Answer:
The 95% confidence interval for the mean number of misspelled words in the population of student essays is between 5.29 and 6.81.
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.95}{2} = 0.025[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]
Now, find the margin of error M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
[tex]M = 1.96*\frac{2.44}{\sqrt{40}} = 0.76[/tex]
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.
