Construct a 90% confidence interval for the population mean, μ. Assume the population has a normaldistribution. A sample of 15 randomly selected math majors has a grade point average of 2.86 with a standarddeviation of 0.78. Round to the nearest hundredth.
A) (2.51, 3.21)
B) (2.41, 3.42)
C) (2.37, 3.56)
D) (2.28, 3.66)

Question

Quizplus · Accepted Answer

To construct a 90% confidence interval for the population mean with a sample size of 15, a mean of 2.86, and a standard deviation of 0.78, we use the formula: $\bar{x} \pm t_{\alpha/2} \times \frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $t_{\alpha/2}$ is the t-score corresponding to the confidence level and degrees of freedom (n-1), $s$ is the sample standard deviation, and $n$ is the sample size. For a 90% confidence interval and 14 degrees of freedom, the t-score is approximately 1.761. Plugging in the values: $2.86 \pm 1.761 \times \frac{0.78}{\sqrt{15}}$ gives an interval of (2.51, 3.21) when rounded to the nearest hundredth.

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