Respuesta :
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 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, [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) = 0.90
P( [tex]-1.676 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]1.676 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.90
P( [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] ) = 0.90
90% 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].
