An educational psychologist wishes to know the - id5671055, montgomerykarloxc24x

Ответ:

JohnBranks3258

22.01.2022

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. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ 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 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $24.9-2.5758 \times {\frac{5.7}{\sqrt{4657} } }$ , $24.9+2.5758 \times {\frac{5.7}{\sqrt{4657} } }$ ]

= [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].

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