1. A company is deciding whether to renew its ad buy with a local TV station. It
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Question
1. A company is deciding whether to renew its ad buy with a local TV station. It will renew the ad if a 2-sided hypothesis test at the 10% level concludes that more than 19% of the local residents remember the ad. They decide to test this by contacting 300 randomly selected local residents. Sixty-seven (67) of the 300 remember the ad.
a. Conduct the requested hypothesis test based on this sample using the normal approximation. (Use Minitab.) Report your conclusions, stated in terms of the data and the sampling scenario.
b. The station advocates that the decision be made based on a 1-sided hypothesis test using the normal approximation at the 10% level using this sample. Why? (Hint: If you conduct the appropriate 1-sided test based on this sample, what will the p-value be? Confirm your answer with Minitab by conducting this 1-sided test.)
Explanation / Answer
Solution:-
a)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P < 0.19
Alternative hypothesis: P > 0.19
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.
Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).
= sqrt[ P * ( 1 - P ) / n ]
= 0.02265
z = (p - P) /
z = 1.47
where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.
Since we have a two-tailed test, the P-value is the probability that the z-score is extreme than 1.47.
Thus, the P-value = 0.1416
Interpret results. Since the P-value (0.146) is greater than the significance level (0.10), we have to accept the null hypothesis.
b)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P < 0.19
Alternative hypothesis: P > 0.19
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.
Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).
= sqrt[ P * ( 1 - P ) / n ]
= 0.02265
z = (p - P) /
z = 1.47
where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.
Since we have a one-tailed test, the P-value is the probability that the z-score is greater than 1.47.
Thus, the P-value = 0.0708
Interpret results. Since the P-value (0.0708) is less than the significance level (0.10), we cannot accept the null hypothesis.
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