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.
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 =
= 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:
![M.E=z_{(\alpha )/(2)} * (\sigma)/(√(n) )](https://img.qammunity.org/2021/formulas/mathematics/college/wahbjtde8p5eyazkjtarb26i4mv1s62ba1.png)
Here,
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:
![1.1=2.807 * (17.6)/(√(n)) \\\\√(n)=2.807 * (17.6)/(1.1)\\\\ n=((2.807 * 17.6)/(1.1) )^(2)\\\\ n=2017.1](https://img.qammunity.org/2021/formulas/mathematics/college/pxmxjdhqpfj0hrmjpupybv8o66o8v3dot7.png)
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.