Chapter 11, Problem 12E

To determine

State whether it is reasonable to conclude that the mean weekly salary of nurses is higher.

Obtain the p-value.

Answer to Problem 12E

No, there is no enough evidence to conclude that the mean weekly salary of nurses is higher.

The p-value is 0.244.

Explanation of Solution

It is expected that the mean weekly salary of nurses is higher.

Therefore, the test hypotheses are given below:

Denote ${\mu }_1$ as the mean weekly salary of nurses and ${\mu }_2$ as the mean weekly salary of school teachers.

Null hypothesis: $H_0:{\mu }_1\le {\mu }_2$.

That is, the mean weekly salary of nurses are at most the mean weekly salary of school teachers.

Alternative hypothesis: $H_a:{\mu }_1>{\mu }_2$

That is, the mean weekly salary of nurses is greater than the mean weekly salary of school teachers.

In this context, the level of significance is 0.05.

Necessary assumptions required for using the formula:

• The sampled populations are approximately normally distributed.
• The two samples are independent.
• The standard deviations for the two populations are equal.

In this context, the two populations are independent and distributed to normal. The population standard deviations are unknown.

Test statistic for two-sample test of means-unknown$\sigma$:

$t=\frac{{\overline{X}}_1-{\overline{X}}_2}{\sqrt{s_P^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}$

Where, $n_1\text{and }{n}_2$ are the sample sizes of two population, $s_1^2\text{and}{s}_2^2$ are the sample variances, and $s_p^2$ is the pooled estimate of ${\sigma }^2$.

Here,

 $s_p^2=\frac{\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2}{n_1+n_2-2}$.

Excel procedure to find the mean and standard deviation for samples school teacher and nurses:

• Enter the data values in the column H and column I.
• Obtain the sample mean 1 (${\overline{X}}_1$) in cell A1, enter the formula “=AVERAGE(H1:H8)”.
• Press “Enter”.
• Obtain the sample variance 1$\left(s_1^2\right)$  in cell B1, enter the formula “=VAR.S(H1:H8)”.
• Press “Enter”.
• Obtain the sample mean 2 (${\overline{X}}_1$) in cell A2, enter the formula “=AVERAGE(I1:I12)”.
• Press “Enter”.
• Obtain the sample variance 2 $\left(s_2^2\right)$ in cell B2, enter the formula “=VAR.S(I1:I12)”.
• Press “Enter”.

Output obtained using EXCEL is given below:

The pooled variance is obtained as given below:

Substitute $s_1^2$ as 1,183.643, $s_2^2$ as 521.66, $n_1$ as 8, and $n_2$ as 12.

$\begin{array}{c}{s}_p^2=\frac{\left(8-1\right)\left(1,183.643\right)+\left(12-1\right)\left(521.66\right)}{8+12-2}\\ =\frac{14,023.761}{18}\\ =779.0978\end{array}$

The test statistic is given below:

Substitute ${\overline{X}}_1$ as 835.75, ${\overline{X}}_2$ as 826.75, $s_p^2$ as 779.0978, $n_1$ as 8, and $n_2$ as 12.

$\begin{array}{c}t=\frac{{\overline{X}}_1-{\overline{X}}_2}{\sqrt{s_P^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}\\ =\frac{835.75-826.75}{\sqrt{779.0978\left(\frac{1}{8}+\frac{1}{12}\right)}}\\ =\frac{9}{12.7402}\\ =0.7064\end{array}$

Rejection region:

In context, the level of the test, $\alpha$, is 0.01.

Here, the alternative is the right-tailed test. Hence, the rejection region will be $t>t_{\alpha }$.

The critical value has to be obtained for $t_{0.01,\left(n_1+n_2-2\right)}=t_{0.01,18}$.

Critical value:

From the “Appendix B, Table B.5 Student’s t Distribution”, the critical value for 18 df for the level of significance 0.01 is 2.552.

Thus, the rejection region under the level of significance of 0.01 is $t>2.552$.

Decision rule:

• If $t>2.552$, reject the null hypothesis.
• Otherwise fail to reject the null hypothesis.

Conclusion:

Here, the test statistic, t, does not lie in the rejection region.

Therefore, by the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence to conclude that the mean weekly salary of nurses is higher.

p-value:

Step-by-step procedure to obtain the p-value using EXCEL:

• In a cell A1, enter the formula “=T.DIST.RT(0.7064,18)”.
• Press “Enter”.

Output obtained using EXCEL is given below:

Thus, the p-value is 0.244.