A recent of a television show had a 10 share, meaning that among 5000 monitored
ID: 3290145 • Letter: A
Question
A recent of a television show had a 10 share, meaning that among 5000 monitored households with TV sets in use, 10% of them were tuned to this program. use a 0.01 significance level to test the claim of an advertiser that among the households with TV sets in use, less than 20% were into the program. Identify the null hypothesis hypothesis, test statistic, P-value conclusion about the null hypothesis, and final that addresses the original claim. Use the P-value method. Use the normal distribution as an approximation of the binomial distribution. Identify the null and hypotheses. Choose the correct answer below. A. H_0: p = 0.20 H_: p notequalto 0.20 H_0: p = 0.20 H_: p 0.80 D. H_0: p = 0.80 H_: p notequalto 0.80 E. H_0L p = 0.80 H_: p 0.20 The test is z = (Round to two decimal places as needed.) The P-value is (Round to four decimal places as needed.) Identify the about the null hypothesis and the final conciliation that addresses the original claim sufficient to support the claim than 20% of the TV sets in use were tuned to the program.Explanation / Answer
Solution:-
The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P = 0.20
Alternative hypothesis: P < 0.20
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 0.01. 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 ] = sqrt [(0.2 * 0.8) / 5000] = 0.00565685
z = (p - P) / = (0.10 - 0.20)/0.00565685 = -17.67768 or -17.68
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 less than -1.75. We use the Normal Distribution Calculator to find P(z < -17.68)
The P-Value is < 0.00001.
The result is significant at p < 0.01.
Interpret results. Since the P-value is less than the significance level (0.01), we cannot accept the null hypothesis.
Conclusion. Reject the null hypothesis. We have sufficient evidence to prove the claim.