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)

5 0

3 years ago

If A = (0,0) and B = (2,5), what is the approximate length of AB?

Bad White [126]

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{0}~,~\stackrel{y_1}{0})\qquad B(\stackrel{x_2}{2}~,~\stackrel{y_2}{5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AB=\sqrt{(2-0)^2+(5-0)^2}\implies AB=\sqrt{2^2+5^2} \\\\\\ AB=\sqrt{29}\implies AB\approx 5.39$

8 0

3 years ago

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