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Natasha2012 [34]
3 years ago
14

Suppose you are planning to sample cat owners to determine the average number of cans of cat food they purchase monthly. The fol

lowing 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?
Mathematics
1 answer:
xxTIMURxx [149]3 years ago
8 0
<h2>Answer with explanation:</h2>

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

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