Construct a 90% confidence interval for the population mean, $\mu.$ Assume the population has a normal distribution. A sample of 15 randomly selected students has a grade point average of 2.86 with a standard
Deviation of 0.78.
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 when the population has a normal distribution and the sample size is small (n < 30), we use the t-distribution. The formula for the confidence interval is given by $\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $s$ is the sample standard deviation, $n$ is the sample size, and $t_{\alpha/2}$ is the t-score that corresponds to the desired confidence level with $n-1$ degrees of freedom.Given: $\bar{x} = 2.86$, $s = 0.78$, $n = 15$, and we are constructing a 90% confidence interval.Degrees of freedom, $df = n - 1 = 15 - 1 = 14$.For a 90% confidence interval and 14 degrees of freedom, the t-score (from t-distribution tables or calculators) is approximately 1.761.The margin of error (ME) is calculated as: $ME = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} = 1.761 \cdot \frac{0.78}{\sqrt{15}} \approx 0.35$.Thus, the confidence interval is: $\bar{x} \pm ME = 2.86 \pm 0.35 = (2.51, 3.21)$.

Construct a 90% Confidence Interval for the Population Mean μ.\mu.μ. Assume the Population Has a Normal Distribution

Construct a 90% Confidence Interval for the Population Mean $\mu.$ Assume the Population Has a Normal Distribution