Question

g what is the 95% confidence interval a researcher wishes to compare the average amount of time spent in extracurricular

Answer 1

Complete question is;

A researcher wished to compare the average amount of time spent in extracurricular activities by high school students in a suburban school district with that in a school district of a large city. The researcher obtained a simple random sample of 60 high school students in a large suburban school district and found the mean time spent in extracurricular activities per week to be six hours with a standard deviation of three hours. The researcher also obtained an independent simple random sample of 40 high school students in a large city school district and found the mean time spent in extracurricular activities per week to be four hours with a standard deviation of two hours. Let x¯1 and x¯2 represent the mean amount of time spent in extracurricular activities per week by the populations of all high school students in the suburban and city school districts, respectively. Assume two-sample t procedures are safe to use?

what is the 95% confidence interval a researcher wishes to compare the average amount of time spent in extracurricular?

Answer:

CI = (0.755, 3.245)

Step-by-step explanation:

For SRS of 60;

Mean: x1¯ = 6

Standard deviation: s1 = 3

For SRS of 40;

Mean: x2¯ = 4

Standard deviation; s2 = 2

Critical value for the confidence interval of 95% is: t = 1.96

Formula for the CI is;

CI = (x¯1 - x¯2) ± t√[(s1²/n1) + ((s2)²/n1)]

Plugging in the relevant values, we have:

CI = (6 - 4) ± 1.96√[(3²/60) + ((4)²/40)]

CI = 2 ± 1.96√[(3²/60) + ((4)²/40)]

CI = 2 ± 1.96√0.55

CI = 2 ± 1.245

CI = [(2 - 1.245), (2 + 1.245)]

CI = (0.755, 3.245)