Question

College dropout study. Refer to the American Economic Review (December 2008) study of college dropouts, Exercise 2.79 (p. 111). Recall that one factor thought to influence the college dropout decision was expected GPA for a student who studied 3 hours per day. In a representative sample of 307 college students who studied 3 hours per day, the mean GPA was$\overline{x}=\mathbf{3}.\mathbf{11}$ and the standard deviation was $\mathit{s}\mathbf{=}\mathbf{0}\mathbf{.66}$. Of interest is, the true mean GPA of all college students who study 3 hours per day.

a. Give a point estimate for $\mathit{\mu }$.

b. Give an interval estimate for $\mathit{\mu }$. Use a confidence coefficient of .98.

c. Comment on the validity of the following statement: “98% of the time, the true mean GPA will fall in the interval computed in part b.”

d. It is unlikely that the GPA values for college students who study 3 hours per day are normally distributed. In fact, it is likely that the GPA distribution is highly skewed. If so, what impact, if any, does this have on the validity of inferences derived from the confidence interval?

Short answer

1. The point estimate for$\mu$is 3.11.
2. Therefore, the 98% confidence interval for$\mu$is $\left(3.0222,3.1978\right)$.
3. Given statement is incorrect. Thecorrect statement is “For 98% confidence, the true mean GPA lies between 3.0222 and 3.1978.”
4. It is not necessary to check whether the distribution of GPA is skewed or not skewed.

Step-by-step solution

Given information

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

Sample size n=307.

Mean $\overline{x}=3.11$, and

the standard deviation $s=0.66$.

Calculating point estimate of μ

Since,

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

Therefore,

The point estimate for$\mu$ is 3.11.

Calculating confidence interval of  μ

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

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

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

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

Let, the confidence interval as,

$\begin{array}{c}\overline{x}\pm z_{\frac{\alpha }{2}}\left({\sigma }_{\overline{x}}\right)=\overline{x}\pm z_{\frac{\alpha }{2}}\left(\frac{s}{\sqrt{n}}\right)\\ =3.11\pm 2.33\left(\frac{0.66}{\sqrt{307}}\right)\\ =3.11\pm 0.0878\end{array}$

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

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

Commenting on the given statement

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

Commenting on the given statement 

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.