Question

You are planning on applying to the Knirhsdaeh institute as a psychology counselor. They hire millions of psychologists around the world. You know the standard deviation for the salaries is $13 k. To find the average, you sample 23 random employees and get an average salary of $81 k. Find an 98% confidence interval for the true average salary of Knirhsdaeh employees.

Answer 1

Answer:

The 98% of the confidence interval for the true average salary of Knirhsdaeh employees as a psychology counselor

(75.4206, 86.5794)

Step-by-step explanation:

Step(i):-

Given that the mean of the sample = $81 k

Given that the size of the sample 'n' = 23

Given that the standard deviation for the salaries is $13 k

Step(ii):-

98% of the confidence interval for the true average salary of Knirhsdaeh employees is determined by

$(x^{-} - t_{0.02} \frac{S.D}{\sqrt{n} } , x^{-} + t_{0.02} \frac{S.D}{\sqrt{n} })$

Degrees of freedom = n-1 = 23-1 =22

$t_{0.02} = 2.5083$

$(81 - 2.0583 \frac{13}{\sqrt{23} } , 81 + 2.0583 \frac{13}{\sqrt{23} } )$

( 81 - 5.57940 , 81 + 5.57940)

(75.4206, 86.5794)

Final answer:-

The 98% of the confidence interval for the true average salary of Knirhsdaeh employees as a psychology counselor

(75.4206, 86.5794)