Question:medium

If a 95% confidence interval for a population mean was reported to be 132 to 160 and sample standard deviation s = 50, then the size of the sample in the study is:
(Given \(Z_{0.025}\) = 1.96)

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The margin of error is the key link between the confidence interval, standard deviation, and sample size. If you are given the interval, you can always calculate the ME and then use its formula to solve for any unknown component.
Updated On: Jan 16, 2026
  • 90
  • 95
  • 50
  • 49
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The Correct Option is D

Solution and Explanation

Step 1: Concept Definition:
A confidence interval indicates a range of likely values for a population parameter. The width of this range is influenced by the confidence level, sample variability, and sample size. The sample size can be determined by rearranging the confidence interval formula.
Step 2: Core Formula:
The formula for a confidence interval for the mean is:\[ \text{CI} = \bar{x} \pm \text{Margin of Error (ME)} \]The Margin of Error is calculated as:\[ \text{ME} = Z_{\alpha/2} \times \frac{s}{\sqrt{n}} \]From the provided interval, both the sample mean (\(\bar{x}\)) and the margin of error can be derived.
Step 3: Calculation Breakdown:
Given the 95% confidence interval: [132, 160].
1. Determine the sample mean (\(\bar{x}\)): This is the midpoint of the interval.\[ \bar{x} = \frac{160 + 132}{2} = 146 \]2. Calculate the Margin of Error (ME): This is half the interval's width.\[ \text{ME} = \frac{160 - 132}{2} = 14 \]3. Solve for n using the ME formula:Known values:- ME = 14- Sample standard deviation, s = 50- For a 95% confidence level, \(Z_{0.025}\) = 1.96.Substitute into the formula:\[ 14 = 1.96 \times \frac{50}{\sqrt{n}} \]Isolate \(\sqrt{n}\):\[ \sqrt{n} = \frac{1.96 \times 50}{14} = 7 \]Square both sides to find n:\[ n = 7^2 = 49 \]Step 4: Conclusion:
The sample size for this study is 49.
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