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

Answer:

90% confidence interval for the mean time per week spent listening to the radio is  $11.5 \pm 1.676 \times \frac{9.2}{\sqrt{51} }$ .

Step-by-step 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. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$  ~ $t_n_-_1$

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 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.676) = 0.90

P( $-1.676 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.676 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.676 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.676 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.676 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.676 \times {\frac{s}{\sqrt{n} } }$ ]

                  = [ $11.5-1.676 \times {\frac{9.2}{\sqrt{51} } }$ , $11.5+1.676 \times {\frac{9.2}{\sqrt{51} } }$ ]

                  = [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].