Respuesta :
Answer:
90% confidence interval for the true population mean textbook weight is [35.79 ounces , 38.21 ounces].
Step-by-step explanation:
We are given that you measure 50 textbooks' weights, and find they have a mean weight of 37 ounces.
Assume the population standard deviation is 5.2 ounces.
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 weight = 37 ounces
[tex]\sigma[/tex] = population standard deviation = 5.2 ounces
n = sample of textbooks = 50
[tex]\mu[/tex] = true population mean textbook weight
Here for constructing 90% confidence interval we have used One-sample z test statistics as we know about 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]37-1.645 \times {\frac{5.2}{\sqrt{50} } }[/tex] , [tex]37+1.645 \times {\frac{5.2}{\sqrt{50} } }[/tex] ]
= [35.79 , 38.21]
Therefore, 90% confidence interval for the true population mean textbook weight is [35.79 ounces , 38.21 ounces].
