It is known that the population variance ($\sigma$²) is 144. At 95% confidence, what size sample should be taken so that the margin of error is does not exceed 5 or less? A)22 B)23 C)24 D)25

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The formula for the margin of error is:$$ME = z * \sqrt{\frac{\sigma^2}{n}}$$where:* ME is the margin of error* z is the z-score for the desired confidence level* σ² is the population variance* n is the sample sizeWe are given that the population variance is 144, the confidence level is 95%, and the margin of error should not exceed 5. The z-score for a 95% confidence level is 1.96. Plugging these values into the formula, we get:$$5 = 1.96 * \sqrt{\frac{144}{n}}$$Solving for n, we get:$$n = \frac{1.96^2 * 144}{5^2} = 23.04$$Since we cannot take a fraction of a sample, we round up to the nearest whole number, which is 23. Therefore, the correct answer is B) 23.

It Is Known That the Population Variance σ\sigmaσ 2) Is 144

It Is Known That the Population Variance $\sigma$ ²) Is 144