Answer to Question #327832 in Statistics and Probability for Vithu

Question

A sample of students from an introductory psychology class were polled regarding the

number of hours they spent studying for the last exam. The data was used to make

inferences regarding the other students taking the course. There data are below:

8, 22.7, 14.5, 9, 9, 3.5, 8, 11, 7.5, 18,

20, 7.5, 9, 10.5,15, 19, 9, 8.5, 14,20

Compute a 99% confidence interval.

Answer 1

The formula to calculate a confidence interval for a population mean is as follows:

$CI=\bar{x}\pm z\cdot\cfrac{s}{\sqrt{n}},$

$CI=\bar{x}\pm z\cdot\cfrac{s}{\sqrt{n}},$

where:

• $\bar{x}-$ sample mean
• z- the chosen z-value, for a 99% confidence interval z = 2.576
• $s-$ sample standard deviation
• n =20:  sample size.

The sample mean:

$\bar{x}=(8+ 22.7+ 14.5+ 9+ 9+ 3.5+ 8+ \\ +11+ 7.5+ 18+20+ 7.5+ 9+ 10.5+\\ +15+ 19+ 9+ 8.5+ 14+20)/ 20=12.19.$

The sample variance:

$s_x^2=\cfrac{\sum(x_i-\mu)^2}{n-1},$

$s^2=((8-12.19)^2+(22.7-12.19)^2+(14.5-12.19)^2+...\\ +...(14-12.19)^2+(20-12.19)^2)/19=28.54.$

The sample standard deviation:

$s=\sqrt{28.54}=5.34.$

So,

$CI=12.19 \pm 2.576\cdot\cfrac{5.34}{\sqrt{20}}=\\= 12.19\pm3.08=(9.11, 15.27).$