Question

Please show steps in the calculation. Please make sure you include checking conditions for using the CLT.

The Human Resources (HR) Department of a certain college has asked all employees who were employed in 2018 to fill out a survey in December 2018. Three items on the survey were: “Your dental expense in 2018”, “Are you in a family with at least three other family members?”, and “Your medical expenses in 2018”. The manager of HR has randomly selected a sample of 169 surveys. Found that the sample average dental expense is $1600 per person with the sample standard deviation is $500. 70 of them are in a family of at least three other members. Also, the sample average medical expense is $2,450 per person and the sample standard deviation is $700.

1. The statistical inference will be made in parts (b), (c) and (d). Please clearly state the three parameters of interest.
2. Provide a 99% interval estimate of the average dental expenses per employee who was employed in the year 2018. please interpret your answer.

Answer 1

Here n = 169

Sample average = = 1600

Sample standard deviation = s = 500

Here population standard deviation is not given so we will use here t interval.

Confidence level = c =0.99

Degrees of freedom = n - 1 = 169 - 1 = 168

99% interval estimate of the average dental expenses per employee who was employed in the year 2018 is

Where tc is t critical for 1- c = 1 - 0.99 = 0.01 and degrees of freedom = 168 can be calculated from excel using command:

=T.DIST.2T(0.01,168)

=2.605                     (Round to 3 decimal)

tc = 2.605                               (From statistical table of t values)

99% interval estimate of the average dental expenses per employee who was employed in the year 2018 is (1499.808,1700.192)

Interpretation:

We are 99% confident that our population parameter will lie in this interval.