significance level of 0.05. Click the icon to view the critical values of the Pearson correlation coefficient r. www The best predicted value ofy is (Round to two decimal places as needed.)

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### Critical Values of the Pearson Correlation Coefficient \( r \)

#### Critical Values of the Pearson Correlation Coefficient \( r \)

| n  | α = 0.05 | α = 0.01 |
|----|----------|----------|
| 4  | 0.950    | 0.990    |
| 5  | 0.878    | 0.959    |
| 6  | 0.811    | 0.917    |
| 7  | 0.754    | 0.875    |
| 8  | 0.707    | 0.834    |
| 9  | 0.666    | 0.798    |
| 10 | 0.632    | 0.765    |
| 11 | 0.602    | 0.735    |
| 12 | 0.576    | 0.708    |
| 13 | 0.553    | 0.684    |
| 14 | 0.532    | 0.661    |
| 15 | 0.514    | 0.641    |
| 16 | 0.497    | 0.623    |
| 17 | 0.482    | 0.606    |
| 18 | 0.468    | 0.590    |
| 19 | 0.456    | 0.575    |
| 20 | 0.444    | 0.561    |
| 25 | 0.396    | 0.505    |
| 30 | 0.361    | 0.463    |
| 35 | 0.335    | 0.430    |
| 40 | 0.312    | 0.402    |
| 45 | 0.294    | 0.378    |
| 50 | 0.279    | 0.361    |
| 60 | 0.254    | 0.330    |
| 70 | 0.236    | 0.305    |
| 80 | 0.220    | 0.286    |
| 90 | 0.207    | 0.269    |
| 100| 0.196
Transcribed Image Text:### Critical Values of the Pearson Correlation Coefficient \( r \) #### Critical Values of the Pearson Correlation Coefficient \( r \) | n | α = 0.05 | α = 0.01 | |----|----------|----------| | 4 | 0.950 | 0.990 | | 5 | 0.878 | 0.959 | | 6 | 0.811 | 0.917 | | 7 | 0.754 | 0.875 | | 8 | 0.707 | 0.834 | | 9 | 0.666 | 0.798 | | 10 | 0.632 | 0.765 | | 11 | 0.602 | 0.735 | | 12 | 0.576 | 0.708 | | 13 | 0.553 | 0.684 | | 14 | 0.532 | 0.661 | | 15 | 0.514 | 0.641 | | 16 | 0.497 | 0.623 | | 17 | 0.482 | 0.606 | | 18 | 0.468 | 0.590 | | 19 | 0.456 | 0.575 | | 20 | 0.444 | 0.561 | | 25 | 0.396 | 0.505 | | 30 | 0.361 | 0.463 | | 35 | 0.335 | 0.430 | | 40 | 0.312 | 0.402 | | 45 | 0.294 | 0.378 | | 50 | 0.279 | 0.361 | | 60 | 0.254 | 0.330 | | 70 | 0.236 | 0.305 | | 80 | 0.220 | 0.286 | | 90 | 0.207 | 0.269 | | 100| 0.196
**Predicting IQ Scores Using a Linear Regression Model**

*Context:*
Suppose IQ scores were obtained for 20 randomly selected sets of twins. The 20 pairs of measurements yield:

- \( \bar{x} = 96.92 \)
- \( \bar{y} = 95.25 \)
- \( r = 0.874 \)
- \( P\text{-value} = 0.000 \)
- \( \hat{y} = 10.98 + 0.87x \)

Where \( x \) represents the IQ score of the twin born first. Use this linear regression model to find the best predicted value of \( \hat{y} \) given that the twin born first has an IQ of 102. Use a significance level of 0.05.

*Details of the Procedure:*

1. **Identify the Linear Regression Equation**:
   - The regression equation provided is \( \hat{y} = 10.98 + 0.87x \).

2. **Substitute the Given IQ**:
   - Here, \( x = 102 \).

3. **Calculate the Predicted Value**:
   - Substitute the value into the equation:
     \[
     \hat{y} = 10.98 + 0.87 \cdot 102
     \]

4. **Compute the Result**:
   - First, multiply 0.87 by 102:
     \[
     0.87 \cdot 102 = 88.74
     \]
   - Add 10.98 to the result:
     \[
     \hat{y} = 10.98 + 88.74 = 99.72
     \]

*Conclusion:*

The best predicted value of \( \hat{y} \) is:

\[
\boxed{99.72}
\]

Please note that the value is rounded to two decimal places as needed.
Transcribed Image Text:**Predicting IQ Scores Using a Linear Regression Model** *Context:* Suppose IQ scores were obtained for 20 randomly selected sets of twins. The 20 pairs of measurements yield: - \( \bar{x} = 96.92 \) - \( \bar{y} = 95.25 \) - \( r = 0.874 \) - \( P\text{-value} = 0.000 \) - \( \hat{y} = 10.98 + 0.87x \) Where \( x \) represents the IQ score of the twin born first. Use this linear regression model to find the best predicted value of \( \hat{y} \) given that the twin born first has an IQ of 102. Use a significance level of 0.05. *Details of the Procedure:* 1. **Identify the Linear Regression Equation**: - The regression equation provided is \( \hat{y} = 10.98 + 0.87x \). 2. **Substitute the Given IQ**: - Here, \( x = 102 \). 3. **Calculate the Predicted Value**: - Substitute the value into the equation: \[ \hat{y} = 10.98 + 0.87 \cdot 102 \] 4. **Compute the Result**: - First, multiply 0.87 by 102: \[ 0.87 \cdot 102 = 88.74 \] - Add 10.98 to the result: \[ \hat{y} = 10.98 + 88.74 = 99.72 \] *Conclusion:* The best predicted value of \( \hat{y} \) is: \[ \boxed{99.72} \] Please note that the value is rounded to two decimal places as needed.
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