Question

9. To test the belief that sons are taller than their fathers, a student randomly selects I3 fathers who have adult male children. She records the height of both the father and son in inches and obtains the following data: Height of father, X 70.3 67.1 709 66.8 Height of son, Y 74.1 69.2 669 692 689 70.2 70.4 728 70.4 71.8 一81-9 | 10 | 11 | 12 | 13 Height of father, 70.1 69.9 70.8 70.2 704 724 Height of son, Y 69.3 a. State the null and alternate hypotheses. b. Draw a normal probability plot and use it to determine whether the differences are approximately normally distributed. c. Construct a boxplot and use it to determine if there are any outliers. d. Determine the P-value (round your answer to four decimal places). Are sons taller than their fathers? Use the α-0.1 level of significance.

Answer 1

# code in R

father <- c(70.3,67.1,70.9,66.8,72.8,70.4,71.8,70.1,69.9,70.8,70.2,70.4,72.4)
son <- c(74.1,69.2,66.9,69.2,68.9,70.2,70.4,69.3,75.8,72.3,69.2,68.6,73.9)
diff <- son - father
qqnorm(diff)
boxplot(diff)

t.test(diff, mu = 0 , alternative = "greater")

#running the code

        One Sample t-test

data:  diff
t = 0.39251, df = 12, p-value = 0.3508
alternative hypothesis: true mean is greater than 0
95 percent confidence interval:
 -1.116693       Inf
sample estimates:
mean of x 
0.3153846 

a)

Ho : $\mu _d$ = 0

Ha: $\mu _d$ > 0

here difference is sons - fathers

b)

Normal Q-Q Plot 寸 1.5 1.0 0.5 0.0 0.5 1.0 Theoretical Quantiles

data seem to lie on a straight line

hence normal distribution assumption is valid

c)

6 4 2 2

There is no outliers

d)

p-value = 0.3508

e)

p-value = 0.3508 > 0.1 (alpha)

hence we fail to reject the null hypothesis

we conclude that there is not sufficient evidence that sons are taller than their fathers