Question

Like father, like son: In 1906, the statistician Karl Pearson measured the heights of 1078 pairs of fathers and sons. The following table presents a sample of 8 pairs, with height measured in inches, simulated from the distribution specified by al Pearson Father's Son's height height 703 65.4 65.7 69.0 736 66.7 66.0 70.9 69.1 749 683 T13 683 681 Send data to Excel Use the P-value method to testo 1-0 versus 10. Can you conclude that father's height is useful in predicting son's height? Use the -- 0.10 level of significance and the TI-84 calculator Part 1 of 4 Compute the least squares regression line for predicting son's height (from father's height(). Round the stape and intercept values to at least four decimal places. Regression line equation: Part: 1/4 Part 2 of 4 Compute the P-value. Round the answer to at least four decimal places P value

Answer 1

> father_height=c(70.8,65.4,65.7,69.0,73.6,66.7,70.1,68.3)
> Son_height=c(69.8,66.0,70.9,69.1,74.9,68.8,73.3,68.3)
> m=lm(Son_height~father_height)
> a=anova(m)
> a
Analysis of Variance Table

Response: Son_height
Df Sum Sq Mean Sq F value Pr(>F)
father_height 1 31.620 31.6198 7.5528 0.03337 *
Residuals 6   25.119 4.1865
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Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
> # p-value for testing H0:β=0 vs H1:β≠0 is 0.03337<0.10 hence we reject H0 and conclude that Father's height is useful in predicting son's height
>
> m

Call:
lm(formula = Son_height ~ father_height)

Coefficients:
(Intercept) father_height
17.8186 0.7616

> # regression equation is ŷ= 17.8186 + 0.7616*x where ŷ= Son_height and x=father_height
> summary(m)

Call:
lm(formula = Son_height ~ father_height)

Residuals:
Min 1Q Median 3Q Max
-1.9368 -1.5557 -0.5402 1.2972 3.0472

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 17.8186 19.0510 0.935 0.3857
father_height 0.7616 0.2771 2.748 0.0334 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 2.046 on 6 degrees of freedom
Multiple R-squared: 0.5573,   Adjusted R-squared: 0.4835
F-statistic: 7.553 on 1 and 6 DF, p-value: 0.03337

> round(a$`Pr(>F)`,4) # p-value
[1] 0.0334 NA