Question

A recent survey of 51 students reported that the average amount of time they spent listening to music was 11.5 hours per week, with a sample standard deviation of 9.2 hours. Which of the following is a 90% confidence interval for the mean time per week spent listening to the radio? (a) 11.5 +1.676 x 9.2 (b) 11.5 +1.282 x 9.2 (c) 11.5 +1.676 x 9.2/51(d) 11.5 +1.282 x 9.2/V51 (e) 11.4 +1.299 x 9.2/51

Answer 1

90% confidence interval for the mean time per week spent listening to the radio is
`11.5 \pm 1.676 * (9.2)/(√(51) )` .

Explanation:

We are given that a recent survey of 51 students reported that the average amount of time they spent listening to music was 11.5 hours per week, with a sample standard deviation of 9.2 hours.

Firstly, the pivotal quantity for 90% confidence interval for the population mean is given by;

P.Q. =
`(\bar X-\mu)/((s)/(√(n) ) )` ~
`t_n_-_1`

where,
`\bar X` = sample average amount of time spent listening to music = 11.5

s = sample standard deviation = 9.2 hours

n = sample of students = 51

`\mu` = population mean per week spent listening to the radio

Here for constructing 90% confidence interval we have used One-sample t test statistics as we know don't about population standard deviation.

So, 90% confidence interval for the population mean,
`\mu` is ;

P(-1.676 <
`t_5_0` < 1.676) = 0.90 {As the critical value of t at 50 degree of

freedom are -1.676 & 1.676 with P = 5%}

P(-1.676 <
`(\bar X-\mu)/((s)/(√(n) ) )` < 1.676) = 0.90

P(
`-1.676 * {(s)/(√(n) ) }` <
`{\bar X-\mu}` <
`1.676 * {(s)/(√(n) ) }` ) = 0.90

P(
`\bar X-1.676 * {(s)/(√(n) ) }` <
`\mu` <
`\bar X+1.676 * {(s)/(√(n) ) }` ) = 0.90

90% confidence interval for
`\mu` = [
`\bar X-1.676 * {(s)/(√(n) ) }` ,
`\bar X+1.676 * {(s)/(√(n) ) }` ]

= [
`11.5-1.676 * {(9.2)/(√(51) ) }` ,
`11.5+1.676 * {(9.2)/(√(51) ) }` ]

= [9.34 hours , 13.66 hours]

Therefore, 90% confidence interval for the mean time per week spent listening to the radio is [9.34 hours , 13.66 hours].