Question

chapter 10 question 009-13

Chapter 10, Testbank, Question 009-013 A sample of S2 female workers and another sample of 64 male workers from a state produced mean weekly earnings of $742.49 for the females and $764.78 for the males. Suppose that the standard deviations of the weekly earnings are $86.88 for the females and $96.38 for the males. The null hypothesis is that the mean weekly earnings are the same for females and males, while the alternative hypothesis is that the mean weekly earnings for females is less than the mean weekly earnings for males. Derive the corresponding 95% confidence interval, rounded to two decimal places, for the difference between the mean weekly earnings for all female and male workers in this state. The lower limit is $ The upper limit is $ The significance level for the test is 1%. What is the critical value of 2? -2.58 -2.17 -2.33 -1.96 What is the value of the test statistic, rounded to three decimal places? What is the p-value for this test, rounded to four decimal places? License Agreement Privacy Policy 1 2000-2020 John Wiley & Sons, Inc. All Rights Reserved. A Division of John Wiley & Sons, Inc. Version 4.24.20.1

100-105

Answer 1

Let
11
be the true mean weekly earnings for females and
112
be the true mean weekly earnings for males

Using this,

Suppose that the standard deviations of the weekly earnings are $86.88 for the females and $96.38 for the males. The null hypothesis is that the mean weekly earnings are the same for females and males, while the alternative hypothesis is that the mean weekly earnings for females is less than the mean weekly earnings for males.

The hypotheses are

Ho:fli = 2 H: M1 <H2 1

We have the following sample information

ni 52 sample size for females n2 = 64 + sample size for males T1 = 742.49 + sample mean weekly earnings for females T2 = 764.78 + sample mean weekly earnings for males 01 = 86.88 + standard deviation of weekly earnings for females 02 = 96.38 + standard deviation of weekly earnings for males

The standard error of the difference between 2 means is

ci ܪܶܘ 80,882ܐ fii-T? = 1 1 06.382 64 170382 101 122 52

The sample sizes are greater than 30 and hence using the central limit theorem, we can say that the sampling distribution of difference between 2 means is normally distributed.

95% confidence level corresponds to level of significance

a=1–95/100 = 0.05

The right tail critical value is

P(Z > 2a/2) = a/2 = 0.05/2 = 0.025 P(Z < 2/2) = 1-0.025 = 0.975

Using the standard normal tables, we get for z=1.96, P(Z<1.96)=0.975

That is, $z_{\alpha/2}=1.96$

The 95% confidence interval is

(21 – 12) + 2a/2011-12 (742.49 – 764.78) = 1.96 x 17.0382 • (-55.68, 11.10)

ans:

• The lower limit is $-55.68
• The upper limit is $11.10

This is a left tailed test (The alternative hypothesis has "<")

The right tail critical value for significance level 1% ($\alpha=0.01$) is

$P(Z>z_c)=\alpha=0.01\\\implies P(Z<z_c)=1-0.01=0.99$

Using the standard normal tables, we get for z=2.33, P(Z<2.33)=0.99

The right tail critical value is 2.33. The left tail critical value is -2.33

ans:

-2.33

The hypothesized value of difference between 2 means is $(\mu_1-\mu_2)_{H_0}=0$

The test statistic is

$z=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1-\mu_2)_{H_0}}{\sigma_{\bar{x}_1-\bar{x}_2}}=\frac{(742.49-764.78)-0}{17.0382}=-1.308$

ans: The value f the test statistic is -1.308

This is a left tailed test. The p-value is the area under the left tail

$\begin{align*} \text{p-value}&=P(Z<-1.308)\\ &=1-P(Z<1.308)\\ &=1-0.9049\quad\text{using the standard normal tables for z=1.31}\\ &=0.0951 \end{align*}$

ans: The p-value for the test is 0.0951

Note: If we use technology using the unrounded -1.308, we get the p-value=0.0954

We will reject the null hypothesis, if the p-value is less than the significance level 0.01.

Here, the p-vale is 0.0951 and it is greater than 0.01. Hence we do not reject the null hypothesis.

We conclude that

Do not reject H₀. There is no sufficient evidence to support the claim that mean weekly earnings for females is less than the mean weekly earnings for males

100-105

Let
Pi
be the true proportion of females satisfied with their jobs and  
P2
be the true proportion of males satisfied with their jobs

The hypotheses are

$\begin{align*} &H_0:p_1=p_2\\ &H_a:p_1<p_2\\ \end{align*}$

We have the following sample information

$\begin{align*} n_1&= 867\leftarrow\text{sample size of female job-holders}\\ x_1&= 536\leftarrow\text{sample number of female job-holders satisfied with their jobs }\\ \hat{p}_1&= 0.6182\leftarrow\text{sample proportion of female job-holders satisfied with their jobs }\\ n_2&= 830\leftarrow\text{sample size of male job-holders}\\ x_2&= 551\leftarrow\text{sample number of male job-holders satisfied with their jobs }\\ \hat{p}_2&= 0.6639\leftarrow\text{sample proportion of male job-holders satisfied with their jobs }\\ \end{align*}$

The pooled sample proportion of job-holders satisfied with their jobs is

$\hat{p}=\frac{x_1+x_2}{n_1+n_2}=\frac{536+551}{867+830}=0.6405$

The standard error of difference between 2 proportions is

$\begin{align*} \hat{\sigma}_{\hat{p}_1-\hat{p}_2}&=\sqrt{\frac{\hat{p}(1-\hat{p})}{n_1}+\frac{\hat{p}(1-\hat{p})}{n_2}}\\ &=\sqrt{\frac{0.6405\times (1-0.6405)}{867}+\frac{0.6405\times (1-0.6405)}{830}}\\ &=0.0233 \end{align*}$

97% confidence level is $\alpha=1-97/100=0.03$ level of significance

The right tail critical value is

$P(Z>z_{\alpha/2})=\alpha/2=0.03/2=0.015\\\implies P(Z<z_{\alpha/2})=1-0.015=0.985$

Using the standard normal tables, for z=2.17, we get P(Z<2.17)=0.985

Hence, $z_{\alpha/2}=2.17$

The 97% confidence interval is

$\begin{align*} &(\hat{p}_1-\hat{p}_2)-z_{\alpha/2}\hat{\sigma}_{\hat{p}_1-\hat{p}_2}\\ \implies &(0.6182-0.6639)\pm 2.17\times 0.0233\\ \implies &(-0.0962, 0.0049) \end{align*}$

ans:

• The lower limit is -0.0962
• The upper limit is  0.0049

This is a left tailed test (The alternative hypothesis has "<")

The right tail critical value for significance level 2.5% ($\alpha=0.025$) is

$P(Z>z_c)=\alpha=0.025\\\implies P(Z<z_c)=1-0.025=0.975$

Using the standard normal tables, we get for z=1.96, P(Z<1.96)=0.975

The right tail critical value is 1.96. The left tail critical value is -1.96

ans:

$1594700849451_image.png$

ans:

The value of the pooled sample proportion is 0.6405

The hypothesized value of the difference between 2 proportion is $(p_1-p_2)_{H_0}=0$

The test statistic is

$z=\frac{(\hat{p}_1-\hat{p}_2)-(p_1-p_2)_{H_0}}{\hat{\sigma}_{\hat{p}_1-\hat{p}_2}}=\frac{(0.6182-0.6639)-0}{0.0233}=-1.958$

ans:

The value of the test statistic,z, is -1.958

We will reject the null hypothesis, if the test statistic is less than the critical value.

Here, the test statistic is -1.958 and it is not less than the critical value -1.96. Hence we do not reject the null hypothesis.

We conclude that

Do not reject H₀. There is no sufficient evidence to support the claim that the proportion of female job holders who are satisfied with their jobs is less than the proportion of male job holders who are satisfied with their jobs