Question

according to the registrars office, the overall gpa of students at a college was normally distributed, but somehow the mean and variance were not available. a statistician collected a random sample of 36 students, and found their mean gpa was 2.88 and s.d. was 0.28. what is the estimated mean gpa of all the students? what is the 90% c.i. for the true mean?

Explanation:

Step1: Estimated mean GPA

The sample mean is used as the point - estimate for the population mean.
$\bar{x}=2.88$

Step2: Calculate the standard error

The formula for the standard error $SE=\frac{s}{\sqrt{n}}$, where $s = 0.28$ and $n = 36$.
$SE=\frac{0.28}{\sqrt{36}}=\frac{0.28}{6}\approx0.0467$

Step3: Find the critical value

For a 90% confidence interval, the significance level $\alpha=1 - 0.90=0.10$, and $\frac{\alpha}{2}=0.05$. The critical value $t_{\frac{\alpha}{2}}$ with $n - 1=36 - 1 = 35$ degrees of freedom (using a t - distribution since the population variance is unknown). From the t - table, $t_{0.05,35}\approx1.690$

Step4: Calculate the margin of error

The margin of error $E=t_{\frac{\alpha}{2}}\times SE$
$E = 1.690\times0.0467\approx0.079$

Step5: Calculate the confidence interval

The confidence interval is $\bar{x}\pm E$
Lower limit: $\bar{x}-E=2.88 - 0.079=2.801$
Upper limit: $\bar{x}+E=2.88+0.079 = 2.959$

Answer:

The estimated mean GPA of all students is $2.88$. The 90% confidence interval for the true mean is $(2.801,2.959)$