Question

Use the following information to complete steps (a) through (d) below. A random sample of ny = 135 individuals results in xy = 40 successes. An independent sample of n2 = 150 individuals results in x2 = 60 successes. Does this represent sufficient evidence to conclude that P, <P2 at the a = 0.10 level of significance? (a) What type of test should be used? A. A hypothesis test regarding the difference between two population proportions from independent samples. B. A hypothesis test regarding the difference of two means using a matched-pairs design. OC. A hypothesis test regarding two population standard deviations. OD. A hypothesis test regarding the difference between two population proportions from dependent samples. (b) Determine the null and alternative hypotheses. Choose the correct answer below. O A. Ho: P1 = P2iH: P1 P2 OB. Ho: P1>P2; H1: P1 P2 OC. Ho: P1 = P2: Hy: P1 P2 OD. Ho: P1 = P2iH: 21*P2 (c) Use technology to calculate the P-value. (Round to three decimal places as needed.) (d) Draw a conclusion based on the hypothesis test. Choose the correct answer below. A. There is not sufficient evidence to reject the null hypothesis because the P-value < c. B. There is not sufficient evidence to reject the null hypothesis because the P-value > a. C. There is sufficient evidence to reject the null hypothesis because the P-value <a. D. There is sufficient evidence to reject the null hypothesis because the P-value >a.

Answer 1

QUESTION 1: (a) A. A hypothesis test regarding the difference between the two population proportions from independent samples. A. HP1=;H:P.<P2

MINITAB procedure: Step 1: Choose Stat > Basic Statistics > 2 Proportions. Step 2: Choose Summarized data. Step 3: In First sample, enter Trials as 135 and Events as 40. Step 4: In Second sample, enter Trials as 150 and Events as 60. Step 5: Select Alternative as Less than. Check Perform hypothesis test. Step 6: Click OK MINITAB output: Test and CI for Two Proportions Sample X N Sample p 1 40 135 0.296296 2 60 150 0.400000 Difference - P(1) - D (2) Estimate for difference: -0.103704 95% upper bound for difference: -0.0114674 Test for difference - 0 (V3 < 0): Z = -1.85 P-Value - 0.032 P-value = 0.032.

(d) C. There is sufficient evidence to reject the null hypothesis because the P-value <a. QUESTION 2: (a) A Ho: 4 = 42 Н. :ut > д. MINITAB Procedure:

Step 1: Choose Stat > Basic Statistics > 2-Sample t. Step 2: Choose Summarized data. Step 3: In first, enter Sample size as 24. Mean as 50.4, Standard deviation as 6.6. Step 4: In second, enter Sample size as 21, Mean as 41.2, Standard deviation as 13.5. Step 5: Select Assume equal variances. Step 6: Choose Options. Step 7: In Confidence level, enter 95. Step 8: In Alternative, select greater than. Step 9: Click OK in all the dialogue boxes. MINITAB output: Two-Sample T-Test and CI Sample N Mean StDev SE Mean 1 24 50.40 6.60 2 21 41.2 13.5 1.3 2.9 Difference - (1) - (2) Estimate for difference: 9.20 95% CI for difference: (2.56, 15.84) I-Test of difference - 0 (v3 »): I-Value - 2.84 P-Value - 0.004 DF - 28 Test Statistic = 2.84

P-value = 0.004. A. Reject H. There is sufficient evidence at the a=0.01 level of significance to conclude that (b) The 95% confidence interval about lly-u, is the range from a lower bound of 2.564 to an upper limit of 15.836.