# 2-3-page paper reflecting your understanding APA

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Question description

https://ippcr.nihtraining.com/handouts/2011/Hypoth...

1. Introduction (25%) Deliver a brief synopsis of the meaning (not a description) of each Chapter and articles you read, in your own words.

What is your reaction to the content of the articles?

What did you learn about Null and research Hypothesis, test statistic and decision rule?

What did you learn about type I and Type II errors and the implications of each?

Did these Chapter and articles change your thoughts about Hypothesis Testing Procedures? If so, how? If not, what remained the same?

3. Conclusion (15%)

Briefly recapitulate your thoughts & conclusion to your critique of the articles and Chapter you read.How did these articles and Chapters impact your thoughts on appropriately interpret results off chi-square test?

Evaluation will be based on how clearly you respond to the above, in particular:

a) The clarity with which you critique the articles;

b) The depth, scope, and organization of your paper; and,

c) Your conclusions, including a description of the impact of these articles and Chapters on any Health Care Setting.

Chapter 7 Hypothesis Testing Procedures Learning Objectives • 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 twosided tests of hypothesis Learning Objectives • Estimate and interpret p-values • Explain the relationship between confidence interval estimates and p-values 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 • Perform chi-square tests by hand • Appropriately interpret the results of chi-square 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. 2. 2. 3. 4. Set up null and research hypotheses, select a Select test statistic Set up decision rule Compute test statistic Draw conclusion & summarize significance Errors in Hypothesis Tests Hypothesis Testing for m • Continuous outcome • 1 Sample H0: m=m0 H1: m>m0, m30 (Find critical s/ n n<30 t= X - μ0 s/ n value in Table 1C, Table 2, df=n-1) Example 7.2. Hypothesis Testing for m 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 Example 7.2. Hypothesis Testing for m 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 the Framingham Heart Study participants. Example 7.2. Hypothesis Testing for m 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) – n, x=# positive responses, x p̂ = n Hypothesis Testing for p • Dichotomous outcome • 1 Sample H0: p=p0 H1: p>p0, p Critical Value from Table 3 Example 7.6. c2 goodness-of-fit test 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 one year later. Number of students 255 None Sporadic 125 90 Regular Example 7.6. c2 goodness-of-fit test 1. H0: p1=0.60, p2=0.25,p3=0.15 H1: H0 is false 2. Test statistic a=0.05 2 (O E) χ2 =  E 3. Decision rule Reject H0 if c2 > 5.99 df=k-1=3-1=2 Example 7.6. c2 goodness-of-fit test 4. Compute test statistic No. students (O) 255 Expected (E) 282 (O-E)2/E 2.59 2 (O E) χ2 =  E None Sporadic Regular 125 90 470 117.5 70.5 470 0.48 5.39 c2 = 8.46 Total Example 7.6. c2 goodness-of-fit test 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 2 – n1 , X1 , s1 (or s1 ), n 2 , X 2 , s 2 (or s 2 ) Two Independent Samples RCT: Set of Subjects Who Meet Study Eligibility Criteria Randomize Treatment 1 Mean Trt 1 Treatment 2 Mean Trt 2 Two Independent Samples 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) • Continuous outcome • 2 Independent Sample H0: m1=m2 (m1m2 = 0) H1: m1>m2, m1m2, m130 and n2> 30 n1<30 or n2<30 Z= t= X1 - X 2 1 1 Sp  n1 n 2 X1 - X 2 Sp 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) 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 SizeMean Std Dev 15 195.9 28.7 15 227.4 30.3 Example 7.2. Hypothesis Testing for (m1m2) 1. H0: m1=m2 H1: m10, md<0, md≠0 Test Statistic n>30 n<30 Z= t= Xd - μ d sd n Xd - μd sd n (Find critical value in Table 1C, Table 2, df=n-1) Example 7.10. Hypothesis Testing for md 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 a=0.05 2. Test statistic t= Xd - μd sd n 3. Decision rule, df=n-1=14 Reject H0 if t > 2.145 or if z < -2.145 Example 7.2. Hypothesis Testing for md 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 • 2 Independent Sample H0: p1=p2 H1: p1>p2, p1 1.96 Example 7.2. Hypothesis Testing for (p1-p2) 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 7.2. Hypothesis Testing for (p1-p2) 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 prevalent CVD between smokers and nonsmokers. Hypothesis Testing for More than 2 Means* • Continuous outcome • k Independent Samples, k > 2 H0: m1=m2=m3 … =mk H1: Means are not all equal Test Statistic F= Σn j (X j  X) 2 /(k  1) 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 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 F = MSB/MSE MS = Mean Square What values of F that 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= # comparison groups N=Total sample size ANOVA Table Source of Variation Sums of Squares Between Treatments SSB = Σ n j (X j - X ) Error SSE = Σ Σ (X - X j) Total SST = Σ Σ (X - X ) df Mean Squares k-1 SSB/k-1 MSB/MSE N-k SSE/N-k F 2 2 2 N-1 Example 7.14. ANOVA Is there a significant difference in mean weight loss among 4 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 1. H0: m1=m2=m3=m4 H1: Means are not all equal a=0.05 2. Test statistic F= Σn j (X j  X) 2 /(k  1) ΣΣ(X  X j ) 2 /(N  k) Example 7.14. ANOVA 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 Summary Statistics on Weight Loss by Treatment Low-Cal Low-Fat Low-Carb N 5 5 5 Mean 6.6 3.0 3.4 1.2 Overall Mean = 3.6 Control 5 Example 7.14. ANOVA 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 SSE = Σ Σ (X - X j) Low-Cal 8 9 6 7 3 Total (X-6.6) 1.4 2.4 -0.6 0.4 -3.6 0 2 (X-6.6)2 2.0 5.8 0.4 0.2 13.0 21.4 Example 7.14. ANOVA SSE = Σ Σ (X - X j) Low-Fat 2 4 3 5 1 Total (X-3.0) -1.0 1.0 0 2.0 -2.0 0 2 (X-3.0)2 1.0 1.0 0 4.0 4.0 10.0 Example 7.14. ANOVA SSE = Σ Σ (X - X j) Low-Carb 3 5 4 2 3 Total (X-3.4) -0.4 1.6 0.6 -1.4 -0.4 0 2 (X-3.4)2 0.2 2.6 0.4 2.0 0.2 5.4 Example 7.14. ANOVA SSE = Σ Σ (X - X j) Control 2 2 -1 0 3 Total (X-1.2) 0.8 0.8 -2.2 -1.2 1.8 0 2 (X-1.2)2 0.6 0.6 4.8 1.4 3.2 10.6 Example 7.14. ANOVA SSE = Σ Σ (X - X j) 2 =21.4 + 10.0 + 5.4 + 10.6 = 47.4 Example 7.14. ANOVA Source of Variation Between Treatments Error Total Sums of Squares df Mean Squares F 75.8 3 25.3 8.43 47.4 16 3.0 123.2 19 Example 7.14. ANOVA 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 4 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 1 2 Factor B 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 1 2 Factor B 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 • 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 3 treatments (A, B, C) EXAMPLE 7.16 Two Factor ANOVA • Mean times to pain relief by treatment and gender Men Women A 14.8 21.4 B 17.4 23.2 C 25.4 32.4 • Is there a difference in mean times to pain relief? Are differences due to treatment? Gender? Or both? EXAMPLE 7.16 Two Factor ANOVA Table Source Of Variation Model Treatment Gender Treatment*Gender Sums of Mean Squares df Squares 967.0 651.5 313.6 1.9 2 F p-value 5 193.4 20.7 0.0001 2 325.7 34.8 0.0001 1 313.6 33.5 0.0001 0.9 0.1 0.9054 Error 224.4 24 Total 1191.4 29 9.4 Hypothesis Testing for Categorical or Ordinal Outcomes* • Categorical or ordinal outcome • 2 or More Samples H0: The distribution of the outcome is independent of the groups H1: H0 is false Test Statistic * c2 test of independence 2 (O E) χ2 =  E (Find critical value in Table 3 df=(r-1)(c-1)) 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 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 90 74 64 42 180 110 25 15 150 39 6 5 50 255 125 90 470 Total Example 7.17. c2 Test of Independence 1. H0: Living arrangement and exercise status are independent H1: H0 is false a=0.05 2. Test statistic 2 (O E) χ2 =  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 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. Compute test statistic Table entries are Observed (Expected) frequencies Exercise Status None Sporadic Dormitory 32 30 (90*255/470=48.8) (23.9) (17.2) On-campus Apt 74 64 (97.7) (47.9) (34.5) Off-campus Apt 110 25 (81.4) (39.9) (28.7) At Home 39 6 5 (27.1) (13.3) (9.6) Total 255 125 90 Regular Total 28 90 42 180 15 150 50 470 Example 7.17. c2 Test of Independence 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 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)

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