Question

a professor at a local university wanted to determine the standard deviation for the age at which a person obtains their bachelors degree. in a sample of 11 people who just obtained their bachelors degree, he found that the standard deviation was 1.96 years. construct and interpret a 95% confidence interval for the standard deviation age at which a person obtains their bachelors degree. note: the data comes from a population that is normally distributed. (make sure to verify the requirements before constructing the confidence interval.)
we are 95% confident that the standard deviation for the age at which a person obtains their bachelors degree will lie between 1.15 and 2.37 years.
we are 95% confident that the standard deviation for the age at which a person obtains their bachelors degree will lie between 1.37 and 3.44 years.
we are 95% confident that the standard deviation for the age at which a person obtains their bachelors degree will lie between 1.43 and 2.96 years.
we are 95% confident that the standard deviation for the age at which a person obtains their bachelors degree will lie between 1.65 and 3.18 years.

Explanation:

Step1: Determine the degrees of freedom

The degrees of freedom \(df=n - 1\), where \(n = 11\). So \(df=11-1 = 10\).

Step2: Find the critical values

For a 95% confidence interval, \(\alpha=1 - 0.95=0.05\). The lower - tail critical value \(\chi_{1-\alpha/2}^2=\chi_{0.975}^2\) and the upper - tail critical value \(\chi_{\alpha/2}^2=\chi_{0.025}^2\).
From the \(\chi^2\) - distribution table, \(\chi_{0.975}^2 = 3.247\) and \(\chi_{0.025}^2=20.483\).

Step3: Calculate the confidence interval for the variance

The formula for the confidence interval for the variance \(\sigma^{2}\) is \(\frac{(n - 1)s^{2}}{\chi_{\alpha/2}^2}\leq\sigma^{2}\leq\frac{(n - 1)s^{2}}{\chi_{1-\alpha/2}^2}\).
Given \(s = 1.96\), then \(s^{2}=(1.96)^{2}=3.8416\) and \(n = 11\).
Substitute the values: \(\frac{(11 - 1)\times3.8416}{20.483}\leq\sigma^{2}\leq\frac{(11 - 1)\times3.8416}{3.247}\).
\(\frac{38.416}{20.483}\leq\sigma^{2}\leq\frac{38.416}{3.247}\).
\(1.876\leq\sigma^{2}\leq11.831\).

Step4: Calculate the confidence interval for the standard deviation

Take the square root of each part of the variance interval. \(\sqrt{1.876}\leq\sigma\leq\sqrt{11.831}\).
\(1.37\leq\sigma\leq3.44\).

Answer:

We are 95% confident that the standard deviation for the age at which a person obtains their Bachelor's degree will lie between 1.37 and 3.44 years.