In a random sample of 63 women at a company, the mean salary is $48,902 with a standard deviation of $5270. In a random sample of 50 men at the company, the mean salary is $53,454 with a standard deviation of $4677. Construct a 95% confidence interval for the difference between the mean salaries of all women and men at the company. show work

($2134.05, $6969.95)

($2715.11, $6388.89)

($3614.81, $5489.19)

($4083.40, $5020.60)

To construct a 95% confidence interval for the difference between the mean salaries of all women and men at the company, we use the following formula:

\( \text{Confidence Interval} = (\bar{x}_1 - \bar{x}_2) \pm \text{critical value} \times \sqrt{\frac{{s_1^2}}{{n_1}} + \frac{{s_2^2}}{{n_2}}} \)

Where:
- \(\bar{x}_1\) and \(\bar{x}_2\) are the means of the women's and men's salaries, respectively.
- \(s_1\) and \(s_2\) are the standard deviations of the women's and men's salaries, respectively.
- \(n_1\) and \(n_2\) are the sample sizes of the women and men, respectively.
- The critical value corresponds to the chosen confidence level.

Given that:
- \(\bar{x}_1 = 48,902\)
- \(s_1 = 5,270\)
- \(n_1 = 63\)
- \(\bar{x}_2 = 53,454\)
- \(s_2 = 4,677\)
- \(n_2 = 50\)
- We want a 95% confidence interval.

Now, we need to find the critical value corresponding to a 95% confidence level. For a two-sample t-test, the critical value is determined by the degrees of freedom, calculated as:

\( \text{Degrees of Freedom} = \frac{{\left(\frac{{s_1^2}}{{n_1}} + \frac{{s_2^2}}{{n_2}}\right)^2}}{{\frac{{\left(\frac{{s_1^2}}{{n_1}}\right)^2}}{{n_1-1}} + \frac{{\left(\frac{{s_2^2}}{{n_2}}\right)^2}}{{n_2-1}}}} \)

Plugging in the values, we get:

\( \text{Degrees of Freedom} = \frac{{\left(\frac{{5270^2}}{{63}} + \frac{{4677^2}}{{50}}\right)^2}}{{\frac{{\left(\frac{{5270^2}}{{63}}\right)^2}}{{63-1}} + \frac{{\left(\frac{{4677^2}}{{50}}\right)^2}}{{50-1}}}} \)

Now, we can look up the critical value for the calculated degrees of freedom in the t-distribution table. For a 95% confidence level with the calculated degrees of freedom, the critical value is approximately 1.985.

Plugging in all the values into the confidence interval formula, we get:

\( \text{Confidence Interval} = (48,902 - 53,454) \pm 1.985 \times \sqrt{\frac{{5270^2}}{{63}} + \frac{{4677^2}}{{50}}} \)

Calculating this expression:

\( \text{Confidence Interval} = -4,552 \pm 1.985 \times \sqrt{\frac{{5270^2}}{{63}} + \frac{{4677^2}}{{50}}} \)

Finally, we can calculate the confidence interval:

\( \text{Confidence Interval} = (-4,552) \pm 1.985 \times 655.2592 \)

\( \text{Confidence Interval} = (-4,552) \pm 1298.7147 \)

\( \text{Confidence Interval} = (-4,552 - 1298.7147, -4,552 + 1298.7147) \)

\( \text{Confidence Interval} = (-5,851.7147, -3,252.2853) \)

Therefore, the correct answer is option: ($ -5,851.71, -3,252.29).