Answer :
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 [tex]\mu[/tex] is given by [tex]T=\frac{\bar{X}-\mu}{S/\sqrt{n}}[/tex] that has a t distribution with (n-1) degrees of freedom. We find a [tex]100(1-\alpha)[/tex]% confidence inverval using [tex]P(-t_{\alpha/2}\leq T \leq t_{\alpha/2}) = 1-\alpha[/tex], where [tex]t_{\alpha/2}[/tex] is the t-value such that there is an area equal to [tex]\alpha/2[/tex] 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 [tex]P(-t_{0.025}\leq T\leq t_{0.025}) = 0.95[/tex] and we find a 90% confidence interval with [tex]P(-t_{0.05}\leq T\leq t_{0.05}) = 0.90[/tex]. Because of [tex]-t_{0.025} < -t_{0.05}[/tex] and [tex]t_{0.05} < t_{0.025}[/tex], the confidence interval would be more narrow if the confidence level were changed to 90%.