vo-Sample $$ \mathbf{Z} $$-Test Problems 31. A company wants to compare the average salaries of male and female employees. A sample of 50 male employees has an average salary of $$ \$ 55,000 $$ with a population standard deviation of $$ \$ 8,000 $$, while a sample of 45 female employees has an average salary of $$ \$ 52,500 $$ with a population standard deviation of $$ \$ 7,500 $$. Test at a $$ 5 \% $$ significance level if there is a salary difference.

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**Step 1. Set up the hypotheses**

We wish to test if there is a salary difference between male and female employees. Let  
$$
\mu_M
$$
be the true average salary of male employees and  
$$
\mu_F
$$
be the true average salary of female employees.

The null and alternative hypotheses are
$$
\begin{aligned}
H_0 &: \mu_M - \mu_F = 0, \
H_a &: \mu_M - \mu_F \neq 0.
\end{aligned}
$$

---

**Step 2. Gather the sample statistics**

For male employees:
- Sample size: $$ n_M = 50 $$
- Sample mean: $$ \bar{x}_M = 55,\!000 $$
- Population standard deviation: $$ \sigma_M = 8,\!000 $$

For female employees:
- Sample size: $$ n_F = 45 $$
- Sample mean: $$ \bar{x}_F = 52,\!500 $$
- Population standard deviation: $$ \sigma_F = 7,\!500 $$

---

**Step 3. Compute the test statistic**

Under the assumption of normality and known population standard deviations, the test statistic is given by
$$
z = \frac{(\bar{x}_M - \bar{x}_F) - 0}{\sqrt{\frac{\sigma_M^2}{n_M} + \frac{\sigma_F^2}{n_F}}}.
$$

Substitute the given values:
$$
z = \frac{55,\!000 - 52,\!500}{\sqrt{\frac{8,\!000^2}{50} + \frac{7,\!500^2}{45}}}.
$$

Compute the difference in sample means:
$$
55,\!000 - 52,\!500 = 2,\!500.
$$

Next, compute the variances:
$$
\frac{8,\!000^2}{50} = \frac{64,\!000,\!000}{50} = 1,\!280,\!000,
$$
$$
\frac{7,\!500^2}{45} = \frac{56,\!250,\!000}{45} \approx 1,\!250,\!000.
$$

The sum is:
$$
1,\!280,\!000 + 1,\!250,\!000 = 2,\!530,\!000.
$$

Now, find the standard error:
$$
\text{SE} = \sqrt{2,\!530,\!000} \approx 1,\!590. \quad (\text{approx})
$$

Thus, the test statistic is:
$$
z \approx \frac{2,\!500}{1,\!590} \approx 1.57.
$$

---

**Step 4. Determine the critical value and make a decision**

At a $$5\%$$ significance level for a two-tailed test, the critical $$z$$ values are approximately $$\pm 1.96$$.

Since
$$
|z| \approx 1.57 < 1.96,
$$
we fail to reject the null hypothesis.

---

**Step 5. Conclusion**

There is insufficient evidence at the $$5\%$$ significance level to conclude that there is a difference in the average salaries of male and female employees.
There is no significant difference in the average salaries of male and female employees at the 5% significance level.

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