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

90% confidence interval for the mean time per week spent listening to the radio is   .

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. =  ~

where, = sample average amount of time spent listening to music = 11.5

             s = sample standard deviation = 9.2 hours

             n = sample of students = 51

              = 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, is ;

P(-1.676 < < 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 < < 1.676) = 0.90

P( < < ) = 0.90

P( < < ) = 0.90

90% confidence interval for = [ , ]

                  = [ , ]

                  = [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].

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step-by-step explanation: