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?

Respuesta :

Answer:  10

Step-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=5

Standard 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 .