Help with homework! You hypothesize that, when both a man and a woman are in the

Question

Help with homework!

You hypothesize that, when both a man and a woman are in the front seat of a car, the man is more likely to be driving than the woman. The null hypothesis is that the man and the woman are driving equally often or the woman is driving more often. You decide to test the hypothesis by observing 20 cars with both a man and a woman in the front, and counting how many of those have a man driving. Say that men are really driving 70% of the time. Using a one tailed test at a= .05, how many men have to be driving in order for you to be able to reject the null hypothesis? (In other words, at least what number of cars must be driven by men for you to reject the null?) Selected Answer: [None Given] Response Incorrect, see the section on Example Calculations in chapter 13, Power. Feedback: You need to use the Binomial Calculator or Excel for this problem. Remember, at this stage you are assuming the null hypothesis is true. Remember this is a one-tailed test with an alpha of .05.

Explanation / Answer

Solution:-

PMen = 0.70

PWomen = 0.30

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: Pmen = PWomen

Alternative hypothesis: Pmen > PWomen

Note that these hypotheses constitute a one-tailed test.

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 = 0.50

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

SE = 0.158

z = (p1 - p2) / SE

z = 2.53

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 one-tailed test, the P-value is the probability that the z-score is more than 2.53. We use the Normal Distribution Calculator to find P(z > 2.53) = 0.0057

Interpret results. Since the P-value (0.0057) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that man is mpore likely to be driving than women.

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