The grade point averages for 10 randomly selected junior college students are listed below. Assume the gradepoint averages are normally distributed. Find a 98% confidence interval for the true mean. Round to the nearesthundredth.2.0 3.2 1.8 2.9 0.9 4.0 3.3 2.9 3.6 0.8
A) (1.55, 3.53)
B) (0.67, 1.81)
C) (2.12, 3.14)
D) (3.11, 4.35)

Question

Quizplus · Accepted Answer

To find a 98% confidence interval for the true mean of the given data, first calculate the sample mean and the sample standard deviation. The sample mean ($\bar{x}$) is $2.44$ and the sample standard deviation (s) is $1.114$. With 10 samples, the degrees of freedom (df) is 9. For a 98% confidence interval and 9 degrees of freedom, the t-value from the t-distribution table is approximately $2.821$. The confidence interval is calculated using the formula: $\bar{x} \pm (t*\frac{s}{\sqrt{n}})$, which gives $(1.55, 3.53)$ when rounded to the nearest hundredth.

The Grade Point Averages for 10 Randomly Selected Junior College