Question

An educational psychologist wishes to know the mean number of words a third grader can read per minute. She wants to ensure that the estimate has an error of at most 0.23 words per minute. A previous study found that the mean was 29.5 words per minute. Assuming that the standard deviation is 3.8, what is the minimum number of third graders that must be included in a sample to construct the 80% confidence interval? Round your answer up to the next integer.

Answer 1

Solution:

We know that confidence interval formula is as follows:

Interval = x(bar) +/- z*s/n^1/2

where x(bar) is mean value, z is the critical value,s is standard deviation, and n is the number of observations

Since, the error has to be of 0.23 words per minute at most, we know that interval must be 0.23 words around the mean

So, |extreme interval value or estimate - x(bar)| <= 0.23

0.23 >= z*s/n^1/2

Finding required critical value for 80% confidence interval: alpha = 1 - 80% = 0.2, so alpha/2 = 0.1

For 0.1, z value = -1.28 (using standard z tables of normal distribution)

So, 0.23 >= -1.28*3.8/n^1/2

n^1/2 >= -1.28*3.8/0.23 = 21.15 (approx)

Squaring on both sides:

n >= 21.15² = 447 (approx)

So, minimum number of third graders that must be included in the sample is 447.