Respuesta :
Answer:
99% confidence interval for the true mean number of words a third grader can read per minute is [35.5 , 35.9].
Step-by-step explanation:
We are given that a sample of 1584 third graders, the mean words per minute read was 35.7. Assume a population standard deviation of 3.3.
Firstly, the pivotal quantity for 99% 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 words per minute read = 35.7
[tex]\sigma[/tex] = population standard deviation = 3.3
n = sample of third graders = 1584
[tex]\mu[/tex] = population mean number of words
Here for constructing 99% confidence interval we have used One-sample z test statistics as we know about the population standard deviation.
So, 99% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-2.58 < N(0,1) < 2.58) = 0.99 {As the critical value of z at 0.5% level
of significance are -2.58 & 2.58}
P(-2.58 < [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < 2.58) = 0.99
P( [tex]-2.58 \times {\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]2.58 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ) = 0.99
P( [tex]\bar X-2.58 \times {\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.58 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ) = 0.99
99% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.58 \times {\frac{\sigma}{\sqrt{n} } }[/tex] , [tex]\bar X+2.58 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ]
= [ [tex]35.7-2.58 \times {\frac{3.3}{\sqrt{1584} } }[/tex] , [tex]35.7+2.58 \times {\frac{3.3}{\sqrt{1584} } }[/tex] ]
= [35.5 , 35.9]
Therefore, 99% confidence interval for the true mean number of words a third grader can read per minute is [35.5 , 35.9].
