Answer:
Explanation:
Let us calculate std dev and std error of two samples
Sample N Mean Std dev Std error
1 65 52.8 9.9 1.22279
2 111 54.1 11.09 1.0526
Assuming equal variances
df =111+65-2=174
Pooled std deviation = combined std deviation = 10.6677
Pooled std error = 10.6677/sqrt 174 = 0.808
difference in means =52.8-54.1 = -1.3
Margin of error = 0.5923
( t critical 1.6660)
Confidence interval
=(-4.4923, 1.8923)
Since 0 lies in this interval we can accept null hypothesis that 10 year old boys and girls have the same average height