Respuesta :
Answer:
With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the 90%.
Step-by-step explanation:
We are given that a random sample of 60 home theater systems has a mean price of$131.00. Assume the population standard deviation is$18.80.
- Firstly, the pivotal quantity for 90% confidence interval for the population mean is given by;
P.Q. = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\bar X[/tex] = sample mean price = $131
[tex]\sigma[/tex] = population standard deviation = $18.80
n = sample of home theater = 60
[tex]\mu[/tex] = population mean
Here for constructing 90% confidence interval we have used One-sample z test statistics as we know about the population standard deviation.
So, 90% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level
of significance are -1.645 & 1.645}
P(-1.645 < [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < 1.645) = 0.90
P( [tex]-1.645 \times {\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]1.645 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ) = 0.90
P( [tex]\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ) = 0.90
90% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }[/tex] , [tex]\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ]
= [[tex]131-1.645 \times {\frac{18.8}{\sqrt{60} } }[/tex] , [tex]131+1.645 \times {\frac{18.8}{\sqrt{60} } }[/tex] ]
= [127.01 , 134.99]
Therefore, 90% confidence interval for the population mean is [127.01 , 134.99].
- Now, the pivotal quantity for 95% confidence interval for the population mean is given by;
P.Q. = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\bar X[/tex] = sample mean price = $131
[tex]\sigma[/tex] = population standard deviation = $18.80
n = sample of home theater = 60
[tex]\mu[/tex] = population mean
Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about the population standard deviation.
So, 95% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level
of significance are -1.96 & 1.96}
P(-1.96 < [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < 1.96) = 0.95
P( [tex]-1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ) = 0.95
P( [tex]\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] , [tex]\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ]
= [[tex]131-1.96 \times {\frac{18.8}{\sqrt{60} } }[/tex] , [tex]131+1.96 \times {\frac{18.8}{\sqrt{60} } }[/tex] ]
= [126.24 , 135.76]
Therefore, 95% confidence interval for the population mean is [126.24 , 135.76].
Now, with 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the 90%.
