Answer:
A sample size of 2018 is needed to be drawn in order to obtain a 99.5% confidence interval with a margin of error equal to 1.1.
Step-by-step explanation:
Missing data:
The value of population standard deviation is 17.6
We have been provided with the following data:
Confidence Level = 99.5%
Margin of error = M.E = 1.1
Population standard Deviation = [tex]\sigma[/tex] = 17.6
What we have to do is find the minimum sample size required to keep the margin of error upto 1.1 with 99.5% confidence level.
Since, the value of population standard deviation is known, we will use the formulas of z-distribution to solve the question.
Formula of Margin of Error is:
[tex]M.E=z_{\frac{\alpha }{2}} \times \frac{\sigma}{\sqrt{n} }[/tex]
Here,
[tex]z_{\frac{\alpha}{2}}[/tex] is the z critical value for the given confidence level. From the z-table the z-value for 99.5% confidence level comes out to be 2.807
Using the values in above formula, we get:
[tex]1.1=2.807 \times \frac{17.6}{\sqrt{n}} \\\\\sqrt{n}=2.807 \times \frac{17.6}{1.1}\\\\ n=(\frac{2.807 \times 17.6}{1.1} )^{2}\\\\ n=2017.1[/tex]
Rounding up to the nearest integer, the minimum sample size required is 2018.
A sample size of 2018 is needed to be drawn in order to obtain a 99.5% confidence interval with a margin of error equal to 1.1.
