In: Statistics and Probability
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:
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.
f) for computing IQR for the Father height follow the steps as shown in images below.
Finally click ok.
g) To compute mean and standard deviation follow the steps as shown below.
Finally click ok.
h ) To test normality follow the steps given below.
Finally click ok.
After following the steps you will get normal probability plot shown below.
H0 : Child's height is normally distributed.
H0 : Child's height is not normally distributed.
From the plot we can clearly see that most of the points lie on the straight line and are closer to straight line. So over variable child height is normal . Also we can see that the p-value is 0.805 > 0.05 so we accept the null hypothesis that our variable is normal.
i) To find 95% C.I. use the steps shown below.
finally click ok.
j) to test is the average child height is different from 140 , follow the steps below.
Here is final output from ( f ) to ( j ):
H0 : Average of child's height is 140 cm.
H1 : Average of child's height is different than 140 cm.
j ) interpretation : Since the p-value is 0.000 < 0.01 we reject the null hypothesis and coclude that the average height of child is different than 140 cm at 1% significance level.