Question

1. In a recent study of 39 ninth-grade students, the mean number of hours per week that they played video games was 86.6. The standard deviation of the sample was 3.8. a. Find the best point estimate of the population mean. b. Find the 90%confidence interval of the mean of the time playing video games. c. Find the 96% confidence interval of the mean of the time playing video games. d. Find the 98% confidence interval of the mean of the time playing video games. e. Find the 99% confidence interval of the mean of the time playing video games.

Answer 1

given that

n=39

sample mean =m=86.6

sample SD=3.8

a)

sample mean is point estimate of population mean so

Point estimate =m=86.6

b)

90% confidence interval is given by

$m \pm t_{38,0.05}*\frac{S}{\sqrt{n}}=86.6 \pm 1.69*\frac{3.8}{\sqrt{39}}=86.6 \pm 1.03$

So interval is (85.57,87.63)

c)

96% confidence interval is given by

$m \pm t_{38,0.02}*\frac{S}{\sqrt{n}}=86.6 \pm 2.13*\frac{3.8}{\sqrt{39}}=86.6 \pm 1.3$

so interval is (85.3,87.9)

d)

98% confidence interval is given by

$m \pm t_{38,0.01}*\frac{S}{\sqrt{n}}=86.6 \pm 2.43*\frac{3.8}{\sqrt{39}}=86.6 \pm 1.48$

so interval is (85.12,88.08)

e)

99% confidence interval is given by

$m \pm t_{38,0.005}*\frac{S}{\sqrt{n}}=86.6 \pm 2.71*\frac{3.8}{\sqrt{39}}=86.6 \pm 1.65$

so interval is (84.95,88.25)