Question

To test the belief that sons are taller than their​ fathers, a student randomly selects 13 fathers who have adult male children. She records the height of both the father and son in inches and obtains the following data. Are sons taller than their​ fathers? Use the

alphaαequals=0.10

level of significance.​ Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers.

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Height of ​Father, Upper X Subscript iXi	Height of​Son, Upper Y Subscript iYi
73.2	78.3
67.8	71.3
69.1	71.7
67.3	69.1
72.8	74.0
71.5	72.1
71.5	71.6
67.1	66.5
72.6	71.3
73.3	71.5
71.9	69.3
73.3	69.7
68.5	63.6

Which conditions must be met by the sample for this​ test? Select all that apply.

A.

The sampling method results in an independent sample.

B.

The sampling method results in a dependent sample.

Your answer is correct.

C.

The sample size is no more than​ 5% of the population size.

Your answer is correct.

D.

The differences are normally distributed or the sample size is large.

Your answer is correct.

E.

The sample size must be large.

Let

d Subscript idiequals=Upper X Subscript iXiminus−Upper Y Subscript iYi.

Write the hypotheses for the test.

Upper H 0H0​:

mu Subscript d Baseline equals 0μd=0

Upper H 1H1​:

mu Subscript d Baseline less than 0μd<0

Calculate the test statistic.

t 0=?

p value= ?

​(Round to two decimal places as​ needed.)

Answer 1

Solution:

Here, we have to use the paired t test for the difference between population means.

The null and alternative hypotheses for this test are given as below:

Null hypothesis: H₀: µ_d = 0

Alternative hypothesis: H_a: µ_d < 0

We will use the level of significance or alpha value for this test as 0.05.

The test statistic formula for this test is given as below:

t = (Dbar - µ_d)/[S_d/sqrt(n)]

Calculation table:

X	Y	Di	(Di - DBar)^2
73.2	78.3	-5.1	25.93159763
67.8	71.3	-3.5	12.19621302
69.1	71.7	-2.6	6.720059172
67.3	69.1	-1.8	3.212366864
72.8	74	-1.2	1.421597633
71.5	72.1	-0.6	0.350828402
71.5	71.6	-0.1	0.00852071
67.1	66.5	0.6	0.369289941
72.6	71.3	1.3	1.710059172
73.3	71.5	1.8	3.267751479
71.9	69.3	2.6	6.800059172
73.3	69.7	3.6	13.01544379
68.5	63.6	4.9	24.08544379

From given data and above table, we have

Dbar = -0.0077

S_d = 2.8736

n = 13

df = n – 1 = 12

α = 0.05

t = (-0.0077 – 0) / [2.8736/sqrt(13)]

t = -0.0077 / 0.7970

t = -0.0097

P-value = 0.4962

(by using t-table)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is insufficient evidence to conclude that sons are taller than their fathers.