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

Explanation:

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

`n=((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=(((2.576)\ 6)/(5))^2=9.55551744\approx10`

Hence, the final sample required to be of 10 .