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 :-

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

Then, the sample size will be :-

$n=(\dfrac{(2.576)\times6}{5})^2\\\\=(3.0912)^2=9.55551744\approx10$

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 :-

$n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2$

Then, the sample size will be :-

$n=0.3(1-0.3)(\dfrac{(2.576)}{5})^2\\\\=(0.21)(0.5152)^2=0.0557405184\approx1$

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