Fluent Essays

Students will critically analyze the readings from Chapter 7 in your textbook. This project is planned to help you review, critique, and apply the readings to your Health Care setting as well as become the foundation for all of your remaining assignments.You need to read the article (in the additional weekly reading resources localize in the Syllabus and also in the Lectures link) assigned for week 2 and develop a 2-3-page paper reflecting your understanding and ability to apply the readings to your Health Care Setting. Each paper must be typewritten with 12-point font and double-spaced with standard margins. Follow APA format when referring to the selected articles and include a reference page.Originality: Turnitin submission required _ch07_slid.pptx Unformatted Attachment Preview Chapter 7 Hypothesis Testing Procedures Learning Objectives (1 of 3) • Define null and research hypothesis, test statistic, level of significance, and decision rule • Distinguish between Type I and Type II errors and discuss the implications of each • Explain the difference between one- and two-sided tests of hypothesis • Estimate and interpret p-values Learning Objectives (2 of 3) • Explain the relationship between confidence interval estimates and pvalues in drawing inferences • Perform analysis of variance by hand • Appropriately interpret the results of analysis of variance tests • Distinguish between one- and two-factor analysis of variance tests Learning Objectives (3 of 3) • Perform chi-square tests by hand • Appropriately interpret the results of chisquare tests • Identify the appropriate hypothesis testing procedures based on type of outcome variable and number of samples Hypothesis Testing • Research hypothesis is generated about unknown population parameter. • Sample data are analyzed and determined to support or refute the research hypothesis. Hypothesis Testing Procedures Step 1 • Null hypothesis (H0): – No difference, no change • Research hypothesis (H1): – What investigator believes to be true Hypothesis Testing Procedures Step 2 • Collect sample data and determine whether sample data support research hypothesis or not. • For example, in test for m, evaluate X . Hypothesis Testing Procedures Step 3 • Set up decision rule to decide when to believe null versus research hypothesis. • Depends on level of significance, a = P(Reject H0|H0 is true) Hypothesis Testing Procedures Steps 4 and 5 • Summarize sample information in test statistic (e.g., Z value). • Draw conclusion by comparing test statistic to decision rule. • Provide final assessment as to whether H1 is likely true given the observed data. p-values • p-values represent the exact significance of the data. • Estimate p-values when rejecting H0 to summarize significance of the data (can approximate with statistical tables, can get exact value with statistical computing package). • p-value is the smallest a where we still reject H0. Hypothesis Testing Procedures 1. Set up null and research hypotheses, select a. 2. Select test statistic. 2. Set up decision rule. 3. Compute test statistic. 4. Draw conclusion and summarize significance. Errors in Hypothesis Tests Hypothesis Testing for m • Continuous outcome • One sample H0: m = m0 H1: m > m0, m < m0, m ≠ m0 Test statistic: n ≥ 30 n < 30 Z= t= X - μ0 s/ n X - μ0 s/ n (Find critical value in Table 1C, Table 2, df = n – 1) Example 7.2. Hypothesis Testing for m (1 of 4) • The National Center for Health Statistics (NCHS) reports the mean total cholesterol for adults is 203. Is the mean total cholesterol in Framingham Heart Study participants significantly different? • In 3310 participants the mean is 200.3 with a standard deviation of 36.8. Example 7.2. Hypothesis Testing for m 1. H0: m = 203 H1: m ≠ 203 a = 0.05 2. Test statistic: Z= X - μ0 s/ n 3. Decision rule: Reject H0 if z ≥ 1.96 or if z ≤ –1.96 (2 of 4) Example 7.2. Hypothesis Testing for m (3 of 4) 4. Compute test statistic: X - μ0 200.3 − 203 Z= = = −4.22 s/ n 36.8 / 3310 5. Conclusion. Reject H0 because –4.22 < –1.96. We have statistically significant evidence at a = 0.05 to show that the mean total cholesterol is different in Framingham Heart Study participants. Example 7.2. Hypothesis Testing for m (4 of 4) • Significance of the findings: Z = –4.22 Table 1C. Critical Values for Two-Sided Tests a Z 0.20 1.282 0.10 1.645 0.05 1.960 0.010 2.576 0.001 3.291 0.0001 3.819 p < 0.0001 New Scenario • Outcome is dichotomous (p = population proportion). – Result of surgery (success, failure) – Cancer remission (yes/no) • One study sample • Data – On each participant, measure outcome (yes/no) x – n, x = number of positive responses, p̂ = n Hypothesis Testing for p • Dichotomous outcome • One sample H0: p = p0 H1: p > p0, p < p0, p ≠ p0 Test statistic: min[np0 , n(1 − p 0 )]  5 Z = p̂ – p 0 p 0 (1 – p 0 ) n (Find critical value in Table 1C) Example 7.4. Hypothesis Testing for p (1 of 3) • The NCHS reports that the prevalence of cigarette smoking among adults in 2002 is 21.1\%. Is the prevalence of smoking lower among participants in the Framingham Heart Study? • In 3536 participants, 482 reported smoking. Example 7.4. Hypothesis Testing for p 1. H0: p = 0.211 H1: p < 0.211 2. Test statistic: a = 0.05 Z = p̂ – p 0 p 0 (1 – p 0 ) n 3. Decision rule: Reject H0 if z ≤ –1.645 (2 of 3) Example 7.4. Hypothesis Testing for p (3 of 3) 4. Compute test statistic: Z = p̂ – p 0 = p 0 (1 – p 0 ) n 0.136 - 0.211 = -10.93 0.211(1- 0.211) 3536 5. Conclusion. Reject H0 because –10.93 < – 1.645. We have statistically significant evidence at a = 0.05 to show that the prevalence of smoking is lower among the Framingham Heart Study participants. (p < 0.0001) Hypothesis Testing for Categorical and Ordinal Outcomes* • Categorical or ordinal outcome • One sample H0: p1 = p10, p2 = p20,…,pk = pk0 H1: H0 is false Test statistic: 2 (O E) χ2 =  E (Find critical value in Table 3, df = k – 1) * c2 goodness-of-fit test Chi-Square Tests • c2 tests are based on the agreement between expected (under H0) and observed (sample) frequencies. Test statistic: (O - E ) χ =Σ E 2 2 Chi-Square Distribution • If H0 is true c2 will be close to 0; if H0 is false, c2 will be large. • Reject H0 if c2 > Critical value from Table 3 Example 7.6. c2 Goodness-of-Fit Test (1 of 4) • A university survey reveals that 60\% of students get no regular exercise, 25\% exercise sporadically and 15\% exercise regularly. The university institutes a health promotion campaign and re-evaluates exercise 1 year later. Number of students None 255 Sporadic 125 Regular 90 Example 7.6. c2 Goodness-of-Fit Test (2 of 4) 1. H0: p1 = 0.60, p2 = 0.25, p3 = 0.15 H1: H0 is false a = 0.05 2. Test statistic: 2 (O E) χ2 =  E 3. Decision rule: df = k – 1 = 3 – 1 = 2 Reject H0 if c2 ≥ 5.99 Example 7.6. c2 Goodness-of-Fit Test (3 of 4) 2 (O E) 2 4. Compute test statistic: χ =  E None Sporadic Regular Total No. students (O) Expected (E) 255 282 125 117.5 90 70.5 (O – E)2/E 2.59 0.48 5.39 c2 = 8.46 470 470 Example 7.6. c2 Goodness-of-Fit Test (4 of 4) 5. Conclusion. Reject H0 because 8.46 > 5.99. We have statistically significant evidence at a = 0.05 to show that the distribution of exercise is not 60\%, 25\%, 15\%. Using Table 3, the p-value is p < 0.005. New Scenario • Outcome is continuous. – SBP, weight, cholesterol • Two independent study samples • Data – On each participant, identify group and measure outcome. 2 1 1 2 2 n1,X ,s (or s1 ),n2 ,X2,s (or s2 ) Two Independent Samples (1 of 2) RCT: Set of Subjects Who Meet Study Eligibility Criteria Randomize Treatment 1 Mean Treatment 1 Treatment 2 Mean Treatment 2 Two Independent Samples (2 of 2) Cohort Study: Set of Subjects Who Meet Study Inclusion Criteria Group 1 Mean Group 1 Group 2 Mean Group 2 Hypothesis Testing for (m1 − m2) (1 of 2) • Continuous outcome • Two independent samples H 0: m 1 = m 2 (m1 − m2 = 0) H1: m1 > m2, m1< m2, m1 ≠ m2 Hypothesis Testing for (m1 − m2) (2 of 2) • Continuous outcome • Two independent samples H0: m1 = m2 H1: m1 > m2, m1 < m2, m1 ≠ m2 Test statistic: n1 ≥ 30 and n2 ≥ 30 n1 < 30 or Z= X1 - X 2 1 1 Sp + n1 n 2 X1 - X 2 t= Sp n2 < 30 1 1 + n1 n 2 (Find critical value in Table 1C, Table 2, df = n1 + n2 – 2) Pooled Estimate of Common Standard Deviation, Sp • Previous formulas assume equal variances (s12 = s22). • If 0.5 ≤ s12/s22 ≤ 2, assumption is reasonable. Sp = (n 1 − 1)s + (n 2 − 1)s n1 + n 2 − 2 2 1 2 2 Example 7.9. Hypothesis Testing for (m1 − m2) (1 of 3) • A clinical trial is run to assess the effectiveness of a new drug in lowering cholesterol. Patients are randomized to receive the new drug or placebo and total cholesterol is measured after 6 weeks on the assigned treatment. • Is there evidence of a statistically significant reduction in cholesterol for patients on the new drug? Example 7.9. Hypothesis Testing for (m1 − m2) New drug Placebo Sample Size 15 15 Mean 195.9 227.4 (2 of 3) Std Dev 28.7 30.3 Example 7.9. Hypothesis Testing for (m1 − m2) 1. H0: m1 = m2 H1: m1 < m2 2. Test statistic: t = a = 0.05 X1 - X 2 1 1 Sp + n1 n 2 3. Decision rule: df = n1 + n2 – 2 = 28 Reject H0 if t ≤ –1.701 (3 of 3) Assess Equality of Variances • Ratio of sample variances: 28.72/30.32 = 0.90 Sp = Sp = (n 1 − 1)s12 + (n 2 − 1)s 22 n1 + n 2 − 2 (15 − 1)28.7 2 + (15 − 1)30.32 15 + 15 − 2 = 870.89 = 29.5 Example 7.9. Hypothesis Testing for (m1 − m2) 4. Compute test statistic: t= X1 - X 2 195.9 − 227.4 = = −2.92 1 1 1 1 Sp + 29.5 + n1 n 2 15 15 5. Conclusion. Reject H0 because –2.92 < –1.701. We have statistically significant evidence at a = 0.05 to show that the mean cholesterol level is lower in patients on treatment as compared to placebo. (p < 0.005) New Scenario • Outcome is continuous. – SBP, weight, cholesterol • Two matched study samples • Data – On each participant, measure outcome under each experimental condition. – Compute differences (D = X1 – X2). n, X d , s d Two Dependent/Matched Samples Subject ID 1 2 . . • Measure 1 55 42 Measure 2 70 60 Measures taken serially in time or under different experimental conditions. Crossover Trial Eligible Participants Treatment Treatment Placebo Placebo R Each participant is measured on treatment and placebo. Hypothesis Testing for md • Continuous outcome • Two matched/paired sample H0: md = 0 H1: md > 0, md < 0, md ≠ 0 Test statistic: X -μ n ≥ 30 n < 30 Z= t= d sd d n Xd - μ d sd n (Find critical value in Table 1C, Table 2, df = n – 1) Example 7.10. Hypothesis Testing for md (1 of 3) • Is there a statistically significant difference in mean systolic blood pressures (SBPs) measured at exams 6 and 7 (approximately 4 years apart) in the Framingham Offspring Study? • Among n = 15 randomly selected participants, the mean difference was –5.3 units and the standard deviation was 12.8 units. Differences were computed by subtracting the exam 6 value from the exam 7 value. Example 7.10. Hypothesis Testing for md 1. H0: md = 0 H1: md ≠ 0 2. Test statistic: (2 of 3) a = 0.05 t= Xd - μ d sd n 3. Decision rule: df = n – 1 = 14 Reject H0 if t ≥ 2.145 or if z ≤ –2.145 Example 7.10. Hypothesis Testing for md (3 of 3) 4. Compute test statistic: Xd - μ d − 5.3 − 0 t= = = −1.60 s d n 12.8 / 15 5. Conclusion. Do not reject H0 because –2.145 < –1.60 < 2.145. We do not have statistically significant evidence at a = 0.05 to show that there is a difference in systolic blood pressures over time. New Scenario • Outcome is dichotomous – Result of surgery (success, failure) – Cancer remission (yes/no) • Two independent study samples • Data – On each participant, identify group and measure outcome (yes/no) n1 , p̂1 , n 2 , p̂ 2 Hypothesis Testing for (p1 – p2) • • Dichotomous outcome Two independent samples H0: p1 = p2 H1: p1 >p2, p1< p2, p1 ≠ p2 Test statistic: min[n1p̂1 , n1 (1 − p̂1 ), n 2 p̂ 2 , n 2 (1 − p̂ 2 )]  5 Z= p̂1 - p̂ 2 1 1  p̂(1 - p̂) +   n1 n 2  (Find critical value in Table 1C) Example 7.12. Hypothesis Testing for (p1 – p2) • (1 of 4) Is the prevalence of CVD different in smokers as compared to nonsmokers in the Framingham Offspring Study? Free of CVD 2757 History of CVD 298 3055 Current smoker 663 81 744 Total 3420 379 3799 Nonsmoker Total Example 7.12. Hypothesis Testing for (p1 – p2) 1. H0: p1 = p2 H1: p1 ≠ p2 2. Test statistic: Z = a = 0.05 p̂1 - p̂ 2  1 1   p̂(1 - p̂) +  n1 n 2  3. Decision rule: Reject H0 if Z ≤ –1.96 or if Z ≥ 1.96 (2 of 4) Example 7.12. Hypothesis Testing for (p1 – p2) (3 of 4) 4. Compute test statistic: Z= Z= p̂1 - p̂ 2  1 1   p̂(1 - p̂) +  n1 n 2  p̂1 = 81 298 = 0.1089, p̂ 2 = = 0.0975 744 3055 0.1089 - 0.0975 1   1 0.0988(1 - 0.0988) +   744 3055  p̂ = 81 + 298 = 0.0988 744 + 3055 = 0.927 Example 7.12. Hypothesis Testing for (p1 – p2) (4 of 4) 5. Conclusion. Do not reject H0 because –1.96 < 0.927 < 1.96. We do not have statistically significant evidence at a = 0.05 to show that there is a difference in prevalence of CVD between smokers and nonsmokers. Hypothesis Testing for More than Two Means* • Continuous outcome • k independent Samples, k > 2 H0: m1 = m2 = m3 … = mk H1: Means are not all equal Test statistic: Σn j (X j − X) 2 /(k − 1) F= ΣΣ(X − X j ) 2 /(N − k) (Find critical value in Table 4) *Analysis of variance Test Statistic: F Statistic • Comparison of two estimates of variability in data – Between treatment variation, is based on the assumption that H0 is true (i.e., population means are equal). – Within treatment, residual or error variation, is independent of H0 (i.e., we do not assume that the population means are equal and we treat each sample separately). F Statistic (1 of 2) Difference between each group mean and overall mean Σn j (X j − X) /(k − 1) 2 F= ΣΣ(X − X j ) /(N − k) 2 Difference between each observation and its group mean (within group variation— error) F Statistic (2 of 2) F = MSB/MSE MS = Mean Square • What values of F indicate H0 is likely true? Decision Rule Reject H0 if F ≥ critical value of F with df1 = k – 1 and df2 = N – k from Table 4 k = Number of comparison groups N = Total sample size ANOVA Table Source of Variation Sums of Squares df Between 2 treatments SSB = Σ n j (X j - X ) MSB/MSE Error Total SSE = Σ Σ (X - X j) SST = Σ Σ (X - X ) Mean Squares k – 1 SSB/k – 1 2 N – k SSE/N – k 2 N–1 F Example 7.14. ANOVA (1 of 12) • Is there a significant difference in mean weight loss among four different diet programs? (Data are pounds lost over 8 weeks) Low-Cal 8 9 6 7 3 Low-Fat 2 4 3 5 1 Low-Carb 3 5 4 2 3 Control 2 2 -1 0 3 Example 7.14. ANOVA (2 of 12) 1. H0: m1 = m2 = m3 = m4 H1: Means are not all equal 2. Test statistic: F= Σn j (X j − X) 2 /(k − 1) ΣΣ(X − X j ) 2 /(N − k) a = 0.05 Example 7.14. ANOVA (3 of 12) 3. Decision rule: df1 = k – 1 = 4 – 1 = 3 df2 = N – k = 20 – 4 =16 Reject H0 if F ≥ 3.24 Example 7.14. ANOVA (4 of 12) Summary Statistics on Weight Loss by Treatment Low-Cal N 5 Mean 6.6 Low-Fat 5 3.0 Overall Mean = 3.6 Low-Carb Control 5 5 3.4 1.2 Example 7.14. ANOVA (5 of 12) SSB = Σ n j (X j - X ) 2 = 5(6.6 – 3.6)2 + 5(3.0 – 3.6)2 + 5(3.4 – 3.6)2 + 5(1.2 – 3.6)2 = 75.8 Example 7.14. ANOVA (6 of 12) SSE = Σ Σ (X - X j) 2 Example 7.14. ANOVA (7 of 12) SSE = Σ Σ (X - X j) 2 Example 7.14. ANOVA (8 of 12) 2 SSE = Σ Σ (X - X j) Example 7.14. ANOVA (9 of 12) SSE = Σ Σ (X - X j) 2 Example 7.14. ANOVA (10 of 12) SSE = Σ Σ (X - X j) 2 = 21.4 + 10.0 + 5.4 + 10.6 = 47.4 Example 7.14. ANOVA (11 of 12) Source of Variation Sums of Squares df Mean Squares F 8.43 Between Treatments Error 75.8 3 25.3 47.4 16 3.0 Total 123.2 19 Example 7.14. ANOVA (12 of 12) 4. Compute test statistic: F = 8.43 5. Conclusion. Reject H0 because 8.43 > 3.24. We have statistically significant evidence at a = 0.05 to show that there is a difference in mean weight loss among four different diet programs. Two-Factor ANOVA • Compare means of a continuous outcome across two grouping variables or factors – Overall test—is there a difference in cell means? – Factor A—marginal means – Factor B—marginal means – Interaction—difference in means across levels of Factor B for each level of Factor A? Interaction Cell Means Factor A Factor B 1 2 1 45 65 2 58 55 3 70 38 75 70 65 60 A1 A2 55 50 45 40 35 1 2 3 No Interaction Cell Means Factor A Factor B 1 2 1 45 38 2 58 55 3 70 65 75 70 65 60 A1 A2 55 50 45 40 35 1 2 3 Example 7.16. Two-Factor ANOVA (1 of 3) • Clinical trial to compare time to pain relief of three competing drugs for joint pain. Investigators hypothesize that there may be a differential effect in men versus women. • Design: N = 30 participants (15 men and 15 women) are assigned to three treatments (A, B, C) Example 7.16. Two-Factor ANOVA (2 of 3) • Mean times to pain relief by treatment and sex • Is there a difference in mean times to pain relief? Are differences due to treatment? Sex? Or both? Example 7.16. Two-Factor ANOVA (3 of 3) Source of Variation Sums of Squares df Mean Square Model Treatment Sex Treatment*Sex 967.0 651.5 313.6 1.9 5 2 1 2 193.4 325.7 313.6 0.9 Error 224.4 24 9.4 Total 1191.4 29 F 20.7 34.8 33.5 0.1 p-value 0.0001 0.0001 0.0001 0.9054 Hypothesis Testing for Categorical or Ordinal Outcomes* • Categorical or ordinal outcome • Two or more samples H0: The distribution of the outcome is independent of the groups H1: H0 is false 2 (O E) 2 χ = Test statistic: E (Find critical value in Table 3: df = (r – 1)(c – 1)) * c2 test of independence Chi-Square Test of Independence • Outcome is categorical or ordinal (2+ levels) and there are two or more independent comparison groups (e.g., treatments). H0: Treatment and outcome are independent distributions of outcome are the same across treatments) Example 7.17. c2 Test of Independence (1 of 6) • Is there a relationship between students’ living arrangement and exercise status? Dormitory On-campus apt. Off-campus apt. At home Total Exercise Status None Sporadic Regular 32 30 28 74 64 42 110 25 15 39 6 5 255 125 90 Total 90 180 150 50 470 Example 7.17. c2 Test of Independence (2 of 6) 1. H0: Living arrangement and exercise status are independent H1: H0 is false a = 0.05 2 (O E) 2. Test statistic: χ 2 =  E 3. Decision rule: df = (r – 1)(c – 1) = 3(2) = 6 Reject H0 if c2 ≥ 12.59 Example 7.17. c2 Test of Independence (3 of 6) 4. Compute test statistic: 2 (O E) χ2 =  E O = Observed frequency E = Expected frequency E = (row total)*(column total)/N Example 7.17. c2 Test of Independence (4 of 6) 4. Compute test statistic: Table entries are Observed (Expected) frequencies None Total Dormitory 32 (90*255/470 = 48.8) On-campus apt. 74 (97.7) Off-campus apt. 110 (81.4) At home 39 (27.1) Total 255 Exercise Status Sporadic Regular 30 (23.9) 64 (47.9) 25 (39.9) 6 (13.3) 125 28 (17.2) 42 (34.5) 15 (28.7) 5 (9.6) 90 90 180 150 50 470 Example 7.17. c2 Test of Independence (5 of 6) 4. Compute test statistic: 2 2 2 2 (32 − 48.8) (30 − 23.9) (28 − 17.2) (5 − 9.6) χ2 = + + + ... + 48.8 23.9 17.2 9.6 χ 2 = 60.5 Example 7.17. c2 Test of Independence (6 of 6) 5. Conclusion. Reject H0 because 60.5 > 12.59. We have statistically significant evidence at a = 0.05 to show that living arrangement and exercise status are not independent. (P < 0.005) ... Purchase answer to see full attachment