Question

A researcher randomly selects 6 fathers who have adult sons and records the​ fathers' and​ sons' heights to obtain the data below. Determine if sons are taller than their fathers at the alpha equals 0.1 level of significance. The normal probability plot and boxplot indicate that the differences are approximately normally distributed with no outliers.Son's Height75.668.268.878.577.676.6Father's height74.568.171.374.974.371.9Using the differences (fathers height)- (son's height) what is your conclusion regarding H0?

Answer 1

Sons are taller than their fathers.

Explanation:

The t-statistic for difference of mean is given by,

`t=\frac{\bar{x_(1)}-\bar{x_(2)}}{\sqrt{(s_(1)^(2))/(n_(1))+(s_(2)^(2))/(n_(2))}}`

Calculating Mean of Son's height =
`\bar{x_(1)}` = 74.22

Mean of Father's height =
`\bar{x_(2)}` = 72.5

Variance of Son's height =
`{s_(1)^(2)}` = 17.155

Variance of Father's height =
`{s_(2)^(2)}` = 5.69

and n₁ = n₂ = 6

⇒
`t=\frac{74.22}-\bar{72.5}{\sqrt{(17.55)/(6)+(5.69)/(6)}`

Using this formula, we get, t = 0.80306.

Using t-table, p-value = 0.220306 with 10 degree of freedom.

and α = 0.10

Since, the p-value is greater than alpha (p > .10), then we fail to reject the null hypothesis, and we say that the result is statistically non-significant.