A sample of 15 recently trained line workers was selected to determine if there is a relationship between the number of hours of training time received by production line
Workers and the time it took (in minutes) for them to trouble shoot their last process
Problem were captured. Use the regression output for the independent variable
Shown below to find the 95% confidence interval for the slope of the regression Equation.
$\text { Training } \quad - 1.8360 \quad 0.1376 \quad - 13.35 \quad 0.000$
A) -4 to 0.32
B) -1.9776 to -1.7224
C) -2.1332 to -1.5388
D) -3.611 to -0.069
E) Can't be determined with the information given.

Question

A sample of 15 recently trained line workers was selected to determine if there is a relationship between the number of hours of training time received by production line
Workers and the time it took (in minutes) for them to trouble shoot their last process
Problem were captured. Use the regression output for the independent variable
Shown below to find the 95% confidence interval for the slope of the regression Equation.
 $\text { Training } \quad - 1.8360 \quad 0.1376 \quad - 13.35 \quad 0.000$
A) -4 to 0.32
B) -1.9776 to -1.7224
C) -2.1332 to -1.5388
D) -3.611 to -0.069
E) Can't be determined with the information given.

Quizplus · Accepted Answer

The 95% confidence interval for the slope of the regression equation is given by:
$$	ext{slope estimate} \pm t_{\alpha/2,n-2}\frac{s}{\sqrt{\sum(x_i - \bar{x})^2}}$$

Where:
slope estimate = -1.8360
$t_{\alpha/2,n-2}$ = t-value for a 95% confidence interval with 13 degrees of freedom (n-2)
s = standard error of the slope estimate = 0.1376
$\sum(x_i - \bar{x})^2$ = sum of squares of deviations of the training time from its mean

Using a t-distribution table or calculator, we find that $t_{\alpha/2,n-2}$ = 2.1604

Plugging the values into the formula, we get:

$$-1.8360 \pm 2.1604\frac{0.1376}{\sqrt{\sum(x_i - \bar{x})^2}}$$

We don't have information about the sum of squares of deviations of the training time from its mean, so we can't calculate the confidence interval. However, we can say that it will be centered around the point estimate of -1.8360 and will have a width determined by the sample standard deviation of the training time.

Looking at the answer choices, we see that only choice C falls within the plausible range of values for the slope.

A Sample of 15 Recently Trained Line Workers Was Selected Training −1.83600.1376−13.350.000\text { Training } \quad - 1.8360 \quad 0.1376 \quad - 13.35 \quad 0.000 Training −1.83600.1376−13.350.000

A Sample of 15 Recently Trained Line Workers Was Selected $\text { Training } \quad - 1.8360 \quad 0.1376 \quad - 13.35 \quad 0.000$