A survey claims that 7 out of 10 customers recommend Company A for pharmacy issu
ID: 3239837 • Letter: A
Question
A survey claims that 7 out of 10 customers recommend Company A for pharmacy issues. To test this claim, a random sample of 100 customers is obtained from the list of rewards customers. Of these 100 customers, 75 indicate that they recommend using Company A for pharmacy needs. We would like to know if the original claim is accurate. State your conclusion using
Alpha = 0.05.
The null hypothesis is supported: there is evidence that 70% of customers recommend Company A for pharmacy issues.
The null hypothesis is not supported: there is evidence that more than 70% of customers recommend Company A for pharmacy issues.
The null hypothesis is not supported: there is evidence that less than 70% of customers recommend Company A for pharmacy issues.
The null hypothesis is not supported: there is evidence that the percentage of customers who recommend Company A for pharmacy issues is not equal to 75%.
A.The null hypothesis is supported: there is evidence that 70% of customers recommend Company A for pharmacy issues.
B.The null hypothesis is not supported: there is evidence that more than 70% of customers recommend Company A for pharmacy issues.
C.The null hypothesis is not supported: there is evidence that less than 70% of customers recommend Company A for pharmacy issues.
D.The null hypothesis is not supported: there is evidence that the percentage of customers who recommend Company A for pharmacy issues is not equal to 75%.
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 = 0.70
Alternative hypothesis: P 0.70
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 ] = sqrt [(0.7 * 0.3) / 100] = 0.04582575694
z = (p - P) / = (0.75 - 0.70)/0.04582575694 = 1.09
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 -1.09 or greater than 1.09.
We use the Normal Distribution Calculator to find P(z < 1.09)
The P-Value is 0.275713.
The result is not significant at p < 0.05.
Interpret results. Since the P-value (0.28) is greater than the significance level (0.05), we cannot reject the null hypothesis.
Conclusion. The null hypothesis is supported: there is evidence that 70% of customers recommend Company A for pharmacy issues.
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