Answer :
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]
### 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]
