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 watched
Television was 19.6 with a standard deviation of 5.8 hours.
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 sample mean ($\bar{x}$) is 19.6, and the sample standard deviation (s) is 5.8, we use the formula: $\bar{x} \pm t_{\alpha/2} \times \frac{s}{\sqrt{n}}$, where $t_{\alpha/2}$ is the t-score for a 90% confidence level with $n-1$ degrees of freedom. For 21 degrees of freedom, the t-score for a 90% confidence interval is approximately 1.721. Plugging in the values, we get: $19.6 \pm 1.721 \times \frac{5.8}{\sqrt{22}}$, which calculates to the interval (17.47, 21.73).

Construct a 90% Confidence Interval for the Population Mean μ.\mu _ { . }μ.​

Construct a 90% Confidence Interval for the Population Mean $\mu _ { . }$