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.
\\( \bar { x } _ { 1 } = \\$ 103,070, n _ { 1 } = 45 \\), and \\( \sigma _ { 1 } = \\$ 9290 ; \bar { x } _ { 2 } = \\$ 85,340, n _ { 2 } = 35 \\), and \\( \sigma _ { 2 } = \\$ 9290 \\)
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 point estimate

The point estimate for \(\mu_1-\mu_2\) is \(\bar{x}_1 - \bar{x}_2\).

$$ \bar{x}_1-\bar{x}_2=103070 - 85340=17730 $$

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

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

Step3: Calculate the margin of error \(E\)

The formula for the margin of error \(E\) when \(\sigma_1\) and \(\sigma_2\) are known is \(E = z_{\alpha/2}\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}\)
Substitute \(\sigma_1 = 9290\), \(n_1 = 45\), \(\sigma_2=9290\), \(n_2 = 35\), and \(z_{\alpha/2}=1.96\)

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Step4: Construct the confidence interval

The confidence interval for \(\mu_1-\mu_2\) is \((\bar{x}_1-\bar{x}_2 - E,\bar{x}_1-\bar{x}_2 + E)\)

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Answer:

\(13436<\mu_1-\mu_2<22024\)