Question

The mean income per year of employees who produce internal and external newsletters and magazines for corporations was reported to be $76300. These editors and designers work on corporate publications but not on marketing materials. As a result of poor economic conditions and oversupply, these corporate communications workers may be experiencing a decrease in salary. A random sample of 40 corporate communications workers revealed that x̄ = $73000. Conduct a hypothesis test to determine whether there is any evidence to suggest that the mean income per year of corporate communications workers has decreased. Assume that the population standard deviation is $6360. Please use the exact value (from R) for all critical values.

What assumptions are required so that you can perform a hypothesis tests and confidence interval for the mean income per year?

Should you use a z distribution or a t distribution in this problem. Explain

Perform the hypothesis test at a 1% significance level.

(0.5 pts.) Calculate the test statistic.

Calculate the p-value. Please write down 5 decimal places.

Calculate and interpret the appropriate 99% bound that is consistent with what you did in part b). You do not need to have any decimal places in the answer; that is, the answer should be an integer.

If the employer also found a cheaper medical insurance that reduced the amount of money that the employees have to pay for health insurance by $1000 per year, in practical terms, are the employees really taking home less money than before? Please explain your answer.

Answer 1

Solution:-

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

Null hypothesis: u > 76300
Alternative hypothesis: u < 76300

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 1005.604
z = (x - u) / SE

z = -3.28

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

The observed sample mean produced a z statistic test statistic of - 3.28.

Thus the P-value in this analysis is 0.00049

Interpret results. Since the P-value (0.00049) is less than the significance level (0.01), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that the mean income per year of corporate communications workers has decreased.

99% confidence interval for the C.I = ( 70410, 75590).

C.I = 73000 + 2.576 × 1005.604

C.I = 73000 + 2590.4367

C.I = ( 70409.56, 75590.44)

C.I = ( 70410, 75590).

Yes, employees really taking home less money than before because both upper and lower limits of 99% confidence interval is 1000 less than $76300.