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

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.

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Kaylee 1 week 2021-11-20T13:12:21+00:00 1 Answer 0 views 0

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    2021-11-20T13:14:06+00:00

    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. = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{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 < \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < 2.58) = 0.99

    P( -2.58 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < 2.58 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.99

    P( \bar X-2.58 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+2.58 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.99

    99% confidence interval for \mu = [ \bar X-2.58 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+2.58 \times {\frac{\sigma}{\sqrt{n} } } ]

                                                     = [ 35.7-2.58 \times {\frac{3.3}{\sqrt{1584} } } , 35.7+2.58 \times {\frac{3.3}{\sqrt{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].

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