A union of restaurant and foodservice workers would like to estimate this year\'

Question

A union of restaurant and foodservice workers would like to estimate this year's mean hourly wage mu for foodservice workers in the U.S. Last year's mean hourly wage was $8.16, and there is reason to believe that this year's value is greater than last year's. The union decides to do a statistical test to see if it can be concluded that the mean has increased. The union chooses a random sample of 75 wages from this year. Suppose that the population of hourly wages of foodservice workers in the U.S. has a standard deviation of $1.18 and that the union performs its hypothesis test using the 0.01 level of significance. Based on this information, answer the questions below. Carry your intermediate computations to at least four decimal places, and round your responses as indicated. What are the null and alternative hypotheses that the union should use for the test? H_0: mu is H_1: mu is Assuming that the actual value of mu is $8.52, what is the probability that the union accepts the null hypothesis? Round your response to at least two decimal places. What is the probability that the union rejects the null hypothesis when, in fact, it is true? Round your response to at least two decimal places. Suppose that the union decides to perform another statistical test using the same population, the same null and alternative hypotheses, and the same sample size, but for this second test the union uses a significance level of 0.1 instead of a significance level of 0.01. Assuming that the actual value of mu is $8.52, how does the power of this second test compare to the power of the original test? The power of the second test is greater than the power of the original test The power of the second test is less than the power of the original test The powers of the two tests are equal

Explanation / Answer

Solution:-

a)

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

Null hypothesis: < 8.16

Alternative hypothesis: > 8.16

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

Formulate an analysis plan. For this analysis, the significance level is 0.01. 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)

S.E = 0.1363

DF = n - 1 = 75 - 1

D.F = 74

t = (x - ) / SE

t = 2.64

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.

The observed sample mean produced a t statistic test statistic of 2.64. We use the t Distribution Calculator to find P(t > 2.64) = 0.0051

b) The probability that uniion accept the null hypothesis is 0.0051.

Interpret results. Since the P-value (0.0051) is less than the significance level (0.01), we have to reject the null hypothesis.

c) The probability that union rejects the null hypothesis when infact it is true is 0.01.

d) The power of the two test are equal.

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