Question

To test the belief that sons are taller than their fathers, a student ran- domly selects 13 fathers who have adult male children. She records the height (in inches) of both the father and the son in the following table. Are sons taller than their fathers? NOTE: A normal probability plot indicated that the differences (X -Y) are approximately normally distributed with no outliers. 70.4 71.8 70.1 70.2 70.4 69.3 eight of Father, Y eight of Son, X eight of Father, | 69.9 | 70.8 | 70.2 | 70.4 | 72.4 Height of Son, X 75.8 72.3 69.2 68.6 73.9 69 1) State the null and alternative hypothesis for this test X - Y. 2) In plain English state what the alternative hypothesis represents. 3) Calculate the mean, standard deviation and the p-value for X - Y the difference in heights between fathers and sons f we are using an a0.1 significance level should we reject or not reject the null hypothesis? State in plain English your confidence in the conclusion of this test.

Answer 1

We have given,

n1 = 13 and n2 = 13

$\alpha =0.10$,

Degree of freedom = 13+ 13- 2 = 24

$H_{0}:\mu_{1}=\mu_{2}$

Vs

$H_{0}:\mu_{1} < \mu_{2}$

Descriptive Statistics for height of father

Sample size:13

Mean (x̄): 70.3

Standard deviation (s): 1.738

Descriptive Statistics for height of Son

Sample size:13

Mean (x̄): 70.615

Standard deviation (s): 2.605

We can use here test statistic formula:

t = [70.3-70.615]/sqrt[(1.738)^2/13+(2.605)^2/13]

t= -0.363

The P-Value is .359891.

The result is not significant at p < .10.

Therefore we do not have sufficient evidence to say that, sons taller than their fathers.

Difference Scores Calculations

Treatment 1

N₁: 13
df₁ = N - 1 = 13 - 1 = 12
M₁: 70.3
SS₁: 36.24
s²₁ = SS₁/(N - 1) = 36.24/(13-1) = 3.02


Treatment 2

N₂: 13
df₂ = N - 1 = 13 - 1 = 12
M₂: 70.62
SS₂: 81.42
s²₂ = SS₂/(N - 1) = 81.42/(13-1) = 6.78


T-value Calculation

s²_p = ((df₁/(df₁ + df₂)) * s²₁) + ((df₂/(df₂+ df₂)) * s²₂) = ((12/24) * 3.02) + ((12/24) * 6.78) = 4.9

s²_M1 = s²_p/N₁ = 4.9/13 = 0.38
s²_M2 = s²_p/N₂ = 4.9/13 = 0.38

t = (M₁ - M₂)/√(s²_M1 + s²_M2) = -0.32/√0.75 = -0.36

The t-value is -0.36316.

BY HELP OF T TABLE AND USING EXCEL WE FOUND EXACT p-value =0.359833. The result is not significant at p < .10.

Since p value is not less than 0.1,we fail to reject the null hypothesis.