Question

To find and interpret a confidence interval , at the specified confidence level$90\%$ for the slope of the population regression line that relates the response variables to the predictor variable.

Short answer

The confidence interval $=[71.0382,85.8079]$

Step-by-step solution

Given information

The given data is

Explanation

Calculate the degree of freedom

$df=n-2$

$=10-2$

$=8$

The value of $t_{0.1}$is $1.860$, according to the student t-test.

Make a confidence interval calculation.

$\overline{x}-t_{0.1}\times \frac{s}{\sqrt{n}}=82.4-1.860\times \frac{3.9497}{\sqrt{10}}$

$=71.0382$

$\overline{x}-t_{0.1}\times \frac{s}{\sqrt{n}}=82.4+1.860\times \frac{3.9497}{\sqrt{10}}$

$=85.8079$

Confidence interval $=[71.0382,85.8079]$