Construct a 90% confidence interval for the population mean, μ. Assume the population has a normaldistribution. In a recent study of 22 eighth graders, the mean number of hours per week that they played videogames was 19.6 with a standard deviation of 5.8 hours. Round to the nearest hundredth.
A) (17.47, 21.73)
B) (18.63, 20.89)
C) (5.87, 7.98)
D) (19.62, 23.12)

Question

Quizplus · Accepted Answer

To construct a 90% confidence interval for the population mean when the population has a normal distribution and the sample size (n) is 22, the mean ($\bar{x}$) is 19.6, and the standard deviation (s) is 5.8, we use the formula: $\bar{x} \pm t_{\alpha/2} \times \frac{s}{\sqrt{n}}$. For a 90% confidence interval and df = n-1 = 21, the t-value from the t-distribution table is approximately 1.721. Plugging in the values: $19.6 \pm 1.721 \times \frac{5.8}{\sqrt{22}}$, we get the interval (17.47, 21.73).

Construct a 90% Confidence Interval for the Population Mean, μ