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The personnel director of a large hospital is interested in determining the relationship (if any) between an employee's age and the number of sick days the employee takes per year. The director randomly selects eight employees and records their age and the number of sick days which they took in the previous year.

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

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 22.

Respuesta :

Answer:

CI ( 90% )  = ( 5.9694 ; 12.2272)

Step-by-step explanation:

The regression line is:

Sick Days = 14.310162 - 0.2369 (Age)

The employee 22 years old will have:

SD = 14.310162 - 0.2369 (22)

SD = 14.310162 - 5.2118     ⇒  SD = 9.0983

SE = 1.682207

And CI 90 %    α = 10%  α =0.1    α/2 = 0.05

sample size  n = 10

degree of freedom   df = n - 2     df = 8

From t-student table we find t(c) = 1.86

CI ( 90% )  =  ( SD[22 years]) ± t(c)*SE

CI ( 90% )  = 9.0983 ± 1.86*1.682207

CI ( 90% )  = 9.0983 ± 3.1289

CI ( 90% )  = ( 5.9694 ; 12.2272)

Using the t-distribution, it is found that the 90% confidence interval for the average number of sick days an employee will take per year, given the employee is 22 is (6, 12.2).

We are given the standard error for the sample, which is why the t-distribution is used to solve this question.

The information to build the interval is:

  • Sample mean of [tex]S(22) = \overline{x} = 14.310162 - 0.2369(22) = 9.1[/tex].

  • Standard error of [tex]s = 1.682207[/tex].
  • Sample size of [tex]n = 10[/tex].

The confidence interval is:

[tex]\overline{x} \pm ts[/tex]

The critical value, using a t-distribution calculator, for a two-tailed 90% confidence interval, with 10 - 1 = 9 df, is t = 1.8331.

Then:

[tex]\overline{x} - ts = 9.1 - 1.8331(1.682207) = 6[/tex]

[tex]\overline{x} + ts = 9.1 + 1.8331(1.682207) = 12.2[/tex]

The 90% confidence interval for the average number of sick days an employee will take per year, given the employee is 22 is (6, 12.2).

A similar problem is given at https://brainly.com/question/15180581

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