The article "plugged in, but tuned out" (usa today, january 20, 2010) summarized data from two surveys of randomly selected kids ages 8 to 18. one survey was conducted in 1999 and the other was conducted in 2009. data on the number of hours per day spent using electronic media, consistent with summary quantities given in the article, are below. 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)find the 99% confidence interval for the difference between the mean number of hours per day spent using electronic media in 2009 and 1999. show all steps of the confidence interval. you may use the formula or r for calculations.

(-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 and

.

and ; and .  

The pooled estimate is given by  

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  

, i.e.,

where is the 0.5th quantile of the t distribution with (15+15-2) = 28 degrees of freedom. So

, 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)

pretty sure 750 each month!

step-by-step explanation: