Homework: Section 10.3 Homework Score: 0 of 1 pt Instructor-created question Sav
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Homework: Section 10.3 Homework Score: 0 of 1 pt Instructor-created question Save 3 of 3 (1 complete) Hw Score: 4.76%, 0.14 of 3 pts Ouestion Help A survey of 1000 adults from a certain region asked Do you enjoy shopping for clothing for yourself?" The results indicated that 56% of the females eng ed shopping for clothing for themselves as compared to 48% of the males. The sample sizes of males and females were not provided Suppose that of 400 females. 224 said that they cnjoyod shopping for clothing for themselves while of 600 males, 288 said that they enjoyed shopping for clothing for themselves. Complete parts a) through (d) below, a. Is there evidence of a difference between males and females in the proportion who enjoy shopping for clothing for themselves at the o.01 level of significance? State the nul and alternative hypotheses, where 1 is the population proportion o females that eno shopping for themselves and 2 is the population proportion of males that enjoy shopping for t Determine the value of the test statistic ZgTAT =(Round to two decimal places as needed.) b. Find the p-value in (a) and interpret its meaning. The p-value in part (a) Round to three decimal places as needed.) What does the p-value mean? 0 A. The p-value is the probability of falsely rejecting the null hypothesis. O B. The p-value is the probability of rejecting the null hypothesis if the test were repeated for different samples O C. The p value is the probability of obtaining a resull at least as extreme as the one oblained wilh this sample given that the null hypolhesis is true 0 D. The p-value is the probability that the nul hypothesis is true. State the conclusion lhe null hypothesis. There is evidence to support lhe claim lhat there is a difference between males and females in the proportion who enjoy shopping for clothing for themselves. c. Construct and interpret a 99% confidence interval estimate or the difference between the proportion of males and em ales who enjoy shopping or clothing or The confidence intervail is to Round to four decirnal places as needed.) What does the confidence interval mean? A researcher can be confident that the difference in the population proportions of males and females that enjoy shopping for themselves is | the interval d. What are the answers to (a) through (c) if 312 males enjoyed shopping for clothing for themselves? ls there evidence of a difference between males and females in the proportion who enjoy shopping for clothing for themselves at the 0.01 level of significance? State the null and altemative hypotheses, where 1 is the population proportion of females that en y shopping for themselves and 's the population proportion of iales that enjoy shopping for themselves B. Detenmine the value of the test statistic. ZgTAT = L (Round to two decimal places as needed.) Find the p-value and interprel its meaing Click to select your answeris) and then click Check Answer Clear All Check AnswerExplanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P1 = P2
Alternative hypothesis: P1 P2
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the proportion from population 1 is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-proportion z-test.
Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).
p = (p1 * n1 + p2 * n2) / (n1 + n2)
p = 0.512
SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = 0.0323
z = (p1 - p2) / SE
z = 2.48
where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.
Since we have a two-tailed test, the P-value is the probability that the z-score is less than -2.48 or greater than 2.48.
Thus, the P-value = 0.0131
Interpret results. Since the P-value (0.0131) is greater than the significance level (0.01), we have to accept the null hypothesis.
From the above test we have sufficient evidence in the favor of the claim that there is no difference betwene the proportions.
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