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

Respuesta :

Answer:

Now if the confidence level increase to 95% then the critical value [tex]t_{\alpha/2}[/tex] 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:

[tex] Width = 2ME[/tex]

And the margin of error is:

[tex]ME=t_{\alpha/2} \frac{s}{\sqrt{n}}[/tex]

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

B. Increase

Step-by-step 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 [tex]\mu[/tex] and for this parameter the confidence interval is given by this formula:

[tex]\bar X  \pm t_{\alpha/2} \frac{s}{\sqrt{n}}[/tex]

The confidence interval at 90% of confidence is given:

[tex] 11.93 \leq \mu \leq 12.47[/tex]

Now if the confidence level increase to 95% then the critical value [tex]t_{\alpha/2}[/tex] 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:

[tex] Width = 2ME[/tex]

And the margin of error is:

[tex]ME=t_{\alpha/2} \frac{s}{\sqrt{n}}[/tex]

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

B. Increase