Question

A coin operated drink machine was designed to discharge a mean of 7 ounces of coffee per cup. In a test of the machine, the discharge amounts in 14 randomly chosen cups of coffee from the machine were recorded. The sample mean and sample standard deviation were 7.02 ounces and 0.24 ounces, respectively. If we assume that the discharge amounts are normally distributed, is there enough evidence, at the 0.05 level of significance, to conclude that the true mean discharge, u, differs from 7 ounces? Perform a two-tailed test. Then fill in the table below. Carry your intermediate computations to at least three decimal places and round your answers as specified in the table o 8 9 P The null hypothesis: x s <a The alternative hypothesis: 8 D- OSO DO The type of test statistics (Choose one) The value of the test statistic (Round to at least three decimal places 0 X ? The p-value: (Round to at least three decimal places 0

Answer 1

The provided sample mean is X = 7.02 and the sample standard deviation is s = 0.24, and the sample size is n = 14. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: u = 7 Ha: j = 7 This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a two-tailed test is te = 2.16. The rejection region for this two-tailed test is R= {t: \t > 2.16} (3) Test Statistics The t-statistic is computed as follows: X - μo t 7.02 7 0.24/V14 0.312 s/n (4) Decision about the null hypothesis = 0.312 <tc = 2.16, it is then concluded that the null Since it is observed that t|| hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.7601, and since p= 0.7601 > 0.05, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean u is different than 7, at the 0.05 significance level.

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