Construct a 90 confidence interval for the population mean m

Construct a 90% confidence interval for the population mean, mu. Assume the population has a normal distribution. In a recent study of 22 eighth graders, the mean number of hours per week that they played video games was 19.6 with a standard deviation of 5.8 hours. Round to the nearest hundredth.

Solution

sample size (n) = 22

mean number of hours per week that they played video games (Xbar) = 19.6

sd = 5.8

Here population standard deviation is unknown and sample size is small (<30) so we use here t-interval.

90% connfidence interval for population mean is,

Xbar - E < mu < Xbar + E

where mu is population mean.

E is the margin of error.

E = (tc*sd) / sqrt(n)

where tc is the critical value for t-distribution.

We can find this value by using EXCEL.

syntax :

=TINV(probability, d.f.)

c = confidence level = 0.90

a = 1 - c = 1 - 0.90 = 0.1

probability = a

d.f. = n - 1 = 22 - 1 = 21

tc = 1.7207

E = (1.7207*5.8) / sqrt(22)

E = 2.1278

lower limit = Xbar - E = 19.6 - 2.1278 = 17.47

upper limit = Xbar + E = 19.6 + 2.1278 = 21.73

Option d) is correct.

90% connfidence interval for population mean is (17.47, 21.73).