Question

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 six employees reveals the following family dental expenses: $180, $260, $60, $40, $100, and $80. It is known that dental expenses follow a normal distribution. Construct a 90% confidence interval for the average family dental expenses for all employees of this corporation.

Answer 1

From the given sample data,

$\bar{x}$ = $\sum$ Xi / n

= 120

S = sqrt( $\sum$ Xi² - n$\bar{x}$² / n-1)

= 83.9047

90% confidence interval for $\mu$ is

$\bar{x}$ - t_$\alpha$/2 * S / sqrt(n) < $\mu$ < $\bar{x}$ + t_$\alpha$/2 * S / sqrt(n)

120 - 2.015 * 83.9047 / sqrt(6) < $\mu$ < 120 + 2.015 * 83.9047 / sqrt(6)

50.9783 < $\mu$ < 189.0217

90% CI for $\mu$ is (50.9783 , 189.0217)