College

Heights (cm) and weights (kg) are measured for 100 randomly selected adult males, and range from heights of 137 to 192 cm and weights of 39 to 150 kg. Let the predictor variable [tex] x [/tex] be the first variable given. The 100 paired measurements yield:

- [tex] \bar{x} = 168.27 \, \text{cm} [/tex]
- [tex] \bar{y} = 81.57 \, \text{kg} [/tex]
- [tex] r = 0.113 [/tex]
- [tex] \text{P-value} = 0.263 [/tex]
- [tex] \hat{y} = -102 + 1.02x [/tex]

Find the best predicted value of [tex] \hat{y} [/tex] (weight) given an adult male who is 182 cm tall. Use a 0.01 significance level.

The best predicted value of [tex] \hat{y} [/tex] for an adult male who is 182 cm tall is [tex] \square [/tex] kg. (Round to two decimal places as needed.)

Answer :

To find the best predicted value of weight [tex]\(\hat{y}\)[/tex] for an adult male who is 182 cm tall, we can use the provided regression equation:

[tex]\[
\hat{y} = -102 + 1.02x
\][/tex]

Here, [tex]\(x\)[/tex] represents the height in centimeters. Let's go through the steps to find the predicted weight:

1. Identify the given height:
- The height of the adult male, [tex]\(x\)[/tex], is 182 cm.

2. Plug the height into the regression equation:
- Substitute [tex]\(x = 182\)[/tex] into the regression equation.
- [tex]\(\hat{y} = -102 + 1.02 \times 182\)[/tex]

3. Calculate the predicted weight:
- First, multiply 1.02 by 182:
[tex]\[
1.02 \times 182 = 185.64
\][/tex]
- Then, add the result to -102:
[tex]\[
\hat{y} = -102 + 185.64 = 83.64
\][/tex]

4. Round the predicted weight:
- The problem asks us to round the answer to two decimal places. The calculated value is already at two decimal places, so [tex]\(\hat{y} = 83.64\)[/tex].

Thus, the best predicted value of [tex]\(\hat{y}\)[/tex] (the weight) for an adult male who is 182 cm tall is 83.64 kg.