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 confidence intervals.
from a random sample of 37 months from january 2006 through december 2020, the mean number of tornadoes per month in the united states was about 101. assume the population standard deviation is 115.
the 90% confidence interval is (round to two decimal places as needed.)
the 95% confidence interval is (round to two decimal places as needed.)
which interval is wider?
the 95% confidence interval
the 90% confidence interval

Explanation:

Step1: Understand confidence interval width

The width of a confidence interval is determined by the margin of error. The formula for the margin of error \(E = z\times\frac{\sigma}{\sqrt{n}}\), where \(z\) is the z - score, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.
For a 90% confidence interval, the z - score \(z_{0.90}\):
The significance level \(\alpha=1 - 0.90=0.10\), and \(\alpha/2 = 0.05\). From the standard normal table, \(z_{\alpha/2}=z_{0.05}\approx1.645\)
For a 95% confidence interval, the z - score \(z_{0.95}\):
The significance level \(\alpha = 1- 0.95=0.05\), and \(\alpha/2=0.025\). From the standard normal table, \(z_{\alpha/2}=z_{0.025}\approx1.96\)
Since \(\frac{\sigma}{\sqrt{n}}\) is constant (because \(\sigma = 115\) and \(n = 37\) are fixed), and \(E\propto z\) (margin of error is proportional to the z - score)

Step2: Compare the z - scores

Because \(1.96>1.645\) (the z - score for 95% confidence interval is larger than the z - score for 90% confidence interval)

Answer:

The 95% confidence interval