Question

8 multiple choice 1.43 points a professor at a local university wanted to determine the standard deviation for the age at which a person obtains their bachelor’s degree. in a sample of 11 people who just obtained their bachelor’s 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 bachelor’s 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 bachelor’s 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 bachelor’s 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 bachelor’s degree will lie between 1.65 and 3.18 years. 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.

Explanation:

Step1: Identify parameters

Sample size $n=11$, sample standard deviation $s=1.96$, confidence level 95% → $\alpha=0.05$. Degrees of freedom $df=n-1=10$.

Step2: Find chi-square values

$\chi^2_{\alpha/2,10}=\chi^2_{0.025,10}=20.483$, $\chi^2_{1-\alpha/2,10}=\chi^2_{0.975,10}=3.247$.

Step3: Calculate interval bounds

Lower bound: $\sqrt{\frac{(n-1)s^2}{\chi^2_{\alpha/2}}}=\sqrt{\frac{10*(1.96)^2}{20.483}}≈\sqrt{\frac{38.416}{20.483}}≈\sqrt{1.875}≈1.369≈1.37$.
Upper bound: $\sqrt{\frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}}=\sqrt{\frac{10*(1.96)^2}{3.247}}≈\sqrt{\frac{38.416}{3.247}}≈\sqrt{11.831}≈3.439≈3.44$.

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.