Exam 4 Name MULTIPLE CHOICE. Choose the one alternative that best completes the

Question

Exam 4 Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question 1) A marketing firm asked a random set of married and single men how much they were ) willing to spend on a vacation. Is there sufficient evidence at is there a difference in the two amounts? 0.05 to conclude that Single Men 70 $825 7900 Married Men 70 Sample size Sample mean 5700 Population variance A) No, because the test value 1.39 is outside the critical region-1961.96. B) No, because the test value 0.28 is inside the critical region-L96·1.96 C) Yes, because the test value 3.95 is outside the critical region-1.96 1 .96. D) Yes, because the test value 1.39 is inside the critical region-1.96z

Explanation / Answer

Solution:-

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

Null hypothesis: 1 - 2 = 0
Alternative hypothesis: 1 - 2 0

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = sqrt[(s12/n1) + (s22/n2)]
SE = 13.94

z = [ (x1 - x2) - d ] / SE

z = 3.95

zcritical = + 1.96

where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a z statistic more extreme than 3.95; that is, less than -3.95 or greater than 3.95.

Thus, the P-value = 0.0001

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we cannot accept the null hypothesis.

Yes, because test value 3.95 lies in the critical region - 1.96 < z < 1.96

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