Chapter 10.2, Problem 50E

a.

To determine

To find: The type of t -statistic involved here.

Answer to Problem 50E

Paired t -static is used here.

Explanation of Solution

Given:

There are two pairs of birds which have been provided with natural caterpillar supply supplemented and control group.

Test statistics for the supplemented and control groups are 0.63 and -2.63 respectively.

In this case, two samples of birds are involved which have been used to make the comparison in two different yeards. The same number of birds are involved in each sample and the condition of before and after is fulfilling here.

Thus, the appropriate test is paired t -statistic.

b.

To determine

To find: The degree of freedom for both the groups.

Answer to Problem 50E

The degrees of freedom are 5 and 6.

Explanation of Solution

Calculation:

The degree of freedom could be computed as:

  $\begin{array}{l}{\text{Degree of freedom}}_{\text{Control group}}=n-1=6-1=5\\ {\text{Degree of freedom}}_{\text{Supplemented group}}=n-1=7-1=5\end{array}$

Hence, the degrees of freedom are 5 and 6.

c.

To determine

To explain: The reason of test statistic not agreeing with the prediction.

Explanation of Solution

For the test statistic t = 0.63, the p -value for one tailed test at 5 degree of freedom is 0.722. Since, the p-value is more than the significance level the null hypothesis does not get rejected.

Similarly, the p -value for one tailed test at 6 degree of freedom is 0.02. Since, the p-value is less than the significance level the null hypothesis does get rejected.