Chapter PVII, Problem 8RE

(a)

To determine

To find: the number of herds of wild horses that studied.

Answer to Problem 8RE

38

Explanation of Solution

In this case, since there are 36 degrees of freedom, 38 herds of wild horses have been studied.

(b)

To determine

To Explain: the conditions significance for the inference satisfied.

Explanation of Solution

It is important to obey the following criteria.

i) Straight enough condition: To attempt linear regression, the scatterplot is straight enough.

(ii) Independence assumption: No pattern is shown in the residual plot.

iii) Does it thicken the plot   Condition: There is a clear distribution of the residuals.

(iv) Almost Natural condition: The residual histogram is unimodal and symmetrical.

(c)

To determine

To construct: the 95 percent confidence interval for slope of the relationship.

Answer to Problem 8RE

(0.131, 0.177)

Explanation of Solution

Formula used:

For the confidence interval

  $b_1\pm t_{n-2}\times SE\left(b_1\right)$

Calculation:

Because the conditions for inference are satisfied, the sampling distribution of the regression slope could be modelled by the student’s t-model with (38-2) = 36 degrees of freedom. using $t_{35}=2.030$ as an estimate.

  $\begin{array}{c}{t}_{35}=2.030\\ b_1\pm t_{n-2}\times SE\left(b_1\right)\\ =0.153969\pm \left(2.030\right)\times 0.0114\\ =\left(0.131,0.177\right)\end{array}$

(d)

To determine

To Explain: the meaning of the slope.

Explanation of Solution

there are 95 percent confident that with each additional adult horse, the mean number of foals in a herd increases by between 0.131 and 0.177 foals.

(e)

To determine

To Calculate: the 90 percent prediction interval, the number of foals that might be born.

Explanation of Solution

Formula used:

  $\widehat{y}\pm t_{n-2}\sqrt{S{E}^2\left(b_1\right)\times {\left(x-\overline{x}\right)}^2+\frac{s^2}{n}+s^2}$

Calculation:

The equation of regression forecasts that herds of 80 adults will have

-1.57835+0.153969(80) = 10.73917 foals. The average size of the herds sampled is 110.237 adult horses. Use $t_{35}=1.6883$ , or use an estimate $t_{35}=1.690$ , from the table.

  $\begin{array}{c}\widehat{y}\pm t_{n-2}\sqrt{S{E}^2\left(b_1\right)\times {\left(x-\overline{x}\right)}^2+\frac{s^2}{n}+s^2}\\ =10.73917\pm \left(1.6883\right)\sqrt{{0.0114}^2\times {\left(80-110.237\right)}^2+\frac{{4.941}^2}{38}+{4.941}^2}\\ =\left(2.26,19.21\right)\end{array}$

there are 95 percent sure that the number of foals will be between 2.26 and 19.21 in a herd of 80 adult horses. This interval of predication is too large to be of much benefit.