Question

when a population is finite, the formula that determines the standard error of the mean $sigma_{x}$ needs to be adjusted. if $n$ is the size of the population and $n$ is the size of the sample (where $n geq 0.05 n$), then the standard error of the mean is $sigma_{x}=\frac{sigma}{sqrt{n}} sqrt{\frac{n - n}{n - 1}}$. the expression $sqrt{\frac{n - n}{n - 1}}$ is called the finite population correction factor. use the finite population correction factor to construct the confidence interval for the population mean described below. $c = 0.99$, $overline{x}=12.5$, $sigma = 9.8$, $n = 400$, $n = 48$. the $99%$ confidence interval for the population mean is (round to two decimal places as needed.)

Explanation:

Step1: Find the z - value

For a \(90\%\) confidence interval, the significance level \(\alpha=1 - 0.90=0.10\), and \(\alpha/2 = 0.05\). Using the standard normal distribution table or a calculator, \(z_{\alpha/2}=z_{0.05}\approx1.645\)

Step2: Calculate the finite - population correction factor

The finite - population correction factor is \(\sqrt{\frac{N - n}{N - 1}}\), where \(N = 400\) and \(n = 49\).

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Step3: Calculate the standard error of the mean \(\sigma_{\bar{x}}\)

The formula for \(\sigma_{\bar{x}}\) with the finite - population correction factor is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\sqrt{\frac{N - n}{N - 1}}\), where \(\sigma = 8.8\) and \(n = 49\)

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

The margin of error \(E = z_{\alpha/2}\sigma_{\bar{x}}\)

$$ E=1.645\times1.18\approx1.94 $$

Step5: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu<\bar{x} + E\), where \(\bar{x}=12.5\)

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Answer:

\((10.56,14.44)\)