The established mean for the number of copies a customer makes per visit at a co

Question

The established mean for the number of copies a customer makes per visit at a copy store is 37.   A store employee claims the actual number of copies made per visit is less. A sample of 71 customers was taken and showed an average of 32 copies per customer. The population standard deviation is known to be 25 copies. At alpha = 0.05, test the employee's claim. Select the alternative hypothesis from choices given. What is your calculated (Z, t,or chi-square) value. Convert your answer to a  P-Value and place it in the space provided. True or False as to the rejection of the null hypothesis.

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: < 37, the actual number of copies made per visit is less than 37
Alternative hypothesis: > 37, the actual number of copies made per visit is not less than 37.

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too large.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n) = 25 / sqrt(71) = 2.967
DF = n - 1 = 71 - 1 = 70
t = (x - ) / SE = (32 - 37)/2.967 = -1.685

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

Here is the logic of the analysis: Given the alternative hypothesis ( > 37), we want to know whether the observed sample mean is large enough to cause us to reject the null hypothesis.

The observed sample mean produced a t statistic test statistic of -1.685.

We use the t Distribution Calculator to find P(t < -1.685) = 0.048221.

Thus the P-value in this analysis is 0.048221.

Interpret results. Since the P-value (0.048221) is smaller than the significance level (0.05), we can reject the null hypothesis.

Conclusion. We fail to accept the null hypothesis, thus accepting the alternate hypothesis that is the actual number of copies made per visit is more than 37.

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