Question

A school district is considering moving the start of the school day at Groveland High from 8:00 a.m. to 9:00 a.m. to allow students to get more sleep at night. They want to estimate how many more hours of sleep teens get if their school starts at 9:00 a.m. compared to teens whose schools start at 8:00 a.m. The district takes random samples of 50 of students from Groveland High and 39 students from Phillips High School, which starts at 9:00 a.m., and surveys them about their sleep habits. The sample mean number of hours of sleep per student at Groveland is 7.1 hours, and the standard deviation is 1.7 hours. At Phillips, the sample mean is 8.3 hours of sleep, with a standard deviation of 1.9 hours. What is the 90% confidence interval for the difference in mean hours of sleep for students at the two high schools

Answer 1

Answer:

The answer is "$\bold{(7.1-8.3) \pm 1.665 \sqrt{\frac{(1.7)^2}{50}+\frac{(1.9)^2}{39}} }$"

Step-by-step explanation:

Given values:

$\bar{x_1}=7.1\\\\\bar{x_2}=8.3\\\\s_1=1.7\\\\s_2=1.9\\\\n_1=50\\\\n_2=39$

Using formula:

$\to (\bar{x_1} -\bar{x_2}) \pm t \sqrt{\frac{(s_1)^2}{n_1}+\frac{(s_1)^2}{n_2}}$

Put the values in the above formula:

$\to (7.1-8.3) \pm 1.665 \sqrt{\frac{(1.7)^2}{50}+\frac{(1.9)^2}{39}}$