Answer to Question #199870 in Statistics and Probability for Delmundo

Question

From our previous exam in Statistics, sample scores from a normal population were taken for evaluation purposes of how the students responded the test. The scores are 38, 27, 34, 50, 33, 36, 21, 27, 22, 29, 27, 27, 22, 29, 31, 28 asssume that the score are normally distributed

Find the value of the point estimate for the population.

Compute for the margin of error at 99% confidence level.

Find 99% confidence interval for the population mean.

Interpret the results.

Answer 1

Point estimate of mean $\bar {X}=\frac{\sum{xi}}{n}=\frac{481}{16}=30.0625$

Margin of error $E=t_{\frac{\alpha}{2},n-1}(\frac{s}{\sqrt{n}})$

Level of significance is 0.01, did= 15, t= 2.9467, s = 7.196932, thus,

$E=2.9467(\frac{7.196932}{\sqrt{16}})=5.3018$

Confidence interval

$CI=\bar{X}\pm E$

$=30.0625\pm5.3018$

$=[24.761,35.364]$

$24.761<\mu<35.364$

We are 99% confident that the population mean lies between 24.761 and 34.364.