Answer:
(-3.0486, -0.2848)
Step-by-step explanation:
Let the number of hours per day spent using electronic media from 1999 be the first population and the number of hours per day spent using electronic media from 1999 the second population.
We have small sample sizes [tex]n_{1} = 15[/tex] and
[tex]n_{2} = 15[/tex].
[tex]\bar{x}_{1} = 5.9333[/tex] and [tex]\bar{x}_{2} = 7.6[/tex]; [tex]s_{1} = 1.0998[/tex] and [tex]s_{2} = 1.5946[/tex].
The pooled estimate is given by
[tex]s_{p}^{2} = \frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(15-1)(1.0998)^{2}+(15-1)(1.5946)^{2}}{15+15-2} = 1.8762[/tex]
The 99% confidence interval for the true mean difference between the mean number of hours per day spent using electronic media in 2009 and 1999 is given by
[tex](\bar{x}_{1}-\bar{x}_{2})\pm t_{0.01/2}s_{p}\sqrt{\frac{1}{15}+\frac{1}{15}}[/tex], i.e.,
[tex](5.9333-7.6)\pm t_{0.005}1.3697\sqrt{\frac{1}{15}+\frac{1}{15}}[/tex]
where [tex]t_{0.005}[/tex] is the 0.5th quantile of the t distribution with (15+15-2) = 28 degrees of freedom. So
[tex]-1.6667\pm(-2.7633)(1.3697)(0.3651)[/tex], i.e.,
(-3.0486, -0.2848)
######
Using R
time1999 <- c(4, 5, 7, 7, 5, 7, 5, 6, 5, 6, 7, 8, 5, 6, 6)
time2009 <- c(5, 9, 5, 8, 7, 6, 7, 9, 7, 9, 6, 9, 10, 9, 8)
n1 <- length(time1999)
n2 <- length(time2009)
(mean(time1999)-mean(time2009))+qt(0.005, df = 28)*sqrt(((n1-1)*var(time1999)+(n2-1)*var(time2009))/(n1+n2-2))*sqrt(1/n1+1/n2)
(mean(time1999)-mean(time2009))-qt(0.005, df = 28)*sqrt(((n1-1)*var(time1999)+(n2-1)*var(time2009))/(n1+n2-2))*sqrt(1/n1+1/n2)
