Managers at a local phone service wireless total counter have a that 74% of the

ID: 3262659 • Letter: M

Question

Managers at a local phone service wireless total counter have a that 74% of the center's customers will have to wait less than five minutes for service. The data below shows the wait stress of a random sample of 25 customers in Does this sample provide evidence that management's is being achieved? Does this sample provide evidence that management's is being achieved? (Round to four decimal places as needed.) A. Yes, because the probability of observing a sample proportion as low as the in this sample given a population presentation of 0.74 is, which is less than 0.05. B. No, because the probability of observing a sample proportion as low as the in this sample given a population presentation of 0.74 is, which is less than 0.05. C. No, because the probability of observing a sample proportion as low as the in this sample given a population presentation of 0.74 is, which is less than D. Yes, because the probability of observing a sample proportion as low as the one in this sample given a population presentation of 0.74 is, which is less than 0.05.

Explanation / Answer

Solution:-

p = 18/25 = 0.72

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

Null hypothesis: P = 0.74

Alternative hypothesis: P 0.74

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

= 0.08773

z = (p - P) /

z = - 0.2279

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 less than - 0.228 or greater than 0.228.

Thus, the P-value = 0.82

Interpret results. Since the P-value (0.82) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Yes, the probability of observing a sample proportion as low as the one in this sample given a population proportion of 0.74 is 0.82 which is more than 0.05.

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