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11.3.16. Regression techniques can be very useful in situations where one variable—say, [tex]\( y \)[/tex]—is difficult to measure but [tex]\( x \)[/tex] is not. Once such an [tex]\( x \)[/tex]-[tex]\( y \)[/tex] relationship has been "calibrated," based on a set of [tex]\( (x_i, y_i) \)[/tex]'s, future values of [tex]\( Y \)[/tex] can be easily estimated using [tex]\( \hat{\beta}_0 + \hat{\beta}_1 x \)[/tex]. Determining the volume of an irregularly shaped object, for example, is often difficult, but weighing that object is likely to be easy.

The following table shows the weights (in kilograms) and the volumes (in cubic decimeters) of eighteen children between the ages of five and eight (15). The estimated regression line has the equation [tex]\( y = -0.104 + 0.988 x \)[/tex], where [tex]\( s = 0.202 \)[/tex].

(a) Construct a [tex]\( 95\% \)[/tex] confidence interval for [tex]\( E(Y \mid 14.0) \)[/tex].

(b) Construct a [tex]\( 95\% \)[/tex] prediction interval for the volume of a child weighing 14.0 kilograms.

[tex]\[
\begin{tabular}{cccc}
\hline
Weight, \( x \) & Volume, \( y \) & Weight, \( x \) & Volume, \( y \) \\
\hline
17.1 & 16.7 & 15.8 & 15.2 \\
10.5 & 10.4 & 15.1 & 14.8 \\
13.8 & 13.5 & 12.1 & 11.9 \\
15.7 & 15.7 & 18.4 & 18.3 \\
11.9 & 11.6 & 17.1 & 16.7 \\
10.4 & 10.2 & 16.7 & 16.6 \\
15.0 & 14.5 & 16.5 & 15.9 \\
16.0 & 15.8 & 15.1 & 15.1 \\
17.8 & 17.6 & 15.1 & 14.5 \\
\hline
\end{tabular}
\][/tex]



Answer :

GinnyAnswer
In this problem, we aim to perform regression analysis to estimate the volume of a child based on their weight using the provided regression equation [tex]\(y = -0.104 + 0.988x\)[/tex]. We will construct both the [tex]\(95\%\)[/tex] confidence interval for the mean volume and the [tex]\(95\%\)[/tex] prediction interval for the volume for a child weighing [tex]\(14.0\)[/tex] kilograms.

### Given Information:
- Regression equation: [tex]\(y = -0.104 + 0.988x\)[/tex]
- Standard deviation of the errors ([tex]\(s\)[/tex]): [tex]\(0.202\)[/tex]
- Number of observations ([tex]\(n\)[/tex]): [tex]\(18\)[/tex]
- Weight to be used for prediction ([tex]\(x\)[/tex]): [tex]\(14.0\)[/tex] kg

The weights and volumes for the [tex]\(18\)[/tex] children are as follows:
[tex]\[ \begin{array}{cccc} \hline \text{Weight, } x & \text{Volume, } y & \text{Weight, } x & \text{Volume, } y \\ \hline 17.1 & 16.7 & 15.8 & 15.2 \\ 10.5 & 10.4 & 15.1 & 14.8 \\ 13.8 & 13.5 & 12.1 & 11.9 \\ 15.7 & 15.7 & 18.4 & 18.3 \\ 11.9 & 11.6 & 17.1 & 16.7 \\ 10.4 & 10.2 & 16.7 & 16.6 \\ 15.0 & 14.5 & 16.5 & 15.9 \\ 16.0 & 15.8 & 15.1 & 15.1 \\ 17.8 & 17.6 & 15.1 & 14.5 \\ \hline \end{array} \][/tex]

### Part (a): Construct a 95% Confidence Interval for [tex]\(E(Y \mid 14.0)\)[/tex]

1. Predict the mean volume [tex]\((\hat{y})\)[/tex] when the weight [tex]\(x = 14.0\)[/tex] kg:
[tex]\[ \hat{y} = -0.104 + 0.988 \cdot 14.0 = 13.728 \][/tex]

2. Calculate the standard error of the mean prediction [tex]\((\text{SE}_{\text{mean}})\)[/tex]:
[tex]\[ \text{SE}_{\text{mean}} = 0.0519 \][/tex]

3. Determine the critical [tex]\(t\)[/tex]-value for a [tex]\(95\%\)[/tex] confidence interval:
Using a [tex]\(t\)[/tex]-distribution table with [tex]\(17\)[/tex] degrees of freedom ([tex]\(n-2 = 18-2 = 16\)[/tex]), we find:
[tex]\[ t_{0.025, 16} \approx 2.12 \][/tex]

4. Construct the [tex]\(95\%\)[/tex] confidence interval:
[tex]\[ \text{CI}_{\text{mean}} = \left(\hat{y} - t \cdot \text{SE}_{\text{mean}}, \hat{y} + t \cdot \text{SE}_{\text{mean}}\right) \][/tex]
[tex]\[ \text{CI}_{\text{mean}} = \left(13.728 - 2.120 \cdot 0.0519, 13.728 + 2.120 \cdot 0.0519\right) = (13.6179, 13.8380) \][/tex]

### Part (b): Construct a 95% Prediction Interval for the Volume of a Child Weighing 14.0 kg

1. Calculate the standard error of the individual prediction [tex]\((\text{SE}_{\text{individual}})\)[/tex]:
[tex]\[ \text{SE}_{\text{individual}} = 0.2086 \][/tex]

2. Construct the [tex]\(95\%\)[/tex] prediction interval:
[tex]\[ \text{PI}_{\text{individual}} = \left(\hat{y} - t \cdot \text{SE}_{\text{individual}}, \hat{y} + t \cdot \text{SE}_{\text{individual}}\right) \][/tex]
[tex]\[ \text{PI}_{\text{individual}} = \left(13.728 - 2.120 \cdot 0.2086, 13.728 + 2.120 \cdot 0.2086\right) = (13.2859, 14.1701) \][/tex]

### Summary:

- 95% Confidence Interval for [tex]\(E(Y \mid 14.0)\)[/tex]: [tex]\((13.6179, 13.8380)\)[/tex]
- 95% Prediction Interval for the volume of a child weighing [tex]\(14.0\)[/tex] kg: [tex]\((13.2859, 14.1701)\)[/tex]

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