Question

Students in a statistics class take a random sample of 100 students at their college and record how many units each student is enrolled in. The students compute a 90% confidence interval for the mean number of units for students at their college and get (11.93, 12.47). Next, the students calculate a 95% confidence interval. As the confidence level increases, which of the following will happen to the interval width?

A. Decrease
B. Increase
C. Stay the same

Answer 1

Now if the confidence level increase to 95% then the critical value
`t_(\alpha/2)` would increase since if we want more confidence the margin of error need's to increase. And since the width for the confidence interval is given by:

`Width = 2ME`

And the margin of error is:

`ME=t_(\alpha/2) (s)/(√(n))`

Then we can conclude that increasing the confidence level from 90% to 95% the width of the interval would:

B. Increase

Explanation:

For this case we can define the variable of interest as the number of units for students at their college and we are interested in a confidence interval for the true mean
`\mu` and for this parameter the confidence interval is given by this formula:

`\bar X  \pm t_(\alpha/2) (s)/(√(n))`

The confidence interval at 90% of confidence is given:

`11.93 \leq \mu \leq 12.47`

Now if the confidence level increase to 95% then the critical value
`t_(\alpha/2)` would increase since if we want more confidence the margin of error need's to increase. And since the width for the confidence interval is given by:

`Width = 2ME`

And the margin of error is:

`ME=t_(\alpha/2) (s)/(√(n))`

Then we can conclude that increasing the confidence level from 90% to 95% the width of the interval would:

B. Increase