Question

questions exercise 08.02 algo (population mean: sigma known)
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a simple random sample of 70 items from a population with \\( \sigma = 7 \\) resulted in a sample mean of 33.
if required, round your answers to two decimal places.
a. provide a \\( 90 \\% \\) confidence interval for the population mean.
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b. provide a \\( 95 \\% \\) confidence interval for the population mean.
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c. provide a \\( 99 \\% \\) confidence interval for the population mean.
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Explanation:

Step1: Calculate the standard error

The formula for the standard error \(SE=\frac{\sigma}{\sqrt{n}}\). Given \(\sigma = 7\) and \(n = 70\), then \(SE=\frac{7}{\sqrt{70}}\approx\frac{7}{8.3666}\approx0.84\).

Step2: Find the z - value for each confidence level

• For a \(90\%\) confidence interval, the significance level \(\alpha=1 - 0.90=0.10\), and \(\alpha/2=0.05\). The \(z\) - value \(z_{\alpha/2}=z_{0.05}\approx1.645\) (from the standard normal distribution table).
• For a \(95\%\) confidence interval, \(\alpha = 1- 0.95=0.05\), \(\alpha/2=0.025\). The \(z\) - value \(z_{\alpha/2}=z_{0.025}\approx1.96\).
• For a \(99\%\) confidence interval, \(\alpha=1 - 0.99=0.01\), \(\alpha/2=0.005\). The \(z\) - value \(z_{\alpha/2}=z_{0.005}\approx2.576\).

Step3: Calculate the confidence interval for each case

The formula for the confidence interval is \(\bar{x}\pm z_{\alpha/2}\times SE\), where \(\bar{x} = 33\) and \(SE\approx0.84\)

• For \(90\%\) confidence interval:

The lower limit \(=33-1.645\times0.84=33 - 1.3818\approx31.62\)
The upper limit \(=33 + 1.645\times0.84=33+1.3818\approx34.38\)

• For \(95\%\) confidence interval:

The lower limit \(=33-1.96\times0.84=33 - 1.6464\approx31.35\)
The upper limit \(=33 + 1.96\times0.84=33+1.6464\approx34.65\)

• For \(99\%\) confidence interval:

The lower limit \(=33-2.576\times0.84=33 - 2.16384\approx30.84\)
The upper limit \(=33 + 2.576\times0.84=33+2.16384\approx35.16\)

Answer:

a. \(31.62\) to \(34.38\)
b. \(31.35\) to \(34.65\)
c. \(30.84\) to \(35.16\)