Aresearcher wishes to estimate the number of households with two cars. how large a sample is needed in order to be 99% confident that the sample proportion will not differ from the true proportion by more than 3%? a previous study indicates that the proportion of households with two cars is 25%.
a) 4 b) 1132 c) 1842 d) 1382

Sample size n = 1382

so correct option is D) 1382

Step-by-step explanation:

given data

confidence level = 99 %

margin of error = 3%

probability = 25 %

to find out

How large a sample size needed

solution

we know here P = 25 %

so 1 - P = 1 - 0.25

1 - P  = 0.75

and we know E margin of error is 0.03 so value of Z for 99%

α = 1 - 99%   = 1 - 0.99

α  = 0.01

and   =

 = 0.005

so Z is here

= 2.576

so

sample size will be

Sample size n =  

put here value

Sample size n = (\frac{2.576}{0.03})^2 * 0.25 * 0.75

Sample size n = 1382

so correct option is D) 1382

stop cheating

step-by-step explanation:

the answer would be 1 1/72

step-by-step explanation:

you can rewrite the equation by separating the parts and then solving individually:

10-9=1

1/8 - 1/9=

find l c d and then combine:

1 1/72