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:  10Step-by-step explanation:The formula to find the sample size is given by :-[tex]n=(\dfrac{z_{\alpha/2}\ \sigma}{E})^2[/tex]Given : Significance level : [tex]\alpha=1-0.99=0.1[/tex]Critical z-value=[tex]z_{\alpha/2}=2.576[/tex]Margin of error : E=5Standard deviation : [tex]\sigma=6[/tex]Now, the required sample size will be :_[tex]n=(\dfrac{(2.576)\ 6}{5})^2=9.55551744\approx10[/tex]Hence, the final sample required to be of 10 .