Respuesta :
Answer:
80% confidence interval of the true average number of hours of TV watched per week is [8.28 hours, 11.02 hours].
Step-by-step explanation:
We are given that a college student is interested in investigating the TV-watching habits of her classmates and surveys 20 people on the number of hours they watch per week. The results are provided below;
Hours of TV per week (X): 6, 14, 13, 6, 16, 10, 19, 4, 5, 5, 18, 8, 7, 14, 8, 8, 9, 12, 6, 5.
Firstly, the Pivotal quantity for 80% confidence interval for the true average 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 mean number of hours of TV watched per week = [tex]\frac{\sum X}{n}[/tex] = 9.65
s = sample standard deviation = [tex]\sqrt{\frac{\sum (X -\bar X)^{2} }{n-1} }[/tex] = 4.61
n = sample of people = 20
[tex]\mu[/tex] = true average number of hours of TV watched per week
Here for constructing 80% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.
So, 80% confidence interval for the true average, [tex]\mu[/tex] is ;
P(-1.33 < [tex]t_1_9[/tex] < 1.33) = 0.80 {As the critical value of t at 19 degrees of
freedom are -1.33 & 1.33 with P = 10%}
P(-1.33 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 1.33) = 0.80
P( [tex]-1.33 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]1.33 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.80
P( [tex]\bar X-1.33 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+1.33 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.80
80% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-1.33 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+1.33 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]9.65-1.33 \times {\frac{4.61}{\sqrt{20} } }[/tex] , [tex]9.65+1.33 \times {\frac{4.61}{\sqrt{20} } }[/tex] ]
= [8.28 hours, 11.02 hours]
Therefore, 80% confidence interval of the true average number of hours of TV watched per week is [8.28 hours, 11.02 hours].
