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

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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

Answers

Answer:90% confidence interval for the mean time per week spent listening to the radio is  [tex]11.5 \pm 1.676 \times \frac{9.2}{\sqrt{51} }[/tex] .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. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex]  ~ [tex]t_n_-_1[/tex]where, [tex]\bar X[/tex] = sample average amount of time spent listening to music = 11.5              s = sample standard deviation = 9.2 hours              n = sample of students = 51              [tex]\mu[/tex] = population mean per week spent listening to the radioHere 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, [tex]\mu[/tex] is ;P(-1.676 < [tex]t_5_0[/tex] < 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 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 1.676) = 0.90P( [tex]-1.676 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]1.676 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.90P( [tex]\bar X-1.676 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+1.676 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.9090% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-1.676 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+1.676 \times {\frac{s}{\sqrt{n} } }[/tex] ]                   = [ [tex]11.5-1.676 \times {\frac{9.2}{\sqrt{51} } }[/tex] , [tex]11.5+1.676 \times {\frac{9.2}{\sqrt{51} } }[/tex] ]                   = [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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