Show Intro/Instructions A magazine is considering the launch of an online editio
ID: 3072927 • Letter: S
Question
Show Intro/Instructions A magazine is considering the launch of an online edition. The magazine plans to go ahead only if it is convinced that more than 24.7% of current readers would subscribe. The magazine contacted a simple random sample of 395 current subscribers, and 118 of those surveyed expressed interest. Test the appropriate hypotheses using a significance level of 0.01 to determine if the magazine should launch an online edition. .Ho: Select an answer ? Hai [Select an answer 1 decision rule: reject Ho if probability ·Test Statistic: z = (Note: round the z-score to two decimal places- carry at least four decimal places throughout all of your calculations) probability probability to four decimal places) decision: Select an answer (Note: round the . Conclusion: At the 0.01 level, there Select an answer v significant evidence to conclude the percentage of current subscribers who would subscribe to an online edition of the magazine is Gelectananswer , 24.7%.Explanation / Answer
SOlution:-
a)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P < 0.247
Alternative hypothesis: P > 0.247
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 (S.D) and compute the z-score test statistic (z).
S.D = sqrt[ P * ( 1 - P ) / n ]
S.D = 0.0217
z = (p - P) / S.D
z = 2.38
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 2.38.
Thus, the P-value = 0.009
Interpret results. Since the P-value (0.009) is less than the significance level (0.01), we cannot accept the null hypothesis.
Reject the null hypothesis.
At 0.01 level, there is significant evidence to conclude the percentage of the current subscibers who sould subscribe to an online edition of the magazine is greater than 24.7%.
b) P-value = 0.582
z = - 0.55
The given test is two tailed test.
P-value = P( - 0.55 > z > 0.55)
P-value = P(z < - 0.55) + P(z > 0.55)
P-value = 0.291 + 0.291
P-value = 0.582