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How large a sample should be selected to provide a 95% confidence interval with a margin of error of 4? Assume that the population standard deviation is 40. Round your answer to next whole number.

2 Answers

6 votes

Final answer:

To achieve a 95% confidence interval with a 4 point margin of error and a population standard deviation of 40, the required sample size is 154, after rounding up to the nearest whole number.

Step-by-step explanation:

To determine how large a sample should be selected to provide a 95% confidence interval with a margin of error of 4, when the population standard deviation is 40, we use the following formula for the sample size (n):

n = (Z* σ / E)^2

Where Z is the z-score corresponding to the desired confidence level, σ (sigma) is the population standard deviation, and E is the margin of error.

For a 95% confidence interval, the Z value is approximately 1.96. Plugging in the values, we have:

n = (1.96 * 40 / 4)^2

n = (49.6 / 4)^2

n = (12.4)^2

n = 153.76

After rounding up to the next whole number, we need a sample size of 154 to achieve our desired margin of error.

User Jay Zamsol
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4.2k points
1 vote

Answer:

The large sample size 'n' = 384.16

Step-by-step explanation:

Explanation:-

Given Margin of error = 4

Given Population standard deviation(σ) = 40

The margin of error is determined by


M.E = Z_(0.05) (S.D)/(√(n) )


4 = 1.96 (40)/(√(n) )

Cross multiplication , we get


√(n) = (40 X 1.96)/(4)

√ n = 19.6

squaring on both sides , we get

n = 384.16

Final answer:-

The large sample size 'n' = 384.16

User JYoThI
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4.6k points