The worldwide market share for a web browser was 20.1 % in a recent month. Suppo
ID: 3221814 • Letter: T
Question
The worldwide market share for a web browser was 20.1 % in a recent month. Suppose that a sample of 200 random students at a certain university finds that 50 use the browser. Complete parts (a) through (d) below. a. At the 0.01 level of significance, is there evidence that the market share for the web browser at the university is greater than the worldwide market share of 20.1 %? Determine the null and alternative hypotheses. A. H0: pi not equals0.201; H1: pi equals0.201 B. H0: pi greater than or equals0.201; H1: pi less than0.201 C. H0: pi less than or equals0.201; H1: pi greater than0.201 D. H0: pi equals0.201; H1: pi not equals0.201 Calculate the test statistic. Upper Z Subscript STAT equalsnothing (Type an integer or a decimal. Round to two decimal places as needed.) What is the p-value? The p-value is nothing . (Type an integer or a decimal. Round to three decimal places as needed.) State the conclusion of the test. Reject Do not reject the null hypothesis. There is sufficient insufficient evidence to conclude that the market share at the university is not equal to less than equal to at least greater than at most the worldwide market share of 20.1 %. b. Suppose that a sample of nequals 400 students at the same university (instead of nequals 200) determines that 25 % of the sample use the web browser. At the 0.01 level of significance, is there evidence that the market share for the web browser at the university is greater than the worldwide market share of 20.1 %? Calculate the test statistic for the second sample. Upper Z Subscript STAT equalsnothing (Type an integer or a decimal. Round to two decimal places as needed.) What is the p-value for the second sample? The p-value is nothing . (Type an integer or a decimal. Round to three decimal places as needed.) State the conclusion of the test using this second sample at the 0.01 level of significance. Do not reject Reject the null hypothesis. There is sufficient insufficient evidence to conclude that the market share at the university is not equal to at least at most less than greater than equal to the worldwide market share of 20.1 %. c. Compare the results of (a) and (b) and discuss the effect that sample size has on the outcome, and, in general, in hypothesis testing. Choose the correct answer below. A. Increasing the sample size had a major effect on being able to reject the null hypothesis. B. Increasing the sample size did not affect being able to reject the null hypothesis. C. Increasing the sample size did not affect not being able to reject the null hypothesis. D. Increasing the sample size had a major effect on not being able to reject the null hypothesis. d. What do you think are your chances of rejecting any null hypothesis concerning a population proportion if a sample size of nequals 20 is used? The likelihood of rejecting a null hypothesis with nequals 20 is relatively high low because sample proportion decreases test statistic increases sample proportion increases test statistic decreases as n decreases. Click to select your answer(s).
Explanation / Answer
Solution:-
a)
p = 50/200
p = 0.25
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P < 0.201
Alternative hypothesis: P > 0.201
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 ]
= 0.02834
z = (p - P) /
z = 1.73
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 more than 1.73. We use the Normal Distribution Calculator to find P(z > 1.73) = 0.0418.
Interpret results. Since the P-value (0.0418) is greater than the significance level (0.01), we have to accept the null hypothesis.
From the above test we do not have sufficient evidence that the market share for the web browser at the university is greater than the worldwide market share of 20.1 %.
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