Question

the data in the table show the average retirement ages for a random sample of workers in country a and a random sample of workers in country b. the population standard deviations are given. complete parts a and b.

country a country b
sample mean 64.2 years 66.2 years
sample size 30 30
population standard deviation 4.0 years 5.0 years

$h_0: mu_1-mu_2 = 0$ $h_1: mu_1-mu_2
eq 0$

calculate the appropriate test statistic.

the test statistic is $-3.01$

(round to two decimal places as needed.)

b. determine the p - value and interpret the results.

the p - value is

(round to three decimal places as needed.)

since the p - value is vs. $h_0$. there is evidence to conclude that the average retirement age in country b is higher than it is in country a.

Explanation:

Part a: Calculating the Test Statistic

We are dealing with a two - sample z - test for the difference in population means (since the population standard deviations are known). The formula for the test statistic \(z\) is:

$$z=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}$$

Here, \(\bar{x}_1 = 64.2\) (sample mean of Country A), \(\bar{x}_2=66.2\) (sample mean of Country B), \(\mu_1-\mu_2 = 0\) (under the null hypothesis \(H_0\)), \(\sigma_1 = 4.0\) (population standard deviation of Country A), \(\sigma_2 = 5.0\) (population standard deviation of Country B), \(n_1=n_2 = 30\) (sample sizes).

Step 1: Calculate the numerator

The numerator is \((\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)=(64.2 - 66.2)-0=- 2.0\)

Step 2: Calculate the denominator

First, calculate \(\frac{\sigma_1^2}{n_1}=\frac{4.0^2}{30}=\frac{16}{30}\approx0.5333\) and \(\frac{\sigma_2^2}{n_2}=\frac{5.0^2}{30}=\frac{25}{30}\approx0.8333\)

Then, \(\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}=\sqrt{0.5333 + 0.8333}=\sqrt{1.3666}\approx1.169\)

Step 3: Calculate the test statistic \(z\)

$$z=\frac{-2.0}{1.169}\approx - 1.71$$

(We can also think of it as \(z=\frac{64.2 - 66.2}{\sqrt{\frac{4^2}{30}+\frac{5^2}{30}}}=\frac{-2}{\sqrt{\frac{16 + 25}{30}}}=\frac{-2}{\sqrt{\frac{41}{30}}}=\frac{-2}{\sqrt{1.3667}}\approx - 1.71\))

Part b: Calculating the p - value

Since the alternative hypothesis is \(H_1:\mu_1-\mu_2
eq0\), this is a two - tailed test. We need to find \(P(Z < - 1.71)+P(Z>1.71)\). But due to the symmetry of the standard normal distribution, \(P(Z < - 1.71)+P(Z > 1.71)=2P(Z < - 1.71)\)

Looking up the value of \(P(Z < - 1.71)\) in the standard normal table, we find that \(P(Z < - 1.71)=0.0436\)

So the p - value is \(2\times0.0436 = 0.0872\approx0.087\) (rounded to three decimal places)

Interpreting the p - value

We compare the p - value with a common significance level \(\alpha = 0.05\) (or other levels like 0.10). If the p - value \(\leq\alpha\), we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.

Since the p - value (\(0.087\)) is greater than \(\alpha = 0.05\), we fail to reject \(H_0\). There is not sufficient evidence to conclude that the average retirement age in Country B is higher than it is in Country A.

Final Answers

Part a

The test statistic is \(\boldsymbol{-1.71}\) (rounded to two decimal places)

Part b

The p - value is \(\boldsymbol{0.087}\) (rounded to three decimal places). Since the p - value is \(\boldsymbol{>}\) \(\alpha\) (assuming \(\alpha = 0.05\)), we \(\boldsymbol{fail\ to\ reject}\) \(H_0\). There is \(\boldsymbol{not\ sufficient}\) evidence to conclude that the average retirement age in Country B is higher than it is in Country A.

Answer:

Part a: Calculating the Test Statistic

We are dealing with a two - sample z - test for the difference in population means (since the population standard deviations are known). The formula for the test statistic \(z\) is:

$$z=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}$$

Here, \(\bar{x}_1 = 64.2\) (sample mean of Country A), \(\bar{x}_2=66.2\) (sample mean of Country B), \(\mu_1-\mu_2 = 0\) (under the null hypothesis \(H_0\)), \(\sigma_1 = 4.0\) (population standard deviation of Country A), \(\sigma_2 = 5.0\) (population standard deviation of Country B), \(n_1=n_2 = 30\) (sample sizes).

Step 1: Calculate the numerator

The numerator is \((\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)=(64.2 - 66.2)-0=- 2.0\)

Step 2: Calculate the denominator

First, calculate \(\frac{\sigma_1^2}{n_1}=\frac{4.0^2}{30}=\frac{16}{30}\approx0.5333\) and \(\frac{\sigma_2^2}{n_2}=\frac{5.0^2}{30}=\frac{25}{30}\approx0.8333\)

Then, \(\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}=\sqrt{0.5333 + 0.8333}=\sqrt{1.3666}\approx1.169\)

Step 3: Calculate the test statistic \(z\)

$$z=\frac{-2.0}{1.169}\approx - 1.71$$

(We can also think of it as \(z=\frac{64.2 - 66.2}{\sqrt{\frac{4^2}{30}+\frac{5^2}{30}}}=\frac{-2}{\sqrt{\frac{16 + 25}{30}}}=\frac{-2}{\sqrt{\frac{41}{30}}}=\frac{-2}{\sqrt{1.3667}}\approx - 1.71\))

Part b: Calculating the p - value

Since the alternative hypothesis is \(H_1:\mu_1-\mu_2
eq0\), this is a two - tailed test. We need to find \(P(Z < - 1.71)+P(Z>1.71)\). But due to the symmetry of the standard normal distribution, \(P(Z < - 1.71)+P(Z > 1.71)=2P(Z < - 1.71)\)

Looking up the value of \(P(Z < - 1.71)\) in the standard normal table, we find that \(P(Z < - 1.71)=0.0436\)

So the p - value is \(2\times0.0436 = 0.0872\approx0.087\) (rounded to three decimal places)

Interpreting the p - value

We compare the p - value with a common significance level \(\alpha = 0.05\) (or other levels like 0.10). If the p - value \(\leq\alpha\), we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.

Since the p - value (\(0.087\)) is greater than \(\alpha = 0.05\), we fail to reject \(H_0\). There is not sufficient evidence to conclude that the average retirement age in Country B is higher than it is in Country A.

Final Answers

Part a

The test statistic is \(\boldsymbol{-1.71}\) (rounded to two decimal places)

Part b

The p - value is \(\boldsymbol{0.087}\) (rounded to three decimal places). Since the p - value is \(\boldsymbol{>}\) \(\alpha\) (assuming \(\alpha = 0.05\)), we \(\boldsymbol{fail\ to\ reject}\) \(H_0\). There is \(\boldsymbol{not\ sufficient}\) evidence to conclude that the average retirement age in Country B is higher than it is in Country A.