Question

Dr. Tease is a professor of English. Recently she counted the number of misspelled words in a group of student essays. She noted the distribution of misspelled words per essay followed the normal distribution with a population standard deviation of 2.44 words per essay. For her 10 a.m. section of 40 students, the mean number of misspelled words was 6.05. Construct a 95% confidence interval for the mean number of misspelled words in the population of student essays.

Answer 1

The 95% confidence interval for the mean number of misspelled words in the population of student essays is between 5.29 and 6.81.

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1-0.95)/(2) = 0.025`

Now, we have to find z in the Ztable as such z has a pvalue of
`1-\alpha`.

So it is z with a pvalue of
`1-0.025 = 0.975`, so
`z = 1.96`

Now, find the margin of error M as such

`M = z*(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

`M = 1.96*(2.44)/(√(40)) = 0.76`

The lower end of the interval is the sample mean subtracted by M. So it is 6.05 - 0.76 = 5.29

The upper end of the interval is the sample mean added to M. So it is 6.05 + 0.76 = 6.81

The 95% confidence interval for the mean number of misspelled words in the population of student essays is between 5.29 and 6.81.