The personnel department of a large corporation wants to estimate the family dental expenses of its employees to determine the feasibility of providing a dental insurance plan. A random sample of 12 employees reveals the following family dental expenses (in dollars). See Attached Excel for Data. Construct a 97% confidence interval estimate for the average family dental expenses for all employees of this corporation.

The data cited is in the attachment.

308.2±106.4

Step-by-step explanation: To construct a confidence interval, first calculate mean (μ) and standard deviation (s) for the sample:

μ = Σvalue/n

μ = 308.2

s = √∑(x - μ)²/n-1

s = 147.9

Calculate standard error of the mean:

=

= 42.72

Find the degrees of freedom:

d.f. = n - 1

d.f. = 12 - 1

d.f. = 11

Find the significance level:

= 0.015

Since sample is smaller than 30, use t-test table and find t-score:

= 2.4907

E = t-score.

E = 2.4907.42.72

E = 106.4

The interval of confidence is: 308.2±106.4, which means that dental insurance plan varies from $201.8 to $414.6.

i cant answer this, theres no table. sorry man

step-by-step explanation: