Question

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 1584 third graders, the mean words per minute read was 35.7. Assume a population standard deviation of 3.3. Construct the confidence interval for the mean number of words a third grader can read per minute. Round your answers to one decimal place.

Answer 1

99% confidence interval for the true mean number of words a third grader can read per minute is [35.5 , 35.9].

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

where,
`\bar X` = sample mean words per minute read = 35.7

`\sigma` = population standard deviation = 3.3

n = sample of third graders = 1584

`\mu` = 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,
`\mu` 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 <
`(\bar X-\mu)/((\sigma)/(√(n) ) )` < 2.58) = 0.99

P(
`-2.58 * {(\sigma)/(√(n) ) }` <
`{\bar X-\mu}` <
`2.58 * {(\sigma)/(√(n) ) }` ) = 0.99

P(
`\bar X-2.58 * {(\sigma)/(√(n) ) }` <
`\mu` <
`\bar X+2.58 * {(\sigma)/(√(n) ) }` ) = 0.99

99% confidence interval for
`\mu` = [
`\bar X-2.58 * {(\sigma)/(√(n) ) }` ,
`\bar X+2.58 * {(\sigma)/(√(n) ) }` ]

= [
`35.7-2.58 * {(3.3)/(√(1584) ) }` ,
`35.7+2.58 * {(3.3)/(√(1584) ) }` ]

= [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].