Question

At a particular academically challenging high school, the average gpa of a high school senior is known to be normally distributed. after a sample of 20 seniors is taken, the average gpa is found to be 2.71 and the variance is determined to be 0.25. find a 90% confidence interval for the population mean gpa.

Answer 1

Answer:

The 90% confidence interval for the population mean gpa is [2.526,2.894].

Step-by-step explanation:

The confidence interval for population mean is

$\mu\pm z^{*}\cdot \frac{\sigma}{\sqrt{n}}$

Where, μ is population mean, σ is standard deviation, z* is the value of z-score and n is number of samples.

The z-score value for 90% confidence interval is 1.645.

The average gpa is found to be 2.71, so μ=2.71. The variance is 0.25, it means

$\sigma^2=0.25$

$\sigma=0.5$

The 90% confidence interval for population mean is

$2.71\pm 1.645\cdot \frac{0.5}{\sqrt{20}}$

$[2.71-1.645\cdot \frac{0.5}{\sqrt{20}},2.71+1.645\cdot \frac{0.5}{\sqrt{20}}]$

$[2.526,2.894]$

Therefore the 90% confidence interval for the population mean gpa is [2.526,2.894].

Answer 2

Solution: The 90% confidence interval for the population mean is:

$\bar{x} \pm t_{\frac{0.1}{2}} \frac{s}{\sqrt{n}}$

Where:

$\bar{x}=2.71$

$s=\sqrt{0.25} =0.5$

$t_{\frac{0.1}{2}}=1.729$ is the critical value at 0.1 significance level for df = n-1 =20-1=19

$2.71 \pm 1.729 \frac{0.5}{\sqrt{20}}$

$2.71 \pm0.19$

$\left( 2.71-0.19,2.71+0.19\right)$

$\left(2.52,2.90 \right)$

Therefore, the 90% confidence interval for the population mean is $\left(2.52,2.90 \right)$