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

Answer:

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

Step-by-step 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 $T=\frac{\bar{X}-\mu}{S/\sqrt{n}}$ 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%.