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MrRissso [65]
3 years ago
15

A researcher wishes to estimate the number of households with two cars. How large a sample is needed in order to be 99% confiden

t 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
Mathematics
1 answer:
solmaris [256]3 years ago
3 0

Answer:

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  \frac{\alpha}{2} = \frac{0.01}{2}

\frac{\alpha}{2}  = 0.005

so Z is here

Z_(\frac{\alpha}{2}) = 2.576

so

sample size will be

Sample size n =  (\frac{(Z_(\frac{\alpha}{2})}{E})^2 * P * (1-P)

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

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<h3>How to solve the mean.</h3>

From the complete information, the mean will be calculated as:

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brainly.com/question/24298037

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