Respuesta :
Answer:
99% confidence interval for the mean number of words a third grader can read per minute is [24.7 , 25.1].
Step-by-step explanation:
We are given that an educational psychologist wishes to know the mean number of words a third grader can read per minute. She wants to make an estimate at the 99% level of confidence. For a sample of 4657 third graders, the mean words per minute read was 24.9. Assume a population standard deviation of 5.7.
So, the pivotal quantity for 99% confidence interval for the mean number of words a third grader can read per minute 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 words per minute read = 24.9
[tex]\sigma[/tex] = population standard deviation = 5.7
n = sample of third graders = 4657
[tex]\mu[/tex] = population mean
So, 99% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-2.5758 < N(0,1) < 2.5758) = 0.99
P(-2.5758 < [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < 2.5758) = 0.99
P( [tex]-2.5758 \times {\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]{\bar X - \mu}[/tex] < [tex]2.5758 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ) = 0.99
P( [tex]\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ) = 0.99
99% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }[/tex] , [tex]\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ]
= [ [tex]24.9-2.5758 \times {\frac{5.7}{\sqrt{4657} } }[/tex] , [tex]24.9+2.5758 \times {\frac{5.7}{\sqrt{4657} } }[/tex] ]
= [24.7 , 25.1]
Therefore, 99% confidence interval for the mean number of words a third grader can read per minute is [24.7 , 25.1].