(a)Step 1

A.95% confidence interval for the mean number of hours it takes for the student to makes course of objectives is, First, compute f-critical value then find confidence interval.

The t critical value for the 95% confidence interval is,

\(\alpha=0.05\)

The sample size is small and two-tailed test. Look in the column headed and the row headed in the fdistribution table by using degree of freedom is,

d.f.=n-1

=45-1=44

The tcritical value for the 95% confidence interval is 2.015.

Step 2

\(95\% C.I.=\overline{x}\pm t_{c}\times s / \sqrt{n}\)

\(=80.50\pm2.015\times51.20/\sqrt{45}\)

\(=80.50\pm 15.38\)

= 65.12 to 95.88

65.12 to 95.884 95% confidence interval for the mean number of hours it takes for the student to makes course of objectives is.

Step 3

Since you have posted a question with multiple sub-parts, we will solve first sub-parts for you. To get remaining sub-part solved please repost the complete question and mention the sub-parts to be solved.

A.95% confidence interval for the mean number of hours it takes for the student to makes course of objectives is, First, compute f-critical value then find confidence interval.

The t critical value for the 95% confidence interval is,

\(\alpha=0.05\)

The sample size is small and two-tailed test. Look in the column headed and the row headed in the fdistribution table by using degree of freedom is,

d.f.=n-1

=45-1=44

The tcritical value for the 95% confidence interval is 2.015.

Step 2

\(95\% C.I.=\overline{x}\pm t_{c}\times s / \sqrt{n}\)

\(=80.50\pm2.015\times51.20/\sqrt{45}\)

\(=80.50\pm 15.38\)

= 65.12 to 95.88

65.12 to 95.884 95% confidence interval for the mean number of hours it takes for the student to makes course of objectives is.

Step 3

Since you have posted a question with multiple sub-parts, we will solve first sub-parts for you. To get remaining sub-part solved please repost the complete question and mention the sub-parts to be solved.