Question

Salaries of postgraduates. The Economics of Education Review (Vol. 21, 2002) published a paper on the relationship between education level and earnings. The data for the research was obtained from the National Adult Literacy Survey of more than 25,000 respondents. The survey revealed that males with a postgraduate degree have a mean salary of \(61,340 (with standard error \(Sx\) = \)2,185), while females with a postgraduate degree have a mean of \(32,227 (with standard error \(Sx\) = \)932).

1. The article reports that a 95% confidence interval for \[{\bf{\mu }}M\] , the population mean salary of all males with post-graduate degrees, is (\(57,050, \)65,631). Based on this interval, is there evidence to say that \[{\bf{\mu }}M\] differs from \(60,000? Explain.
2. Use the summary information to test the hypothesis that the true mean salary of males with postgraduate degrees differs from \)60,000. Use \(\alpha \) =.05.
3. Explain why the inferences in parts a and b agree.
4. The article reports that a 95% confidence interval for \(\mu F\) , the population mean salary of all females with post-graduate degrees, is (\(30,396, \)34,058). Based on this interval, is there evidence to say that \(\mu F\)differs from \(33,000? Explain.
5. Use the summary information to test the hypothesis that the true mean salary of females with postgraduate degrees differs from \)33,000. Use \(\alpha \) =.05.
6. Explain why the inferences in parts d and e agree.

Short answer

a. Substantiation for the provided information

b. \(t = 0.613\)

c.

d. Insufficient evidence to conclude

e. \(z = - 0.829\)

f.

Step-by-step solution

Given information

Referring to exercise 7.4 and the texts 404,405, the following steps can be reduced.

Step 2:

a.

The 95% confidence interval is between 57050 and 65631.

This range comprises 60000, so there is no indication that the \(\mu M\) differs from 60000 based on the confidence interval.

Step 3:

b.

The test statistics formula is as follows:

\(\begin{array}{l}t = \frac{{\bar x - \mu M}}{{SE}}\\t = \frac{{61340 - 60000}}{{2185}}\\t = 0.613\end{array}\)

The associated p-value is 0.539876. At P = 0.05, the result is not significant.

Step 4:

d.

The \(95\% \) Confidence Interval range is 30396 to 304058, and the interval involves 33000, so it is not sure that the \(\mu M\) differs from 33000, so there is insufficient evidence.

Step 5:

e.

The sample is so massive that it is favorable to use the z test rather than the t-test when the population standard deviation is unknown.

The following information is provided:

\(\begin{array}{l}\bar x = 32,227\\Sx = 932\\\end{array}\)

Test hypothesis is as follows.

\(\)\(\begin{array}{l}H0:\mu = \$ \,33,000\\H\alpha :\mu \ne \,\,\$ \,33,000\end{array}\)

The test statistics formula is as follows:

\(\begin{array}{l}z = \frac{{\bar x - \mu M}}{{Sx}}\\z = \frac{{32227 - 33000}}{{932}}\\z = - 0.829\end{array}\)

So the test is two-tailed; the critical values of the z test for the alpha of 0.05 are -1.96 and 1.96, respectively. The null hypothesis is not rejected if the z value is between -1.96 and 1.96. Based on the evidence presented, it is impossible to conclude that the actual mean salary of females with postgraduate degrees differs from $ 33000.