Problem 4)The National Sleep Foundation asked a random sample of U.S. adults que

Question

Problem 4)The National Sleep Foundation asked a random sample of U.S. adults questions about their sleep habits. One of the questions was about snoring. The Foundation was interested in whether there is a difference inthe frequency of snoring between older and younger adults. Of the 184 younger people (younger was less than 30), 48 reported snoring. Of the 811 older people, 316 reported snoring. A significanc test was conducted to determine if there was a differencein the percentage of people who snored in the two age groups. The printout is below.a)Is this an observational study or an experiment?Circle one.b)Can we do inference based on a normal distribution? For the younger group, is the count of “successes” 10?Is the count of “failures” 10?Show how you arrived at your answer.For the older group, is the count of “successes” 10?Is the count of “failures” 10?Show how you arrived at your answer.random samples? AST 311 Homework 8List of Problems, Page 3of 4c)What is the response variable?d)Group 1 is (circle one) younger peopleolder peoplee)What is the sample proportion of younger people who reported snoring?f)What are the hypotheses for the significance test?g)What is the conclusion of the significance test in the context of the problem?h)Interpret the confidence interval in the context of the problem.

Explanation / Answer

Solution:-

PYounger = 48/184

PYounger = 0.2609

POlder = 316/811

POlder = 0.3896

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 = 0.3658

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }

SE = 0.03933

z = (p1 - p2) / SE

z = - 3.27

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 - 3.27 or greater than 3.27

Thus, the P-value = 0.001

Interpret results. Since the P-value (0.001) is less than the significance level (0.05), we cannot accept the null hypothesis.

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