699 372 582 Repetitions (using Vitamin C) 145 185 387 593 248 245 349 902 Participant 9 10 11 12 13 14 15 Repetitions (using placebo) 363 258 288 526 180 172 278 Repetitions (using Vitamin C) 159 122 264 1052 218 117 185 6 3. A company claims that its consumer product ratings (0 -10) have changed from last year to this year. The table below shows the company’s product ratings from the same eight consumers for last year and this year. At alpha = .05, test the claim. Assume populations are normally distributed. Consumer 1 2 3 4 5 6 7 8 Rating (last year) 5 7 2 3 9 10 8 7 Rating (this year) 5 9 4 6 9 9 9 8 4. A sleep disorder specialist wants to test the effectiveness of a new drug that is reported to increase the number of hours of sleep patients get during the night. To do so, the specialist randomly selects 16 patients and records the number of hours of sleep each gets with and without the new drug. The table below shows the results of the two-night study. Construct a 90\% confidence interval for . Assume populations are normally distributed. Patient 1 2 3 4 5 6 7 8 Hours of sleep (without the drug) 1.8 2.0 3.4 3.5 3.7 3.8 3.9 3.9 Hours of sleep Using the drug) 3.0 3.6 4.0 4.4 4.5 5.2 5.5 5.7 Patient 9 10 11 12 13 14 15 16 Hours of sleep (without the drug) 4.0 4.9 5.1 5.2 5.0 4.5 4.2 4.7 Hours of sleep Using the drug) 6.2 6.3 6.6 7.8 7.2 6.5 5.6 5.9 µd 7 11.3 and 10.3 2 Prop Z Test and 2 Prop Z Interval 1) In a 4-week study about the effectiveness of using magnetic insoles to treat plantar heel pain, 54 subjects wore magnetic insoles and 41 subjects wore nonmagnetic insoles. The results are shown below. At alpha = .01, test the claim that there is a difference in the proportion of subjects who feel all or mostly better between the two groups. Do you feel all or mostly better? Magnetic Insoles Nonmagnetic Insoles Yes 17 18 No 37 23 2) In a survey of 200 males ages 18 to 24, 39\% were enrolled in college. In a survey of 220 females ages 18 to 24, 45\% were enrolled in college. At alpha = .05, test the claim that the proportion of males ages 18 to 24 who enrolled in college is less than the proportion of females ages 18 to 24 enrolled in college. 3) In a survey of 480 drivers from the South, 408 wear a seatbelt. In a survey of 360 drivers from the Northeast, 288 wear a seat belt. At alpha = .05, test the claim that the proportion of drivers who wear seat belts is greater in the South than in the Northeast. 4) In a survey of 10,000 students taking the SAT, 6\% were planning to study education in college. In another study of 8000 students taken 10 years before, 9\% were planning to study education in college. Construct a 95\% confidence interval for p1 – p2 where p1 is the proportion from the recent study and p2 is the proportion from the survey taken 10 years ago. 8 11.4 2 Sample F Test 1. The table below shows a sample of the waiting times (in days) for a heart transplant for two age groups. At alpha = .05, test the claim that the variances of the waiting times differ between the two age groups. 18-34 35-49 158 170 212 209 213 173 162 194 196 200 169 210 2.A state school administrator claims that the standard deviations of science assessment test scores for eighth-grade students are the same in Districts 1and 2. A sample of 12 test scores from District 1 has a standard deviation of 36.8 points, and a sample of 14 test scores from District 2 has a standard deviation of 32.5 points. At alpha = .10, test the administrator’s claim. 3. An employment information service claims that the standard deviation of the annual salaries for actuaries is greater in New York than in California. You select a sample of actuaries from each state. The results of each survey are shown below. At alpha = .05, test the claim. Actuaries in New York Actuaries in California S1 = $39, 700 S2= $29,000 N1 = 41 N2=61 4.1, 4.2 13.1 Linear Correlation and Regression 1. Two variables have a positive linear correlation. Does the dependent variable increase or decrease as the independent variable increases? 2. Two variables have a negative linear correlation. Does the dependent variable increase or decrease as the independent variable increases? 3. Describe the range of values for the correlation coefficient. 4. What the sample correlation coefficient r measure? Which value indicates a stronger correlation r = .918 or r = -.932? Explain your reasoning. 5. Discuss the difference between r and . 6. In your own words, what does it mean to say “correlation does not imply causation” ? ρ 9 In the following Exercises (a) display the data in a scatter plot, (b) calculate the sample correlation r, and (c) describe the type of correlation and interpret the correlation in the context of the data. 15. The ages (in years) of 10 men and their systolic blood pressures (in millimeters of mercury): Age, x 16 25 39 45 49 64 70 29 57 22 Systolic blood pressure, y 109 122 143 132 199 185 199 130 175 118 10 16. The maximum weights (in kilograms) for which one repetition of a half squat can be performed and the times (in seconds) to run a 10-meter sprint for 12 international soccer players: Max weight, x 175 180 155 210 150 190 185 160 190 180 160 170 Time, y 1.80 1.77 2.05 1.42 2.04 1.61 1.70 1.91 1.60 1.63 1.98 1.90 17. The earnings per share (in dollars) and the dividends per share (in dollars) for 6 medical supply companies in a recent year: Earnings per share, x 2.79 5.10 4.53 3.06 3.70 2.20 Dividends per share, y .52 2.40 1.46 .88 1.04 .22 18. The weights (in pounds) of eight vehicles and the variability of their braking distances (in feet) when stopping on a dry surface are shown below in the table. At alpha = .01, test the claim that there is a significant linear correlation between vehicle weight and variability in braking distance on a dry surface. Weight, x 5940 5340 6500 5100 5850 4800 5600 5890 Variability, y 1.78 1.93 1.91 1.59 1.66 1.50 1.61 1.70 Regression In the exercises below, find the equation of the regression line for the data. Then construct a scatter plot of the data and draw the regression line. {Each pair of data has a significant correlation.} Then use the regression equation to predict the value of y for each of the x-values, if meaningful. If the x-value is not meaningful to predict the value of y, explain why not. 1. The height (in feet) and the numbers of stories of nine notable buildings in Atlanta. Height, x 869 820 771 696 692 676 656 492 486 Stories, y 60 50 50 52 40 47 41 39 26 (a) x = 800 feet (b) x = 750 feet (c) x = 400 feet (d) x = 625 feet 11 2. The number of hours 9 students spent studying for a test and their scores on that test. Hours spent studying, x 0 2 4 5 5 5 6 7 8 Test scores, y 40 51 64 69 73 75 93 90 95 (a) x = 3 hours (b) x = 6.5 hours (c) x = 13 hours (d) x = 4.5 hours 3. The heart rates (in beats per minute) and QT intervals (in milliseconds) for 13 males (the figure below shows the QT interval of a heartbeat in an electrocardiogram). 12 Heart rate, x 60 75 62 68 84 97 66 65 86 78 93 75 88 QT interval, y 403 363 381 367 341 317 401 384 342 377 329 377 349 (a) x = 120 beats per min (b) x = 67 beats per min (c) x = 90 beats per min (d) x = 83 beats per min Coefficient of Determination In the exercises below, use the value of the correlation coefficient to calculate the coefficient of determination . What does this tell you about the explained variation of the data about the regression line? About the unexplained variation? 1. r = .465 2. r = -.957 13.3 Multiple Regression In the exercises below, use the multiple regression equation to predict the y-values for the values of the independent variable. 1. The equation used to predict the annual cauliflower yield (in pounds per acre) is , where x1 is the number of acres planted and x2 is the number of acres harvested. r2 y ^ = 24,791+ 4.508x1 − 4.723x2 (a)x1 = 36,500 x2 = 36,100 (b)x1 = 38,100 x2 = 37,800 (c)x1 = 39,000 x2 = 38,800 (d)x1 = 42,200 x2 = 42,100 13 2.The volume (in cubic feet) of a black cherry tree can be modeled by the equation , where x1 is the tree’s height (in feet) and x2 is the tree’s diameter (in inches). 12.1 Goodness of Fit 1. A researcher claims that the ages of people who go to movies at least once a month are distributed as shown in the figure below. You randomly select 1000 people who go to the movies at least once a month and record the age of each. The table shows the results. At alpha = .10, test the researcher’s claim. Survey results Age Frequency, f 2-17 240 18-24 214 25-39 183 40-49 156 50+ 207 y ^ = −52.2+ .3x1 + 4.5x2 (a)x1 = 70 x2 = 8.6 (b)x1 = 65 x2 =11.0 (c)x1 = 83 x2 =17.6 (d)x1 = 87 x2 =19.6 14 2. A research firm claims that the distribution of the days of the week that people are most likely to order food for delivery is different from the distribution shown in the figure below. You randomly select 500 people and record which day of the week each is most likely to order food for delivery. The table below shows the results. At alpha = .01, test the research firm’s claim. Survey results Day Frequency, f Sunday 43 Monday 16 Tuesday 25 Wednesday 49 Thursday 46 Friday 168 Saturday 153 15 12.2 Tests of Independence 1. The contingency table below shows the results of a random sample of students by the location of school and the number of those students achieving basic skill levels in three subjects. At alpha = .01, test the hypothesis that the variables are independent. Subject Location of school Reading Math Science Urban 43 42 38 Suburban 63 66 65 2. The contingency table below shows the results of a random sample of former smokers by the number of times they tried to quit smoking before they were habit- free and gender. At alpha = .05, test the claim that the number of times they tried to quit before they were habit-free is related to gender. Number of times tried to quit before habit-free Gender 1 2-3 4 or more Male 271 257 149 Female 146 139 80 14.1 ANOVA 1. The table below shows the costs per ounce (in dollars) for a sample of toothpastes exhibiting very good stain removal, good stain removal, and fair stain removal. At alpha = .05, test the claim that at least one mean cost per ounce is difference from the others. Very good .47 .49 .41 .37 .48 .51 Good .60 .64 .58 .75 .46 Fair .34 .46 .44 .60 16 2. The well-being index is a way to measure how people are faring physically, emotionally, socially, and professionally, as well as to rate the overall quality of their lives and their outlooks for the future. The table below shows the well-being index scores for a sample of states from four regions of the United States. At alpha = .10, test the claim the mean scores is the same for all regions. Northeast Midwest South West 67.6 66.6 64.2 66.1 67.3 67.6 64.1 67.4 68.4 65.6 65.8 69.7 66.2 68.9 66.1 68.5 66.5 65.5 62.7 65.2 68.6 68.5 68.0 66.7 67.4 63.6 67.1 68.0 65.2 68.8 65.2 67.7 64.0 66.6 15.1 Non-Parametric Statistics and the Sign Test 1. What is a nonparametric test? How does a nonparametric test differ from a parametric test? What are the advantages and disadvantages of using a nonparametric test? 2. When the sign test is used, what population parameter is being tested? 3. Describe the test statistic for the sign test when the sample size n is less than or equal to 25 and when n is greater than 25. 4. In your own words, explain why the hypothesis test discussed in this section is called the sign test. 5. Explain how to use the sign test to test a population median. 6. List the two conditions that must be met in order to use the paired-sample sign test. 7. A financial service accountant claims that the median amount of new credit card charges for the previous month was more than $300. You randomly select 12 credit card accounts and record the amount of new charges for each account for the previous month. The amounts (in dollars) are listed below. At alpha = .01, test the accountant’s claim. 346.71 382.59 255.03 202.17 309.80 265.88 299.41 270.38 296.54 318.46 245.92 309.47 17 8. A real estate agent claims that the median sales price of new privately owned one-family homes sold in a recent month is $193,000 or less. The sales prices (in dollars) of 10 randomly selected homes are listed below. At alpha = .05, test the agent’s claim. 200,800 229,500 205,900 190,700 140,200 193,900 249,000 170,900 184,500 207,500 9. A physician claims that lower back pain intensity scores will decrease after receiving acupuncture treatment. The table below shows the lower back pain intensity scores for eight patients before and after receiving acupuncture for eight weeks. At alpha= .05, test the physician’s claim. Patient 1 2 3 4 5 6 7 8 Intensity (before) 59.2 46.3 65.4 74.0 79.3 81.6 44.4 59.1 Intensity (after) 12.4 22.5 18.6 59.3 70.1 70.2 13.2 25.9 10. A tutoring agency claims that by completing a special course, students will improve their critical reading SAT scores. In part of a study, 12 students take the critical reading part of the SAT, complete the special course, then take the critical reading part of the SAT again. The students’ scores are shown below. At alpha = .05, test the agency’s claim. Student 1 2 3 4 5 6 7 8 9 10 11 12 1st score 300 450 350 430 300 470 530 200 200 350 360 250 2nd score 300 520 400 410 300 480 700 250 390 350 480 300 15.2 and 15.3 Wilcoxon Signed Rank and Wilcoxon Rank Sum 1. How do you know whether to use a Wilcoxon signed-rank test or a Wilcoxon rank sum test. 2. In a study testing the effects of calcium supplements on blood pressure in men, 12 men were randomly chosen and given supplements for 12 weeks. The table below shows the measurements for each subject’s diastolic blood pressure taken before and after the 12-week treatment period. At alpha = .01, test the claim that there was no reduction in diastolic blood pressure. Patient 1 2 3 4 5 6 7 8 9 10 11 12 Before 108 109 120 129 112 111 117 135 124 118 130 115 After 99 115 105 116 115 117 108 122 120 126 128 106 18 3. A college administrator claims that there is a difference in the earnings of people with bachelor’s degrees and those with advanced degrees. The table below shows the earnings (in thousands of dollars) of a random sample of 11 people with bachelor’s degrees and 10 people with advanced degrees. At alpha = .05, test the claim. Bachelor’s 56 52 65 78 72 52 46 58 62 54 56 Advanced 84 87 95 81 86 86 93 93 90 82 4. A teacher’s union representative claims that there is a difference in the salaries earned by teachers in Wisconsin and Michigan. The table below shows the salaries (in thousands of dollars) of a random sample of 11 teachers from Wisconsin and 12 teachers from Michigan. At alpha = .05, test the claim. Wisconsin 55 59 49 56 51 61 55 61 53 47 52 Michigan 64 68 58 65 60 70 64 70 62 56 61 79 Kruskal-Wallis Test 1. What are the conditions for using a Kruskal - Wallis test? 2. The table below shows the annual premiums for a random sample of home insurance policies in Connecticut, Massachusetts, and Virginia. At alpha = .05 test the claim that the distribution of the annual premiums in at least one state is different from the others. State Annual premium (in dollars) CONN 1053 848 1013 1163 1288 929 1070 MA 1132 1052 1007 1322 1137 916 784 VA 885 800 616 695 982 688 605 3. The table below shows the annual salaries for a random sample of private industry workers in Kentucky, North Carolina, South Carolina, and West Virginia. At alpha = .10, test the claim that the distribution of the annual salaries of private industry workers in at least one state is different from the others. State Hourly pay rate (in dollars) KT 35.3 37.0 45.9 57.5 33.7 28.3 35.3 NC 43.5 41.9 36.6 54.3 35.5 39.6 43.5 SC 29.8 37.4 43.5 42.9 34.7 36.1 29.8 WV 31.6 42.7 33.4 41.9 47.1 34.9 31.6