A random sample of 150 students has a grade point average with a mean of 2.86. Assume the population standard deviation is 0.78. Construct the confidence interval for the population mean, $\mu ,$ if c = 0.98.
A)(2.71, 3.01)
B)(2.51, 3.53)
C)(2.43, 3.79)
D)(2.31, 3.88)

Question

Quizplus · Accepted Answer

To construct a 98% confidence interval for the population mean, we use the formula for the confidence interval: $\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $Z$ is the Z-score corresponding to the confidence level, $\sigma$ is the population standard deviation, and $n$ is the sample size. Given $\bar{x} = 2.86$, $\sigma = 0.78$, and $n = 150$, and knowing that the Z-score for a 98% confidence level is approximately 2.33 (from Z-tables or standard normal distribution tables), we calculate the margin of error as $2.33 \times \frac{0.78}{\sqrt{150}}$, which gives approximately 0.15. Therefore, the confidence interval is $2.86 \pm 0.15$, which results in the interval (2.71, 3.01).

A Random Sample of 150 Students Has a Grade Point μ,\mu ,μ,

A Random Sample of 150 Students Has a Grade Point $\mu ,$