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? If only 30 percent of households have a cat, what is the initial number of households that need to be contacted?

Answer 1

Answer with explanation:

Given : Significance level :
`\alpha: 1-0.99=0.01`

Critical value :
`z_(\alpha/2)=2.576`

Margin of error :
`E=5`

Standard deviation :
`\sigma=6`

The formula to find the sample size :-

`n=((z_(\alpha/2)*\sigma)/(E))^2`

Then, the sample size will be :-

`n=(((2.576)*6)/(5))^2\\\\=(3.0912)^2=9.55551744\approx10`

The minimum final size sample required is 10.

If only 30 percent of households have a cat, then the proportion of households have a cat = 0.3

The formula to find the sample size :-

`n=p(1-p)((z_(\alpha/2))/(E))^2`

Then, the sample size will be :-

`n=0.3(1-0.3)(((2.576))/(5))^2\\\\=(0.21)(0.5152)^2=0.0557405184\approx1`

Hence, the initial number of households that need to be contacted =1