Question

The estimated regression line and the standard error are given. Sick Days=14.310162−0.2369(Age) se=1.682207 Find the 90% confidence interval for the average number of sick days an employee will take per year, given the employee is 28. Round your answer to two decimal places.

Employee 1 2 3 4 5 6 7 8 9 10
Age 30 50 40 55 30 28 60 25 30 45
Sick Days 7 4 3 2 9 10 0 8 5 2

Answer 1

To find the 90% confidence interval for the average number of sick days an employee will take per year, given the employee is 28, use the estimated regression line and the standard error.

Step-by-step explanation:

To find the 90% confidence interval for the average number of sick days an employee will take per year, given the employee is 28, we need to use the estimated regression line and the standard error. We know that the estimated regression line is Sick Days = 14.310162 - 0.2369(Age) and the standard error is 1.682207. To calculate the confidence interval, we use the formula: Mean ± (t)*(Standard Error), where t is the critical value from the t-distribution table corresponding to the desired confidence level.

Substituting the values, we have X = 14.310162 - 0.2369*(28) = 7.643774. The critical value for a 90% confidence level with 8 degrees of freedom is about 1.860. So the confidence interval is 7.643774 ± (1.860)*(1.682207) = [4.484856, 10.802692]. Therefore, the average number of sick days an employee will take per year, given the employee is 28, is estimated to be between 4.48 and 10.80.

Answer 2

To find the 90% confidence interval for the average number of sick days an employee of age 28 will take per year, we used the given regression line and standard error, substituting the age into the regression equation to obtain the point estimate, and then used a t-value to find the margin of error and calculate the confidence interval. The resulting 90% confidence interval is (4.91, 10.45) sick days.

Step-by-step explanation:

To find the 90% confidence interval for the average number of sick days an employee will take per year given that the employee is 28, we need to use the estimated regression line for the sick days, which is given by:

Sick Days = 14.310162 - 0.2369(Age)

First, we substitute the Age (28) into the regression equation to find the predicted average number of sick days:

Sick Days = 14.310162 - 0.2369(28) = 14.310162 - 6.6332 = 7.676962

The point estimate is therefore 7.68 since we round to two decimal places.

Next, we construct a confidence interval using the standard error (se) of 1.682207 and the value of t for the 90% confidence level which is not provided directly but can be found from a t-distribution table or statistical software. Assuming the value of t for this level is approximately 1.645 for a large sample size, we can calculate the margin of error (ME):

ME = t * se = 1.645 * 1.682207 = 2.767427

The confidence interval is then calculated by subtracting and adding this margin of error from the point estimate:

Lower limit: 7.68 - 2.767427 = 4.912573 (rounded to 4.91)

Upper limit: 7.68 + 2.767427 = 10.447427 (rounded to 10.45)

Therefore, the 90% confidence interval for the average number of sick days is (4.91, 10.45).