Question

Data.

Child_Height	Fheight	Gender
145.0	136.5	1
125.5	121.0	1
125.5	119.5	1
133.5	128.0	1
126.0	135.5	1
142.5	131.0	1
127.5	134.5	1
133.5	132.5	1
130.5	133.5	1
131.0	126.0	1
136.0	124.0	1
132.5	146.5	1
132.0	134.5	1
121.5	142.0	1
131.0	119.0	1
136.0	133.5	1
134.5	118.5	1
138.5	134.0	2
128.5	132.0	2
140.5	131.5	2
134.5	141.5	2
132.5	136.5	2
126.0	136.5	2
143.0	136.0	2
130.0	143.0	2
123.5	134.0	2
143.5	146.0	2
140.0	137.0	2
132.5	136.5	2
138.5	144.0	2
135.0	139.0	2
135.0	136.0	2

In the given dataset (0. STAT101_Project.xlsx) or (STAT101_Project.MPJ), there are three variables; Child height (Child_Height ), Father height (Fheight) and Gender for the child. Please do the following using the Minitab software and interpret all the results

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

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

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

10. 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

7)MINITAB OUTPUT:

Descriptive Statistics: child height

Variable Mean StDev
child height 133.30 6.08

8)

large p-value indicates that the assumption of normality is appropriate.

9)95% CI=(131.183, 135.410)

10)MINITAB OUTPUT:

One-Sample Z: child height

Test of mu = 140 vs not = 140
The assumed standard deviation = 6.1

Variable N Mean StDev SE Mean 95% CI Z P
child height 32 133.297 6.083 1.078 (131.183, 135.410) -6.22 0.000

p-value<0.10 hence we reject the null and strongly conclude that there is sufficient evidence to support the claim that the mean is significantly differ from 140cm.