Question

Suppose you are planning to sample cat owners to determine the average number of cans of cat food they purchase monthly. The following standards have been set: a confidence level of 99 percent and an error of less than 5 units. Past research has indicated that the standard deviation should be 6 units. What is the final sample required?

Answer 1

Answer:  10

Step-by-step explanation:

The formula to find the sample size is given by :-

$n=(\dfrac{z_{\alpha/2}\ \sigma}{E})^2$

Given : Significance level : $\alpha=1-0.99=0.1$

Critical z-value=$z_{\alpha/2}=2.576$

Margin of error : E=5

Standard deviation : $\sigma=6$

Now, the required sample size will be :_

$n=(\dfrac{(2.576)\ 6}{5})^2=9.55551744\approx10$

Hence, the final sample required to be of 10 .