Chapter PVII, Problem 4RE

(a)

To determine

To construct: the 95 percent interval and the meaning on the basis of given data.

Explanation of Solution

Given:

  

Formula used:

  $SE=\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

For the interval

  ${\overline{y}}_1-{\overline{y}}_2\pm SE$

Calculation:

  

  $\begin{array}{c}SE=\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =\sqrt{\frac{3.74}{15}+\frac{3.52}{15}}\\ =1.325\end{array}$

Now, use the following for the interval

  $\begin{array}{c}{\overline{y}}_1-{\overline{y}}_2\pm SE\\ 13.047-8.51\pm 1.325\\ 4.53\pm 1.325\\ \text{The}\text{interval}\text{is}\left(3.2,5.9\right)\end{array}$

Based on the provided data, with 95% confidence interval 5-year yields arebetween 3.2% and 5.9% larger than the 3-year yields onaverage

(b)

To determine

To Explain: the indication that there is an association between 3 and 5 year rates of return.

Explanation of Solution

Given:

  

Calculation:

The regression line using the given coefficients is the following

  $5-\text{Year=6}\text{.93+0}\text{.719}\times \text{3-Year}$

To test whether there is proof that of an association, using the following hypothesis

  $\begin{array}{c}{H}_0:{\beta }_1=0\\ vs\\ H_1:{\beta }_1\ne 0\end{array}$

Calculating t

  $t=\frac{\left({\overline{y}}_1-{\overline{y}}_2\right)}{SE\left({\overline{y}}_1-{\overline{y}}_2\right)}$

Hence, for t = 4.27 using a table or Excel, the P-value is 0.0009.

This P-value is very small; therefore, rejecting the null hypothesis. There is strong proof of an association.