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W6: Hypothesis testing I and II; Testing differences between means, variances and proportions Subscribe A town official claims that the average vehicle in their area sells for more than the 40th percentile of your data set. Using the data, you obtained in week 1, as well as the summary statistics you found for the original data set (excluding the super car outlier), run a hypothesis test to determine if the claim can be supported.  Make sure you state all the important values, so your fellow classmates can use them to run a hypothesis test as well.  Use the descriptive statistics you found during Week 2 NOT the new SD you found during Week 4.  Because again, we are using the original 10 sample data set NOT a new smaller sample size.  Use alpha = .05 to test your claim. (Note: You will want to use the function =PERCENTILE.INC  in Excel to find the 40th percentile of your data set.  Hopefully this Excel function looks familiar to you from Week 2.) First determine if you are using a z or t-test and explain why. Then conduct a four-step hypothesis test including a sentence at the end justifying the support or lack of support for the claim and why you made that choice. I encourage you to review the Week 6 Hypothesis Testing PDF at the bottom of the discussion.  This will give you a step by step example on how to calculate and run a hypothesis test using Excel. I DO NOT recommend doing this by hand.  Let Excel do the heavy lifting for you.  You can also use this PDF in Quizzes section. There were 5 additional PDFs that were created to help you with the Homework, Lessons and Tests in Quizzes section.  While they wont be used to answer the questions in the discussion, they are just as useful and beneficial.  I encourage you to review these ASAP!  These PDFs are also located at the bottom of the discussion. Once you have posted your initial discussion, you must reply to at least two other learners post. Each post must be a different topic. So, you will have your initial post from one topic, your first follow-up post from a different topic, and your second follow-up post from one of the other topics. Of course, you are more than welcome to respond to more than two learners.” Instructions: Make sure you include your data set in your initial post as well.  You must also respond to at least 2 other students. Responses may include direct questions. In your first peer response post, look at the hypothesis test results of one of your classmates and explain what a type 1 error would mean in a practical sense. Looking at your classmates outcome, is a type 1 error likely or not? What specific values indicated this? In your second peer response post, using your classmates values, run another hypothesis test using this scenario:  A town official claims that the average vehicle in their area Does Not sell for 80th percentile of your data set.  Conduct a four-step hypothesis test including a sentence at the end justifying the support or lack of support for the claim and why you made that choice.  Note: this test will be different than the initial post, starting with the hypothesis scenario. Use alpha = .05 to test your claim.This week will continue to discuss hypothesis testing and confidence interval, but now we will discuss 2 samples. Just like with 1 – sample hypothesis testing there are 4 steps we will follow. To review those 4 steps please review the Week 6 Hypothesis Testing PDF. But the conclusion will still be the same. If the p-value is < alpha, you Reject Ho and state this test is significant. If the p-value is > alpha, you Do Not Reject Ho and state this test is not significant. In this document we will discuss 2 – sample proportion hypothesis testing and confidence intervals. There are still 3 different hypothesis scenarios with a 2 – sample proportion hypothesis test. Lower Tail Test (1 tail): Ho: 𝑝1 − 𝑝2 = 0 Ha: 𝑝1 − 𝑝2 < 0 Upper Tailed Test (1 tail): Ho: 𝑝1 − 𝑝2 = 0 Ha: 𝑝1 − 𝑝2 > 0 Two Tailed Test: Ho: 𝑝1 − 𝑝2 = 0 Ha: 𝑝1 − 𝑝2 ≠ 0 The hypothesized value is 0 and the same key words apply from a 1 – sample hypothesis test to determine which scenario to use. The Z – Test Statistic = 𝑝1− 𝑝2 √𝑝∗𝑞( 1 𝑛1 + 1 𝑛2 ) Where 𝑝 = 𝑥1+𝑥2 𝑛1+𝑛2 𝑞 = 1 − 𝑝 We will then use =NORM.S.DIST function in Excel to find the p-value. This Excel function should look familiar from Week 4. Example: In a developing section of a district 50 people were surveyed and 38 were in favor of the new proposal. For the rest of the district 100 people were surveyed and only 65 people were in favor of the new proposal. Is there evidence that the number of people favoring the new proposal is greater in the developing section than the rest of the district? Use alpha = .05 First step is to state the hypothesis scenario. Because the key word says greater this means it is an upper tailed test. Ho: 𝑝1 − 𝑝2 = 0 Ha: 𝑝1 − 𝑝2 > 0 The first proportion is favoring the new proposal in the developing district and the second proportion is favoring the new proposal in the rest of the district. 𝑝1 = 38 50 = .76 𝑞1 = 1 − .76 = .24 𝑝2 = 65 100 = .65 𝑞2 = 1 − .65 = .35 𝑝 = 38 + 65 50 + 100 = .68667 𝑞 = 1 − .68667 = .31333 Now that we have these values we can plug them in to find the Test Statistic. Z – Test Statistic = .76−.65 √.68667∗.31333( 1 50 + 1 100 ) = 1.369 Now that we have the Z-Test Statistics we can use the =NORM.S.DIST function to find the p-value. And yes, we can have a negative Z- Test Statistic, if we do that is fine. You DO NOT have to take the absolute value of anything. Use the Test Stat. as is in the Excel function. In Excel input =NORM.S.DIST(1.369,TRUE) We will write out TRUE because this test is cumulative. We see this p-value = .9145 BUT remember when we use this function in Excel, this function is in the less than form. This means if we were running a Lower Tailed test, this would be our p-value. BUT since we are running an Upper Tailed Test we need to take 1 - .9145 to get the p-value for our test. P-valHypothesis Testing is a decision-making process called a Test of Significance. Below are the 4 unique parts to Hypothesis Testing. 1) The Hypothesis Scenario. This includes the Null and Alternative scenarios. a. Ho: Null Hypothesis Ha or H1: Alternative Hypothesis 2) T- Test Statistic 𝑇𝑆 = �̅� − 𝜇 𝑠 √𝑛 ⁄ Where �̅� is the sample mean, 𝑠 is the sample standard deviation, 𝑛 is the sample size, and 𝜇 is the population mean defined in the null hypothesis of step 1 3) P-value. The p-value tells you if there is evidence to suggest that your null hypothesis is incorrect. You will either reject your null hypothesis, or you fail to reject the null hypothesis. *NOTE: we NEVER Accept null hypotheses, rather find that there was not enough evidence to reject a null hypothesis. To find the p-value we will use =T.DIST(), T.DIST.RT() or T.DIST.2T depending on the direction of the alternative hypothesis. The degrees of freedom (DF) is n – 1. 4) Conclusion: a. State whether your p-value is greater than alpha (𝛼) or less than alpha (𝛼). *NOTE, the value of alpha (𝛼) is defined prior to conducting a hypothesis test. The most common choices for the value of alpha (𝛼) are 0.05, 0.01, 0.10 and 0.005. Alpha (𝛼) is the acceptable Type 1 error. Type 1 error occurs when a null hypothesis is rejected, though in reality the null hypothesis should not have been rejected because the null hypothesis is true. Type 1 error is also called the “false positive” rate. b. Decide to Reject the null hypothesis (Ho) if the p-value is less than alpha (𝛼), or Fail to reject the null hypothesis (Ho) if the p-value is more than alpha (𝛼) c. If you reject Ho, state that there is evidence for the alternative hypothesis. If you fail to reject Ho, state that there is not enough evidence to support the alternative hypothesis. *NOTE: we NEVER Accept or Reject alternative hypotheses, rather we say that our sample provide evidence to support the alternative hypothesis, or our sample does not provide enough evidence to support the alternative hypothesis. The alternative hypothesis is not directly being tested and therefore can not directly be accepted or rejected! There are 3 different directions of the alternative hypothesis 1) A Left sided, or a Left Tailed Test. We use a left sided test to examine if the population mean is lower than the mean defined in the null hypothesis (Ho). A left sided test uses the following hypothesis scenario: 𝐻0: 𝜇 = 𝑐 𝐻𝑎 : 𝜇 < 𝑐 Where 𝑐 is the number we are testing whether the population mean, 𝜇, is less than. For example if we wanted to know if the population mean, 𝜇, is less than 15, then 𝑐 would be 15 in both the null and alternative hypothesis. To find the p-value for a left sided test, we use =T.DIST() in excel. 2) A Right sided, or a Right Tailed Test. We use a right sided test to examine if the population mean is higher than the mean defined in the null hypothesis (Ho). A right sidedIn this document we will discuss 2 – sample T- Paired or Matched hypothesis testing and confidence intervals that used a mean and sample SD. There are still 3 different hypothesis scenarios with a 2 – Sample Paired Hypothesis Test. Lower Tail Test (1 tail): Ho: �̅�𝑑𝑖𝑓𝑓 = 0 Ha: �̅�𝑑𝑖𝑓𝑓 < 0 Upper Tailed Test (1 tail): Ho: �̅�𝑑𝑖𝑓𝑓 = 0 Ha: �̅�𝑑𝑖𝑓𝑓 > 0 Two Tailed Test: Ho: �̅�𝑑𝑖𝑓𝑓 = 0 Ha: �̅�𝑑𝑖𝑓𝑓 ≠ 0 The hypothesized value is 0 and the same key words apply from a 1 – sample hypothesis test to determine which scenario to use. �̅�𝑑𝑖𝑓𝑓 is the average of the difference column. Paired samples are samples that share something in common. They are dependent on one another. Since they share something in common the samples go from 2 to 1. This will be very similar to the 1-sample T hypothesis test in the discussion forum. The T – Test Statistic = �̅�𝑑𝑖𝑓𝑓−0 𝑆𝐷𝑑𝑖𝑓𝑓 √𝑛 Where 𝑆𝐷𝑑𝑖𝑓𝑓 this is SD of the difference column We can use =T.DIST, =T.DIST.RT and =T.DIST.2T to find the p-values. These should look familiar from the discussion forum. Example: Blood plasma cancer is characterized by increased blood vessel formulation in the bone marrow that is a predictive factor in survival. One treatment approach used for blood plasma cancer is stem cell transplantation with the patient’s own stem cells. Measurements were taken immediately prior to the stem cell transplant and at the time complete response was determined. The estimate of the mean difference, in bone marrow microvessel density before and after the stem cell transplant. There are 7 patients that are sampled. Is the blood vessel formulation in the bone marrow different after the stem cell transplant? Use alpha = .05. Before After 187 168 199 183 175 155 177 160 193 180 188 175 179 166 Here we see that these sample “share” something in common because it is the same patient and measurements were taken before and after treatment. First step is to state the hypothesis scenario. Because the key word says different this means it is a two tailed test. Ho: �̅�𝑑𝑖𝑓𝑓 = 0 Ha: �̅�𝑑𝑖𝑓𝑓 ≠ 0 Before we start calculating anything by hand and because we are given the raw data set, we can actually run this hypothesis test in Excel. And since you installed the Data Analysis Toolpak it is easy to do. Go to Data -> Data Analysis -> and scroll to where it says t-Test Paired Two Samples for Means and click OK Under Input: Variable 1 Range: you will highlight the Before column and make sure you include the top row where the Label is located. Variable 2 Range: you will highlight the After column and make sure you include the top row where the Label is located. Check the “Labels” box because we did include the first row of labels. For Alpha out 0.05 but this can be change depending on what significance level you use. Then make sure the bubble for New Workbook Ply: highlight and click OK. ItIn this document we will discuss 2 – sample Z- hypothesis testing and confidence intervals that uses a mean’s and known population S’s. This PDF discusses Z-Critical Value and you are discussing a sample mean and a population S. There are still 3 different hypothesis scenarios with a 2 – Sample Z Hypothesis Test. Lower Tail Test (1 tail): Ho: �̅�1 − �̅�2 = 0 Ha: �̅�1 − �̅�2 < 0 Upper Tailed Test (1 tail): Ho: �̅�1 − �̅�2 = 0 Ha: �̅�1 − �̅�2 > 0 Two Tailed Test: Ho:�̅�1 − �̅�2 = 0 Ha: �̅�1 − �̅�2 ≠ 0 The hypothesized value is 0 and the same key words apply from a 1 – sample hypothesis test to determine which scenario to use. 𝜇1 − 𝜇2 is the difference between the average in the first sample and the average in the second sample. The Z – Test Statistic = �̅�1− �̅�2−0 √ 𝑆1 2 𝑛1 + 𝑆2 2 𝑛2 Where S is the population standard deviation, 𝜇1 𝑎𝑛𝑑 𝜇2 are averages and n1 and n2 are the sample sizes. We can use =NORM.S.DIST to find the p-values. These should look familiar from the discussion forum. Example: A dietitian has developed a diet that is low in fats, carbs, and cholesterol. The dietitian wishes to examine the effects this diet has on the weights of obese people. Two random samples of 30 obese each are selected, and one group of 30 people is places on the low-fat diet. The other 30 people are places on a diet that contains approximately the same quantity of food, but has is not low in fats, carbs, and cholesterol. For each person the amount of weight lost (or gained) in a three-week period is recorded. There is a difference in the population mean weight losses for the two diets? The population S1 = 4.67 and the population S2 = 4.04. Use alpha = .05. Here we see we are given the Raw Data set. WL Low Diet WL Regular Diet 8 6 21 14 13 4 8 6 11 13 4 11 3 11 6 8 16 14 5 8 10 6 8 4 8 12 12 2 7 1 3 2 12 6 14 1 16 0 11 9 10 5 9 10 10 6 8 6 14 9 3 8 7 3 14 1 11 7 14 8 First step is to state the hypothesis scenario. Because the key word says difference this means it is a two tailed test. Ho: �̅�1 − �̅�2 = 0 Ha: �̅�1 − �̅�2 ≠ 0 Before we start calculating anything by hand and because we are given the raw data set, we can actually run this hypothesis test in Excel. And since you installed the Data Analysis Toolpak it is easy to do. First you need to input this Raw Data into Excel. Then go to Data -> Data Analysis -> and scroll to where it says z-Test Two Sample for Means and click OK Under Input: Variable 1 Range: you will highlight the WL Low Diet column and make sure you include the top row where the Label is located. Variable 2 Range: you will highlight the WL Regular Diet column and make sure you include the top row where the Label is located. Hypothesize Mean Difference: can be left as 0 Variance 1 Variance (known): Here is where you will put the Known Variance In this document we will discuss 2 – sample T- hypothesis testing and confidence intervals that uses a mean’s and unequal unknown Sample SD’s. This PDF discusses T-Critical Value and you are assuming unequal variances and discussing a sample mean and sample SD. There are still 3 different hypothesis scenarios with a 2 – Sample T Hypothesis Test. Lower Tail Test (1 tail): Ho: �̅�1 − �̅�2 = 0 Ha: �̅�1 − �̅�2 < 0 Upper Tailed Test (1 tail): Ho: �̅�1 − �̅�2 = 0 Ha: �̅�1 − �̅�2 > 0 Two Tailed Test: Ho: �̅�1 − �̅�2 = 0 Ha: �̅�1 − �̅�2 ≠ 0 The hypothesized value is 0 and the same key words apply from a 1 – sample hypothesis test to determine which scenario to use. 𝜇1 − 𝜇2 is the difference between the average in the first sample and the average in the second sample. The T – Test Statistic = �̅�1− �̅�2−0 √ 𝑆𝐷1 2 𝑛1 + 𝑆𝐷2 2 𝑛2 Where SD is the sample standard deviation, 𝜇1 𝑎𝑛𝑑 𝜇2 are averages and n1 and n2 are the sample sizes. We can use =T.DIST, =T.DIST.RT and =T.DIST.2T to find the p-values. These should look familiar from the discussion forum. Example: Suppose you wish to compare a new method of teaching reading to “slower learners” with the current standard method. You decide to base your comparison on the results of a reading test given at the end of a learning period of 6 months. A random sample of 23 “slower learners”, 11 are taught by the new method and 12 are taught by the standard method. All 23 children are taught by qualified instructors under similar conditions for 6 months. Does the new reading method increase test scores when compared to the standard reading method? Use Alpha = .05. New Method Standard Method 80 79 76 73 70 72 80 62 79 76 66 68 85 70 71 86 81 75 76 68 75 73 66 First step is to state the hypothesis scenario. Because the key word says increase this means it is an upper tailed test. Ho: �̅�1 − �̅�2 = 0 Ha:�̅�1 − �̅�2 > 0 Before we start calculating anything by hand and because we are given the raw data set, we can actually run this hypothesis test in Excel. And since you installed the Data Analysis Toolpak it is easy to do. First you need to input this Raw Data into Excel. Then go to Data -> Data Analysis -> and scroll to where it says t-Test: Two-Sample Assuming Unequal Variances and click OK Under Input: Variable 1 Range: you will highlight the New Method column and make sure you include the top row where the Label is located. Variable 2 Range: you will highlight the Standard Method column and make sure you include the top row where the Label is located. Hypothesize Mean Difference: 0 Check the “Labels” box because we did include the first row of labels. For Alpha out 0.05 but this can be change depending on what significance level you use. Then make sure the bubble for New Workbook Ply: highlight and click OK. It should look similar to the screenshHypothesis Testing is a decision-making process called a Test of Significance. There are 4 unique parts to Hypothesis Testing. 1) The Hypothesis Scenario. This includes the Null and Alternative scenarios. a. Ho: Null Hypothesis Ha or H1: Alternative Hypothesis 2) Z- Test Statistic Z- Test Stat = �̂�−𝑝0 (√ 𝑝0∗𝑞0 𝑛 ) Where “𝑝0” is the hypothesized value and 𝑞0 = 1 − 𝑝0. 3) P- value. The p-value tells you if something will be significant or not and if you can Accept or Reject the claim. You will use the p-value to draw a conclusion regarding the hypothesis test. a. We will use =NORM.S.DIST function to find the p-value. It should look familiar from Week 4. 4) Conclusion: a. If the p-value is less than alpha (< α) then Reject Ho/Accept Ha. b. If the p-value is greater than alpha (> α) then We Do Not Reject Ho. c. The most common alpha value is .05. If no, alpha value is given it will default to .05 but do note that alpha can also be, .10, .01, and .005 to name a few. Essentially alpha can be any value the statistician deems fit, but the most common values are .05, .01 and .10. One last thing before we get to an example. There are 3 different scenarios that are associated with the Hypothesis Scenario. 1) There is a Lower tailed (one tailed) Test or a Left Tailed Test. If the problem asks if there a significant decrease or less than or lower than or fewer than, then the problem is a lower tailed test. The “<” sign corresponds with the Ha. The hypothesis scenario will look like: a. Ho: �̂� = 𝑝0 Ha: �̂� < 𝑝0 (Here we see that “𝑝0” is the hypothesized value and the Less Than Sign “<” lines up with the Ha) 2) There is an Upper tailed (one tailed) Test or a Right Tailed Test. If the problem asks is there a significant increase or more than or greater than or higher than, then the problem is an upper tailed test. The “>” sign corresponds with the Ha. The hypothesis scenario will look like: a. Ho: �̂� = 𝑝0 Ha: �̂� > 𝑝0 (Here we see that “𝑝0” is the hypothesized value and the Greater Than Sign “>” lines up with the Ha) 3) There is a Two tailed Test. If the problem asks is there a significant difference or statistical evidence or asks if it is not the same, then the problem is a two-tailed test. The “≠” sign corresponds with the Ha. The hypothesis scenario will look like: a. Ho: �̂� = 𝑝0 Ha: �̂� ≠ 𝑝0 (Here we see that “𝑝0” is the hypothesized value and the Greater Than Sign “≠” lines up with the Ha) The hypothesized value is what we think should happen or what has been found to be true in the past. Now let’s continue to look at our car price data from Week 3. In Week 3, I asked you to calculate the average and then find how many data points fell below the average. We called this value p and then we found q. If we look back at my data set, we see that p = .70 and q = .30. We will call this �̂� = .70 and �̂� = .30. We want to run Sheet1 Vehicle type/class Year Make Model Price MPG (City) MPG (hwy) Max Seating SUV 2020 BMW X3 $42,945 25 29 5 Sedan 2019 Chevrolet Cruze $18,870 28 38 5 SUV 2021 Chevrolet Tahoe $55,095 16 20 9 Coupe 2020 Chevrolet Camaro $26,495 20 30 4 SUV 2020 Ford Edge $32,345 21 29 5 SUV 2020 Ford Explorer $34,010 21 28 7 Luxury 2020 Porsche 911 $111,550 18 24 4 Van/Minivan 2020 Toyota Sienna $32,760 19 26 7 Sedan 2020 Dodge Charger $31,390 19 30 5 Van/Minivan 2019 Dodge Grand Caravan $27,595 17 25 7 Qualitative Qualitative Qualitative Qualitative Quantitative Quantitative Quantitative QuantitativeSean Hi everyone, Well, this week I needed more than just the .pdfs and textbook, I have been spending a lot of time reviewing differences in Z & T- tests. Since I spent so much time on YouTube and reading the book over and over again, I have determined to use a T-test based upon a sample size of less than 30. The assignment was to determine the sell of vehicles for more than 40th percentile of the vehicles from week 1 which I computed in excel as =PERCENTILE.INC(A1:A10,0.4), with A1:10 as my prices, 0.4 is percentage = $41665.40 Based upon the request to determine a “more than” function, I used a RIGHT SIDED TEST Step 1: 𝐻0: 𝜇 = 41665.40 𝐻𝑎: 𝜇 > 41665.40 Step 2: & Step 3 Mean (x-bar): 44280.2 Standard Error (SE): 6264.048 Standard Deviation (s): 19808.66 Count (n): 10 𝐻0: 𝜇 = 41665.40 𝐻𝑎: 𝜇 > 41665.40 Sample Error 6264.048 TS: =(44280.20-41665.40)/6264.048 =0.41743 P-Value: =T.DIST.RT(D10,D4-1) = (.41743, 10-1) =0.343073 Step 4: Conclusion a.0.343 > .05, Therefore our p-value is greater than alpha. b.We fail to reject the null hypothesis (Ho) because our p-value is more than alpha c.There is not enough evidence to suggest that the average vehicle from the type of car you chose during week 1 sells for more than $41665.40. The WK6 tab on my spreadsheet contains all formulas and data, I can only hope it is correct. I hope everyone has a good week. Take care, Sean Dylan Hello class, For this week’s assignment we were tasked to run a hypothesis test using the data from week 1. The scenario that we were presented with was “A town official claims that the average vehicle in their area sells for more than the 40th percentile of your data set.” The pdf in Handy Helpers was useful in completing this assignment. First, I determined that I was going to use a t-test. I used a t-test because I was comparing a sample mean with the population mean and it is a small sample group. The next step was to determine the value for the 40th percentile of my week 1 data set. I utilized the excel function =PERCENTILE.INC and highlighted my original data set of 10 cars. This value came to $28,080. I used a Right Sided Test for my hypothesis. I was testing the claim that the average vehicle in their area sells for more than $28,080. Our professor set the alpha to 0.05. This is the value that will be used to test the following hypotheses: Ho:u= 28,080- null hypothesis Ha:u< 28,080- alternative hypothesis Next, I needed to calculate the test statistic. I started with using excel to find the standard error, SE. I used the following formula: = Sample SD/ SQRT(n) My SE was 4961.37. I used the TS formula from the pdf. The test statistic formula was also performed in excel with the formula =(32,069.50-28,080)/4961.47 with a value of 0.8041. After that I needed to determine the p-value. The excel function =T.DIST.RT(0.8041,10-1) with a value of 0.22. Based on this calculation we determine that the p-value 0.22 is more than the alpha set todescriptivestats sample size 10 Descri Mean 32069.5 Standard Error 4961.3723151967 Median 28920 Mode ERROR:#N/A Standard Deviation 15689.2368361243 Sample Variance 246152152.5 Kurtosis 8.2076699903 Skewness 2.7509552142 Range 55505 Minimum 19650 Maximum 75155 Sum 320695 Count 10 Confidence Level(95.0\%) 11223.4039201309 T Critical Value 2.2621571628 SE 4961.3723151967 ME 11223.4039201309 descriptivestats sample size 11 Price Mean 192790.454545455 Standard Error 160783.596649566 Median 28940 Mode ERROR:#N/A Standard Deviation 533258.862530453 Sample Variance 284365014467.273 Kurtosis 10.9770526255 Skewness 3.3119504866 Range 1780350 Minimum 19650 Maximum 1800000 Sum 2120695 Count 11 Confidence Level(95.0\%) 358248.178456988 Data Vehicle type/class Year Make Model Price MPG (city) MPG (highway) # of cylinders WEEK 3 WEEK 4 SUV 2021 GMC Yukon $ 75,155.00 15 22 8 P 0.7 less than 500$ Compact SUV 2021 Hyundai Santa Fe $ 26,850.00 25 28 4 Q 0.3 New SD 7844.6184180622 Compact SUV 2021 Ford Escape $ 24,885.00 44 37 4 P(X=4) 3.68\% p(x<31569.50) 47\% SUV 2021 Kia Sorento $ 29,390.00 24 29 4 P(X<5)/ P(X<=4) 4.73\% Pickup Truck 2021 Ford F150 $ 28,940.00 25 26 8 P(X>6)/ 1-P(X<=6) 64.96\% 1000$ above Pickup Truck 2021 Chevy Silverado 1500 $ 28,900.00 22 28 8 P(X>=4)/ 1-P(X<=3) 98.94\% New SD 7844.6184180622 Compact Car 2021 Hyundai Elantra $ 19,650.00 33 43 4 p(x>33069.50) 100\% Compact Car 2021 Chevy Malibu $ 22,140.00 29 36 4 Pickup Truck 2021 Dodge Ram 1500 $ 32,595.00 22 32 8 equal to mean $32069.50 SUV 2021 KIA Telluride $ 32,190.00 20 26 6 New SD 7844.6184180622 p(x=32069.50) 0\% Mean: $ 32,069.50 Median: $ 28,920.00 within $1500 SD: 15689.2368361243 New SD 7844.6184180622 Sample Size: 10 p(30569.50<x<33569.50) 15\% Week 2 Outlier Vehicle type/class Year Make Model Price MPG (city) MPG (highway) # of cylinders WEEK 5 Classmate 1- Week 5 SUV 2021 GMC Yukon $ 75,155.00 15 22 8 T Critical Value 2.2621571628 T-Critcal Value 1.8331129327 Compact SUV 2021 Hyundai Santa Fe $ 26,850.00 25 28 4 SE 4961.3723151967 SE 5206.771494 Compact SUV 2021 Ford Escape $ 24,885.00 44 37 4 ME 11223.4039201309 ME 10724.2086330932 SUV 2021 Kia Sorento $ 29,390.00 24 29 4 Mean $ 32,069.50 Mean $ 55,882.20 Pickup Truck 2021 Ford F150 $ 28,940.00 25 26 8 equation from pdf 11223.4039201309 Pickup Truck 2021 Chevy Silverado 1500 $ 28,900.00 22 28 8 55882.20-10724.2086 $ 45,157.99 Compact Car 2021 Hyundai Elantra $ 19,650.00 33 43 4 32069.50-11223.40 $ 20,846.10 $ 22,446.81 55882.20+10724.2086 $ 66,606.41 $ 21,448.42 Compact Car 2021 Chevy Malibu fail to reject null hypothesis because p-value is greater than alpha 29 36 4 32069.50+11223.40 $ 43,292.90 WK6 $11,787.00 Mean (x-bar): 44280.2 $15,795.00 Standard Error (SE): 6264.047794624 $37,703.00 Standard Deviation (s): 19808.6584031664 >40\% 41665.4 $40,421.00 Count (n): 10 $42,495.00 $44,028.00 𝐻0: 𝜇 = 41665.40 $54,984.00 𝐻𝑎: 𝜇 > 41665.40 $55,175.00 $66,660.00 Sample Error 6264.047794624 $73,754.00 TS: 0.4174297652 P-Value: 0.3430729332 Conclusion a. 0.343 > .05, Therefore our p-value is greater than alpha. b. We fail to reject the null hypothesis (Ho) because our p-value is more than alpha c. There is not enough evidence to suggest that the average vehicle from the type of car you chose during week 1 sells for more than $41665.40. Original Data Vehicle Type/Class Year Make Model Price MPG (city) MPG (highway) Engine Size (Quantitative) Hatchbacks 2021 Kia Rio $11,787.00 33 41 V4, 1.6 Liter Sedan 2021 Hyundai Accent $15,795.00 33 41 V4, 1.6 Liter Truck 2021 Toyota Tacoma $37,703.00 19 24 V6, 3.5 Liter Minivan 2021 Toyota Sienna $40,421.00 36 36 V4, 2.5 Liter SUV 2021 Mercedes Benz GLA $42,495.00 25 34 V4, 2.0 Liter Offroad Vehicle 2021 Jeep Wrangler $44,028.00 17 23 V6, 3.6 Liter Coupe 2021 Toyota GR Supra $54,984.00 22 30 V4, 2.0 Liter Truck 2021 Chevy Silverado 2500 $55,175.00 18 20 V8, 6.6Liter SUV 2021 Chevy Tahoe $66,660.00 16 20 V8, 5.3 Liter Sports 2021 Porshe 718 Spyder $73,754.00 17 23 V6, 4.0 Liter Mean $44,280.20 Median $43,261.50 Standard Deviation $19,808.66 Count 10 Week 3 Exactly 4 - P(x=4) 0.111476736 11.15\% Fewer than 5 - P(x < 5); P(x<j) = P (x ≤ 5-1) = P(x ≤ 4) 0.1662386176 16.62\% More than 6 - P(x > 6) P(x>j) = 1 - P (x ≤ 6) 0.3822806016 38.23\% At least 4 - P( x ≥ 4) same as 1 – P(x ≤ 4 - 1) = 1-P(x ≤ 3) 0.9452381184 94.52\% Super Car Data Vehicle Type/Class Year Make Model Price MPG (city) MPG (highway) Engine Size (Quantitative) Hatchbacks 2021 Kia Rio $11,787.00 33 41 V4, 1.6 Liter Sedan 2021 Hyundai Accent $15,795.00 33 41 V4, 1.6 Liter Truck 2021 Toyota Tacoma $37,703.00 19 24 V6, 3.5 Liter Minivan 2021 Toyota Sienna $40,421.00 36 36 V4, 2.5 Liter SUV 2021 Mercedes Benz GLA $42,495.00 25 34 V4, 2.0 Liter Offroad Vehicle 2021 Jeep Wrangler $44,028.00 17 23 V6, 3.6 Liter Coupe 2021 Toyota GR Supra $54,984.00 22 30 V4, 2.0 Liter Truck 2021 Chevy Silverado 2500 $55,175.00 18 20 V8, 6.6Liter SUV 2021 Chevy Tahoe $66,660.00 16 20 V8, 5.3 Liter Sports 2021 Porshe 718 Spyder $73,754.00 17 23 V6, 4.0 Liter Luxury 2021 Bugatti Chiron Pur Sport $3,500,000.00 8 13 V16/8.0 Liter Mean $358,436.55 Median $44,028.00 Standard Deviation $1,042,108.17 Count 11 Wk3 Bi Dist Price Observation 1 $11,787.00 Observation 2 $15,795.00 Observation 3 $37,703.00 Observation 4 $40,421.00 Observation 5 $42,495.00 Observation 6 $44,028.00 Observation 7 $54,984.00 Observation 8 $55,175.00 Observation 9 $66,660.00 Observation