Answer:
96
Explanation:
From the given information:
At 95% Confidence interval level,Level of significance
0.05, the value of Z from the standard normal tables = 1.96
Margin of Error = 0.10
Let assume that the estimated proportion = 0.5
therefore; the sample size n can be determined by using the formula:
![n =((Z)/(E))^2 * p* (1-p)](https://img.qammunity.org/2021/formulas/mathematics/college/185rckimiz60vrxl037w3wshfaswbjehc6.png)
![n =((1.96)/(0.1))^2 * 0.5* (1-0.5)](https://img.qammunity.org/2021/formulas/mathematics/college/sg9ec0n11ukbtrj38cc0b16469xwwjv9q8.png)
![n =(19.6)^2 * 0.5* (0.5)](https://img.qammunity.org/2021/formulas/mathematics/college/404mr7z3dy9f4ndq8hkvx92iexpfmf5te3.png)
n = 96.04
n
96