Provide an appropriate response.
-The grade point averages for 10 randomly selected junior college students are listed below. Assume the grade point averages are normally distributed. Find a $98 \%$ confidence interval for the true mean.
$\begin{array} { l l l l l l l l l l } 2.0 & 3.2 & 1.8 & 2.9 & 0.9 & 4.0 & 3.3 & 2.9 & 3.6 & 0.8 \end{array}$
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, first calculate the sample mean ($\bar{x}$) and the sample standard deviation (s). For the given data:- $\bar{x} = \frac{2.0 + 3.2 + 1.8 + 2.9 + 0.9 + 4.0 + 3.3 + 2.9 + 3.6 + 0.8}{10} = 2.54$- $s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$, which calculates to approximately 1.16.Since the sample size is small (n=10) and we assume a normal distribution, we use the t-distribution for the confidence interval. The degrees of freedom (df) is $n-1 = 9$. For a 98% confidence interval and 9 degrees of freedom, the t-value from t-distribution tables is approximately 2.821.The confidence interval formula is $\bar{x} \pm t \times \frac{s}{\sqrt{n}}$, which calculates to:- $2.54 \pm 2.821 \times \frac{1.16}{\sqrt{10}}$- $2.54 \pm 1.02$- $(1.52, 3.56)$The correct answer, considering rounding and the values provided in the options, is closest to option A) $(1.55, 3.53)$.

Provide an Appropriate Response 98%98 \%98% Confidence Interval for the True Mean

Provide an Appropriate Response $98 \%$ Confidence Interval for the True Mean