Question

In a random sample of 90 students in first grade, the mean hours of sleep per night was 9.2 with a standard deviation of 0.37 in a random sample of 70 students in the 12th grade, the mean hours of sleep per night was 7.3 with a standard deviation of 0.58. Construct a 99% confidence interval for the difference between the mean hours of sleep.

A. (1.82 h, 1.98 h)
B. (1.79 h, 2.01 h)
C. (1.74 h, 2.06 h)
D. (1.69 h, 2.11 h)

Answer 1

Option C is right

Explanation:

Given are two samples

I group has sample size 90, mean = 9.2 and std dev =0,.37

II group has sample size 70, mean = 7.3 and std dev = 0.58

we find that mean difference i.e. difference between mean hours of sleep between two groups

= 9.2-7.3 = 1.9

Std error for difference = 0.0620

Since sample size is sufficiently large we can use Z critical for 99%

i.e. 2.58

Margin of error =±2.58(0.062)

=0.16

Hence confidence interval 99% for difference in hours of sleep

= (1.9-0.16,1.9+0.16)

=(1.74, 2.06)