Question

To test the belief that sons are taller than their​ fathers, a student randomly selects 6 fathers who have adult male children. She records the height of both the father and son in inches and obtains the accompanying data. Are sons taller than their​ fathers? Use the alpha equals 0.1 level of significance. Note that a normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers.

Observation	1	2	3	4	5	6
Height of father​ (in inches), Xi	69.8	66.3	71.8	67.1	71.9	70.5
Height of son​ (in inches), Yi	73.2	68.5	68.0	67.4	68.0	76.4

Answer 1

Here we want to test,

Ho: The average height of father and son is same

Vs

H1: The average height of father is less than his son's height (ie. son are taller than their father)

x: father's height , y: son's height

Here we will use the R software to test which is free software, the R-codes as follows also i will provide the procedure of test by theory.
> father_Ht=c(69.8,66.3,71.8,67.1,71.9,70.5)
> son_Ht=c(73.2,68.5,68.0,67.4,68.0,76.4)
> xbar=mean(father_Ht)
> ybar=mean(son_Ht)
> xbar
[1] 69.56667
> ybar
[1] 70.25
> sx=sqrt(var(father_Ht))
> sx
[1] 2.371216
> sy=sqrt(var(son_Ht))
> sy
[1] 3.683341
> sp=(5*sx+5*sy)/10
> sp
[1] 3.027279
> k=sqrt((1/6)+(1/6))
> k
[1] 0.5773503
> t_stat=(xbar-ybar)/(sp*k)
> t_stat
[1] -0.3909677

> ttab=qt(0.99,10)
> ttab
[1] 2.763769

xbar= 69.56667 ybar= 70.25

The test statistic under Ho

t-stat=(xbar-ybar)/(sp*k)

= -0.3909677

and table value is ttab=qt(0.99,10)=2.763, here t-stat < -ttab, therefor we may accept ho at the 1% level of significance. and conclude that the average of father's height and son's height is same, there is no statistical evidance that the son are taller than the father.

OR

> tcal=t.test(father_Ht,son_Ht,var.equal=T,level=0.99)
> tcal

   Two Sample t-test

data: father_Ht and son_Ht
t = -0.3821, df = 10, p-value = 0.7104
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-4.668077 3.301410
sample estimates:
mean of x mean of y
69.56667 70.25000

table value is ttab=qt(0.99,10)=2.763, here t-stat < -ttab, also P-value > 0.01, therefore we may accept Ho at the 1% level of significance. and conclude that the average of father's height and son's height is same, there is no statistical evidance that the son are taller than the father.