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

$n=20$

$$mean=\bar{x}=\dfrac{1}{20}(8+ 22.7+ 14.5+ 9+9$$

$$+ 3.5+ 8+ 11+ 7.5+ 18+20+ 7.5 +9$$

$$+ 10.5 +15+ 19+ 9+ 8.5+ 14 +20)=12.185$$

$$s^2=\dfrac{1}{20-1}((8-12.185)^2+(22.7-12.185)^2$$

$$+(14.5-12.185)^2+(9-12.185)^2$$

$$+(9-12.185)^2+(3.5-12.185)^2$$

$$+(8-12.185)^2+(11-12.185)^2$$

$$+(7.5-12.185)^2+(18-12.185)^2$$

$$+(20-12.185)^2+(7.5-12.185)^2$$

$$+(9-12.185)^2+(10.5-12.185)^2$$

$$+(15-12.185)^2+(19-12.185)^2$$

$$+(9-12.185)^2+(8.5-12.185)^2$$

$$+(14-12.185)^2+(20-12.185)^2)$$

$$=28.542395$$

$$s=\sqrt{28.542395}\approx5.3425083$$

The critical value for $\alpha = 0.01$ and $df = n-1 = 19$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} =2.860935.$ The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(12.185-2.860935\times\dfrac{5.3425083}{\sqrt{20}},$$

$$12.185+2.860935\times\dfrac{5.3425083}{\sqrt{20}}$$

$$=(8.767,15.603)$$

Therefore, based on the data provided, the 99% confidence interval for the population mean is $8.767 < \mu < 15.603,$ which indicates that we are 99% 

confident that the true population mean $\mu$ is contained by the interval $(8.767, 15.603).$