Question

questions exercise 08.03 algo (population mean: sigma known)
a simple random sample of 20 items resulted in a sample mean of 10. the population standard deviation is \\( \sigma=20 \\).
a. compute the \\( 95 \\% \\) confidence interval for the population mean. round your answers to one decimal place.

b. assume that the same sample mean was obtained from a sample of 90 items. provide a \\( 95 \\% \\) confidence interval for the population mean. round your answers to two decimal places

c. what is the effect of a larger sample size on the interval estimate?
larger sample provides a margin of error.

Explanation:

Step1: Find the z - value for 95% confidence level

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

Step2: Calculate the margin of error for part (a)

The formula for the margin of error \(E=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\)
Given \(n = 20\), \(\sigma=20\), \(z_{\alpha/2}=1.96\)
\(E=1.96\times\frac{20}{\sqrt{20}}\)
\(=\frac{1.96\times20}{4.472}\)
\(=\frac{39.2}{4.472}\approx8.8\)
The confidence interval is \(\bar{x}\pm E\), with \(\bar{x} = 10\)
The lower limit \(=10 - 8.8=1.2\)
The upper limit \(=10+8.8 = 18.8\)

Step3: Calculate the margin of error for part (b)

Given \(n = 90\), \(\sigma=20\), \(z_{\alpha/2}=1.96\)
\(E=1.96\times\frac{20}{\sqrt{90}}\)
\(=\frac{1.96\times20}{9.487}\)
\(=\frac{39.2}{9.487}\approx4.13\)
The confidence interval is \(\bar{x}\pm E\), with \(\bar{x}=10\)
The lower limit \(=10 - 4.13 = 5.87\)
The upper limit \(=10+4.13=14.13\)

Answer:

a. \(1.2\), \(18.8\)
b. \(5.87\), \(14.13\)
c. Larger sample provides a smaller margin of error.