Question

11) A marketing research company is estimating the average total compensation of CEOs in the service industry. Data were randomly collected from 18 CEOs and the 95% confidence interval for the mean was calculated to be ($2,181,260, $5,836,180). What would happen to the confidence interval if the confidence level were changed to 90%?

Answer 1

The confidence interval would be more narrow if the confidence level were changed to 90%

Explanation:

For a small sample size of n = 18 a pivotal quantity that we can use to form a confidence interval for
`\mu` is given by
`T=\frac{\bar{X}-\mu}{S/√(n)}` that has a t distribution with (n-1) degrees of freedom. We find a
`100(1-\alpha)`% confidence inverval using
`P(-t_(\alpha/2)\leq T \leq t_(\alpha/2)) = 1-\alpha`, where
`t_(\alpha/2)` is the t-value such that there is an area equal to
`\alpha/2` above this t-value and below the curve of the density of the t distribution with n-1 df. We find a 95% confidence interval with
`P(-t_(0.025)\leq T\leq t_(0.025)) = 0.95` and we find a 90% confidence interval with
`P(-t_(0.05)\leq T\leq t_(0.05)) = 0.90`. Because of
`-t_(0.025) < -t_(0.05)` and
`t_(0.05) < t_(0.025)`, the confidence interval would be more narrow if the confidence level were changed to 90%.