Question

you are given the sample mean and the population standard deviation. use this information to construct the 90% and 95% confidence intervals for the population mean. interpret the results and compare the widths of the confide intervals. from a random sample of 35 months from january 2006 through december 2020, the mean number of tornadoes month in the united states was about 101. assume the population standard deviation is 106. the 90% confidence interval is (□,□) (round to two decimal places as needed.)

Explanation:

Step1: Find the z - score for 90% confidence level

For a 90% confidence level, the significance level \(\alpha=1 - 0.90=0.10\), and \(\alpha/2 = 0.05\). The z - score \(z_{\alpha/2}\) is such that \(P(Z>z_{\alpha/2})=\alpha/2 = 0.05\). From the standard normal table, \(z_{\alpha/2}=z_{0.05}\approx1.645\)

Step2: Calculate the margin of error \(E\)

The formula for the margin of error when the population standard deviation \(\sigma\) is known is \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\)
Given \(n = 35\), \(\sigma=106\), and \(z_{\alpha/2}=1.645\)
\(E=1.645\times\frac{106}{\sqrt{35}}\)
First, \(\sqrt{35}\approx5.916\)
\(E = 1.645\times\frac{106}{5.916}\)
\(\frac{106}{5.916}\approx17.92\)
\(E=1.645\times17.92\approx29.48\)

Step3: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is given by \(\bar{x}-E<\mu <\bar{x}+E\)
Given \(\bar{x} = 101\)
The lower limit is \(101-29.48 = 71.52\)
The upper limit is \(101 + 29.48=130.48\)

Answer:

\((71.52,130.48)\)