Question

Child height (cm)	Father height (cm)	Gender
145.0	136.5	M
125.5	121.0	M
125.5	119.5	M
133.5	128.0	M
126.0	135.5	M
142.5	131.0	M
127.5	134.5	M
133.5	132.5	M
130.5	133.5	M
131.0	126.0	M
136.0	124.0	M
132.5	146.5	M
132.0	134.5	M
121.5	142.0	M
131.0	119.0	M
136.0	133.5	M
134.5	118.5	M
138.5	134.0	F
128.5	132.0	F
140.5	131.5	F
134.5	141.5	F
132.5	136.5	F
126.0	136.5	F
143.0	136.0	F
130.0	143.0	F
123.5	134.0	F
143.5	146.0	F
140.0	137.0	F
132.5	136.5	F
138.5	144.0	F
135.0	139.0	F
135.0	136.0	F

using the Minitab software:

e. Construct the Box plot and Stem-Leave for the child’s height.

f. Compute the upper and lower bounds of the IQR for the Father height.

g. Compute the mean and standard deviation for the child’s height.

h. Test the normality for the child’s height.

i. Estimate a 95% confidence interval for the child’s height. Assume the population standard deviation is 6.1. Interpret the results.

j. Do the dataset provide sufficient evidence that the average of the child’s height is different from 140 cm? Using 1% significance level.

Answer 1

Que.e

Box plot:

Go to graph ---> Boxplot ---> Select 'Simple' and the select child height as graph variable.

Stem and leaf plot:

Go to graph ---> Stem and leaf plot---> select child height as graph variable.

Stem-and-leaf of Child height (cm) N = 32
Leaf Unit = 1.0
N* = 13

1 12 1
2 12 3
4 12 55
7 12 667
8 12 8
12 13 0011
(6) 13 222233
14 13 4455
10 13 66
8 13 88
6 14 00
4 14 233
1 14 5

Que.f

Select 'Stat' ---> 'Basic statistics' ---> 'Descriptive Statistics' ---> In 'Statistic' option select Q1, Q3 and IQR

output:

Descriptive Statistics: Father height (cm)

Variable N N* Minimum Q1 Median Q3 Maximum IQR
Father height (cm) 32 0 118.50 131.13 134.50 136.88 146.50 5.75

Thus

Lower bound = Q1 - 1.5 IQR = 131.13 - 1.5 * 5.75 = 122.505

Upper bound = Q3 + 1.5 IQR = 136.88 + 1.5 * 5.75 = 145.505

Que.g

Select 'Stat' ---> 'Basic statistics' ---> 'Descriptive Statistics'

Descriptive Statistics: Child height (cm)

Variable Mean StDev
Child height (cm) 133.30 6.08

Mean = 133.30

Standard deviation = 6.08

Que.8

Select 'Graph' ---> 'Probability plot' ---> Select variable child's height

Since p-value is greater than 0.05, we conclude that child's height are normally distributed.

Que.i

Select 'Stat' ---> 'Basic Statistic' ---> '1 Sample z test'

One-Sample Z: Child height (cm)

The assumed standard deviation = 6.1

Variable N Mean StDev SE Mean 95% CI
Child height (cm) 32 133.30 6.08 1.08 (131.18, 135.41)

Que. j

Select 'Stat' ---> 'Basic Statistic' ---> '1 Sample z test' ---> Refer following screen shot and enter information accordingly.

Output:

One-Sample Z: Child height (cm)

Test of μ = 140 vs ≠ 140
The assumed standard deviation = 6.1

Variable N Mean StDev SE Mean 99% CI Z P
Child height (cm) 32 133.30 6.08 1.08 (130.52, 136.07) -6.22 0.000

Since p-value is less than 0.01, we reject null hypothesis and conclude that the average of the child’s height is different from 140 cm.