A professor was curious about her students' grade point averages (GPAs). She took a random sample of 15 students and found a mean GPA of 3.01 with a standard deviation of 0.534. Which of the following formulas gives a 99% confidence interval for the mean GPA of the professor's students?
A) $ 3.01 \pm 2.576 \frac{0.534}{\sqrt{15}} $

B) $ 3.01 \pm 2.977 \frac{0.534}{\sqrt{14}} $

C) $ 3.01 \pm 2.947 \frac{0.534}{\sqrt{15}} $

D) $ 3.01 \pm 2.977 \frac{0.534}{\sqrt{15}} $

E) $ 3.01 \pm 2.947 \frac{0.534}{\sqrt{14}} $

Question

A professor was curious about her students' grade point averages (GPAs). She took a random sample of 15 students and found a mean GPA of 3.01 with a standard deviation of 0.534. Which of the following formulas gives a 99% confidence interval for the mean GPA of the professor's students? 
A) $ 3.01 \pm 2.576 \frac{0.534}{\sqrt{15}} $

B) $ 3.01 \pm 2.977 \frac{0.534}{\sqrt{14}} $

C) $ 3.01 \pm 2.947 \frac{0.534}{\sqrt{15}} $

D) $ 3.01 \pm 2.977 \frac{0.534}{\sqrt{15}} $

E) $ 3.01 \pm 2.947 \frac{0.534}{\sqrt{14}} $

Quizplus · Accepted Answer

We can use the formula: $\bar{x} \pm t_{\frac{\alpha}{2},n-1} \frac{s}{\sqrt{n}}$ to find the 99% confidence interval for the mean GPA of the professor's students.

Substituting the given values, we get: $3.01 \pm t_{\frac{0.01}{2},14} \frac{0.534}{\sqrt{15}}$

Using a t-table with 14 degrees of freedom, the t-value for a 99% confidence interval and a two-tailed test is 2.977.

Substituting this value, we get: $3.01 \pm 2.977 \frac{0.534}{\sqrt{15}}$

Therefore, the answer is D.

A Professor Was Curious About Her Students' Grade Point Averages 3.01±2.5760.53415 3.01 \pm 2.576 \frac{0.534}{\sqrt{15}} 3.01±2.57615​0.534​

A Professor Was Curious About Her Students' Grade Point Averages $3.01 \pm 2.576 \frac{0.534}{\sqrt{15}}$