Question

9.6 in order to compare the means of two populations, inde- NW pendent random samples of 400 observations are selected from each population, with the following results Sample 1 Sample 2 $.240 s2 200 5,275 1150 a. Use a 95% confidence interval to estimate the dif- ference between the population means (μ,-μ Interpret the confidence interval. b. Test the null hypothesis Ho (μι-μ)--0 versus the c. Suppose the test in part b were conducted with the d. Test thenull hypothesis Ho: (ANR)-25 versus the alternative hypothesis H 2(μι-μ2)-0. Give the p-value of the test, and interpret the result. alternative hypothesis Ha (A H2) a 0. How would your answer to part b change? alternative H, (μ1-μ2) pri 25. Give the p-value, and interpret the result. Compare your answer with that obtained from the test conducted in part b. e. What assumptions are necessary to ensure the validity of the inferential procedures applied in parts a-d?

Answer 1

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95% CI for μ 1-42 There are 3 ways to calculate this, as different professors prefer different ones: 1. Using known variance assumption: This is justified because given sample sizes are large and we can assume sample standard deviations to be almost equal to population standard deviations (si σ2 and s2 σ2). Typically sample size is considered large when ni,n2 2 30. 2. Using equal variance assumption: In this method we assume that the pop- ulation variances are equal (but unknown) for both populations. This is also called pooled variance method 3. Without equal variance assumption: This method has no assumption on variance Normality and independence between two populations is necessary assumption in all cases. 95% CI for μ1-μ2 with known variance We want to construct confidence interval of mean difference for independent samples from data sample 1sample 2 527525240 n400 n2400 Significance level α-1-0.95-0.05 Standard Error: ơ7,-b- The Critical Value2a/20.025 12.5 n n2 1.96 (From z table Critical Values , iza/2- 11.96 0T + σ2-24.5 Margin of Error-E-h/2 x TL Point Estimate of difference: T1-235 Limits of 95% confidence interval are given by: Lower limit -2- E-35 -24.5 10.5 Upper limit -i -E- 35+ 24.5- 59.5 95% confidence interval is:--22 - (10.5, 59.5) E-35 24.5 With 95% CI using normal dist

95% CI for μ1-μ2 using equal variance assumption We want to construct confidence interval of mcan difference for independent samples from data: sample 1 sample 2 5275 | #2 : : 5240 81-150 | 82-200 400400 Significance level-α-1-0.95-0.05 Since Variances are assumed equal, but known, we use pooled variance Degrees of freedom: = df = n1 + n2-2-798 The pooled cstimate of Variance is: s2= (ni-1 )81 + (n2-1)83 = 31250 (ni-1) + (n2-1) 倍. + 12.5 n n2 The standard error is s-- Critical Value = ta/2df = t0.025.df=798 = 1.963 (from t-table, two-tails, d.f. = 798) + TP Margin of Error = E = to/2.df × 2 1.963 × 12.5 E 24.538 Margin of Error, E 24.538 Point Estimate of difference:-1-T2-35 Limits of 95% confidence interval are given by: Lower lin it-a-1-T2-E-35-24.5375 = 10.4625 Upper limit -x2 + E-35 + 24.5375 = 59.5375 95% confidence interval is:--T2 E-35 245375 (10.4625, 59.5375) 95% CI using t-dist: 10.463 < ,41-42 < 59.538 Note that exact answer using technology is: (10.4632350897, 59.5367649103) 95% CI for μ1-μ2 without equal variance assumption

samples from data sample1sample 2 5275 | 2-5240 81= 150 | 82-200 n1-400 | ng 400 nificance level 1-0.95 0.05 Since Variances are not equal, and are unknown, we use 2 sample t-distribution method: Degrees of freedom :-df-smaller of ni-l or n2-1 = 399 Standard Error: σί,-i T 2 Critical Value -ta/2,too25,3 399 ) 1.966 from t-table, two-tails, d.f. - Margin of Error-E to2,df X 2250040000 1.966 × 400 E 1.966 x 12.5 24.575 Margin of Error, E 24.575 Point Estimate of difference: -2-35 Limits of 95% confidence interval are given by: Lower limit = -x2ーE-35-24.575-10.425 Upper limit 2 + E-35 + 24.575 59.575 95% confidence interval is: 2-1-7, E 35 24.575 (10.425, 59.575) 95% CI using t-dist: 10.425 < μ1-μ2 < 59.575 Note that exact answer using technology is: (10.4259088010, 59.5740911990) Note that some books teach Welch's formula, where approrimate degrees of free- n2 739.985163 dom are:VA And t-critical would have been: ±1.963 Note that eract answer for Welch's formula is (10.4602591367, 59.5397408633) Thank youDe

Excel add-in (Mega Stat) procedure: Step 1: In EXCEL, Select Add-Ins > Mega Stat > Hypothesis Tests Step 2: Choose Compare Two Independent Groups. Step 3: Choose Summary Input. Step 4: In Group 1, enter B1:B4 (Select ranges with label mean, SD, n). Step 5: In Group 2, enter D1:D4 (Select ranges with label mean, SD, n). Step 6: Enter 0 Under Hypothesized difference. Step 7: Check t-test (unequal variance), enter Confidence level as 95.0 Step 8: Choose not equal in alternative. Step 9: Click OK. Excel add-in (Mega Stat) output:

Hypothesis Test: Independent Groups (t-test, unequal variance) Sample 1 Sample 2 5240 mean 5275 150 400 200 std. dev 400 n 739 df 35.000 difference (Sample 1 Sample 2) 12.500 standard error of difference 0 hypothesized difference 2.80 t 0052 p-value (two-tailed) Thus the p-value is 0.0052 Conclusion Use a significance level, a0.05 Here, p-value is 0.0052, which is lesser than the value of α-005 That is, p-value (-0.0052)< α(-0.05) Therefore, by the rejection rule, reject the null hypothesis Ho

Interpretation There is sufficient evidence at level of significance α-: 0.05 That is, there is a statistically significant difference between the population means. Excel add-in (Mega Stat) procedure: Step 1: In EXCEL, Select Add-Ins > Mega Stat > Hypothesis Tests Step 2: Choose Compare Two Independent Groups. Step 3: Choose Summary Input. Step 4: In Group 1, enter Bl:B4 (Select ranges with label mean, SD, n). Step 5: In Group 2, enter D1:D4 (Select ranges with label mean, SD, n). Step 6: Enter 0 Under Hypothesized difference. Step 7: Check t-test (unequal variance), enter Confidence level as 95.0. Step 8: Choose greater than in alternative. Step 9: Click OK. Excel add-in (Mega Stat) output:

Hypothesis Test: Independent Groups (t-test, unequal variance) Sample 1 Sample 2 5240 mean 5275 150 400 200 std. dev 400 n 739 df 35.000 difference (Sample 1 Sample 2) 12.500 standard error of difference 0 hypothesized difference 2.80 t 0026 p-value (one-tailed, upper) Thus the p-value is 0.0052 Conclusion Use a significance level, a 0.05 Here, p-value is 0.0026, which is lesser than the value of α-0.05 That is, p-value(-0.0026)< α(-0.05) Therefore, by the rejection rule, reject the null hypothesis Ho

Interpretation There is sufficient evidence at level of significance α 0.05 That is, there is a statistically significant difference between the population means There is no change between the results of part (b) and part (c) Excel add-in (Mega Stat) procedure: Step 1: In EXCEL, Select Add-Ins> Mega Stat> Hypothesis Tests. Step 2: Choose Compare Two Independent Groups Step 3: Choose Summary Input Step 4: In Group 1, enter B1:B4 (Select ranges with label mean, SD, n). Step 5: In Group 2, enter D1:D4 (Select ranges with label mean, SD, n). Step 6: Enter 25 Under Hypothesized difference. Step 7: Check t-test (unequal variance), enter Confidence level as 95.0 Step 8: Choose not equal in alternative. Step 9: Click OK. Excel add-in (Mega Stat) output:

Hypothesis Test: Independent Groups (t-test, unequal variance) Sample 1 Sample 2 5275 150 400 5240mean 200 std. dev 400 n 739 df 35.000 difference (Sample 1 Sample 2) 12.500 standard error of difference 25 hypothesized difference 0.80 t 4240 p-value (two-tailed) Thus the p-value is 0.4240 Conclusion Use a significance level, α 0.05 Here, p-value is 0.4240, which is greater than the value of α 0.05 That is, p-value(-0.4240)> α(-0.05) Therefore, by the rejection rule, do not reject the null hypothesis Ho

Interpretation There is no sufficient evidence at level of significance α-0.05 That is, the difference between the population means is 25 The result of part (d) is changed from part(b). Assumptions The populations from which the samples have been drawn should be nomal The standard deviation of the populations should be equal when σ is unknown Samples have to be randomly drawn independent of each other

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