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

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].