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svetlana [45]
3 years ago
9

An educational psychologist wishes to know the mean number of words a third grader can read per minute. She wants to make an est

imate 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.
Mathematics
1 answer:
zhannawk [14.2K]3 years ago
6 0

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. = \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

<u>99% confidence interval for </u>\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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