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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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Therefore x = 1/3 or 3

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Read 2 more answers
Find the sum of a finite geometric sequence from n = 1 to n = 7, using the expression −4(6)n − 1.
Verizon [17]

Answer:

<h2>-223,948</h2>

Step-by-step explanation:

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a₁ - first term

r - common ratio

We have

a_n=-4(6)^{n-1}

Calculate a₁. Put n = 1:

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Calculate the common ratio:

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7 0
3 years ago
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