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)
C. (0,0) which is the origin
The answer is D)60.
You do L×W×H which is 4×3×6
that equals 60.
Hope this helps
She will need 8 pints
2 cups in 1 pint so
16/2=8
