91影视

Let X is a GPA for a student who studied 3 hours per day.

Sample size n=307.

Mean $\stackrel{炉}{x}=3.11$, and

the standard deviation $s=0.66$.

Since,

The sample mean$\left(\stackrel{炉}{x}\right)$ is a single number estimator, which is the point estimator of the target parameter $\left(渭\right)$.

Therefore,

The point estimate for$渭$ is 3.11.

For the 98% confidence interval, the level of significance is 0.02.

$\begin{array}{c}\left(1鈭�\mathrm{伪}\right)=0.98\\ \mathrm{伪}=0.02\\ \frac{\mathrm{伪}}{2}=0.01\end{array}$

From table, the value of$z_{\frac{伪}{2}}$ is given below:

$\begin{array}{c}{z}_{\frac{伪}{2}}=z_{0.01}\\ =2.33\end{array}$

Let, the confidence interval as,

$\begin{array}{c}\stackrel{炉}{x}卤z_{\frac{伪}{2}}\left({蟽}_{\stackrel{炉}{x}}\right)=\stackrel{炉}{x}卤z_{\frac{伪}{2}}\left(\frac{s}{\sqrt{n}}\right)\\ =3.11卤2.33\left(\frac{0.66}{\sqrt{307}}\right)\\ =3.11卤0.0878\end{array}$

That is $\left(3.11鈭�0.0878,3.11+0.0878\right)=\left(3.0222,3.1978\right)$

Therefore, the 98% confidence interval for$渭$ is $\left(3.0222,3.1978\right)$.

The statement 鈥�98% of the time, the true mean GPA will fall in the interval computed in part b.鈥� is incorrect, because there is no probability concerned in the interval after the computation of the confidence interval. Thus, the correct statement is 鈥渇or 98% confidence, the true mean GPA lies between 3.0222 and 3.1978鈥�.

Here, the central limit theorem is applied since the sample size is 307(>30). Thus, it is not necessary to check whether the distribution of GPA is skewed or not skewed.