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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%99% level of confidence. For a sample of 46574657 third graders, the mean words per minute read was 24.924.9. Assume a population standard deviation of 5.75.7. Construct the confidence interval for the mean number of words a third grader can read per minute. Round your answers to one decimal place.

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Answer:

99% confidence interval for the mean number of words a third grader can read per minute is [24.7 , 25.1].

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. =
(\bar X - \mu)/((\sigma)/(√(n) ) ) ~ N(0,1)

where,
\bar X = sample mean words per minute read = 24.9


\sigma = population standard deviation = 5.7

n = sample of third graders = 4657


\mu = population mean

So, 99% confidence interval for the population mean,
\mu is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99

P(-2.5758 <
(\bar X - \mu)/((\sigma)/(√(n) ) ) < 2.5758) = 0.99

P(
-2.5758 * {(\sigma)/(√(n) ) } <
{\bar X - \mu} <
2.5758 * {(\sigma)/(√(n) ) } ) = 0.99

P(
\bar X-2.5758 * {(\sigma)/(√(n) ) } <
\mu <
\bar X+2.5758 * {(\sigma)/(√(n) ) } ) = 0.99

99% confidence interval for
\mu = [
\bar X-2.5758 * {(\sigma)/(√(n) ) } ,
\bar X+2.5758 * {(\sigma)/(√(n) ) } ]

= [
24.9-2.5758 * {(5.7)/(√(4657) ) } ,
24.9+2.5758 * {(5.7)/(√(4657) ) } ]

= [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].

User Derferman
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