The grade point averages for 10 randomly selected high school students are listed below. Assume the grade point averages are normally distributed. $\begin{array} { l l l l l l l l l l }
2.0 & 3.2 & 1.8 & 2.9 & 0.9 & 4.0 & 3.3 & 2.9 & 3.6 & 0.8
\end{array}$ Find a 98% confidence interval for the true mean.
A)(1.55, 3.53)
B)(0.67, 1.81)
C)(2.12, 3.14)
D)(3.11, 4.35)

Question

The grade point averages for 10 randomly selected high school students are listed below. Assume the grade point averages are normally distributed. $\begin{array} { l l l l l l l l l l } 
2.0 & 3.2 & 1.8 & 2.9 & 0.9 & 4.0 & 3.3 & 2.9 & 3.6 & 0.8
\end{array}$ Find a 98% confidence interval for the true mean.
A)(1.55, 3.53)
B)(0.67, 1.81)
C)(2.12, 3.14)
D)(3.11, 4.35)

Quizplus · Accepted Answer

To find a 98% confidence interval for the true mean of a normally distributed dataset, we first calculate the sample mean ($\bar{x}$) and the sample standard deviation (s). For the given data:$\bar{x} = \frac{2.0 + 3.2 + 1.8 + 2.9 + 0.9 + 4.0 + 3.3 + 2.9 + 3.6 + 0.8}{10} = \frac{25.4}{10} = 2.54$$s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$, where $x_i$ are the data points, $\bar{x}$ is the sample mean, and $n$ is the number of data points.$s \approx 1.16$ (calculation of s is simplified here for brevity).For a 98% confidence interval and $n = 10$, degrees of freedom $df = n - 1 = 9$. The t-value from the t-distribution table for 98% confidence and 9 degrees of freedom is approximately 2.821.The formula for the confidence interval is $\bar{x} \pm t \cdot \frac{s}{\sqrt{n}}$.Plugging in the values: $2.54 \pm 2.821 \cdot \frac{1.16}{\sqrt{10}}$, we get the interval $(1.55, 3.53)$.Therefore, the correct choice is (A)(1.55, 3.53).

The Grade Point Averages for 10 Randomly Selected High School 2.03.21.82.90.94.03.32.93.60.8\begin{array} { l l l l l l l l l l } 2.0 & 3.2 & 1.8 & 2.9 & 0.9 & 4.0 & 3.3 & 2.9 & 3.6 & 0.8\end{array}2.0​3.2​1.8​2.9​0.9​4.0​3.3​2.9​3.6​0.8​

The Grade Point Averages for 10 Randomly Selected High School $\begin{array} { l l l l l l l l l l } 2.0 & 3.2 & 1.8 & 2.9 & 0.9 & 4.0 & 3.3 & 2.9 & 3.6 & 0.8\end{array}$