Question

the descriptive statistics for the annual salaries from a random sample of microbiologists from two regions are shown below. construct a 95% confidence interval for the difference between the mean annual salaries.
\\( \overline { x } _ { 1 } = \\$ 101,850, n _ { 1 } = 35 \\), and \\( \sigma _ { 1 } = \\$ 8870 ; \overline { x } _ { 2 } = \\$ 85,730, n _ { 2 } = 44 \\), and \\( \sigma _ { 2 } = \\$ 8875 \\)
complete the 95% confidence interval for \\( \mu _ { 1 } - \mu _ { 2 } \\) below.
\\( \\$ \square < \mu _ { 1 } - \mu _ { 2 } < \\$ \square \\)
(round to the nearest dollar as needed )

Explanation:

Step1: Calculate the difference in sample means

The difference in sample means is \(\bar{x}_1-\bar{x}_2 = 101850 - 85730=16120\)

Step2: Find the critical value \(z_{\alpha/2}\)

For a 95% confidence interval, \(\alpha=1 - 0.95=0.05\), and \(\alpha/2=0.025\). The critical value \(z_{\alpha/2}=z_{0.025}\). From the standard normal distribution table, \(z_{0.025} = 1.96\)

Step3: Calculate the standard error \(SE\)

The formula for the standard error when the population variances \(\sigma_1^{2}\) and \(\sigma_2^{2}\) are known is \(SE=\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}\)
Substitute \(\sigma_1 = 8870\), \(n_1 = 35\), \(\sigma_2=8875\), \(n_2 = 44\)

$$
LATEXBLOCK0
$$

Step4: Calculate the margin of error \(E\)

The margin of error \(E = z_{\alpha/2}\times SE\)
Substitute \(z_{\alpha/2}=1.96\) and \(SE\approx2009.26\)
\(E=1.96\times2009.26\approx3938.15\)

Step5: Calculate the confidence interval

The confidence interval for \(\mu_1-\mu_2\) is \((\bar{x}_1 - \bar{x}_2)-E<\mu_1-\mu_2<(\bar{x}_1 - \bar{x}_2)+E\)
Substitute \(\bar{x}_1-\bar{x}_2 = 16120\) and \(E\approx3938.15\)
\(16120- 3938.15<\mu_1-\mu_2<16120 + 3938.15\)
\(12181.85<\mu_1-\mu_2<20058.15\)

Answer:

\(12182<\mu_1-\mu_2<20058\)