If a sample of size of 30 has a confidence interval of approximately 10.58

Question

If a sample of size of 30 has a confidence interval of approximately 10.58 < μ < 16.42, what would the most likely standard deviation be?

Quizplus · Accepted Answer

The confidence interval is calculated using the formula: $\text{mean} \pm \text{Z} * (\frac{\text{standard deviation}}{\sqrt{n}})$, where Z is the Z-score corresponding to the confidence level (not given here, but not needed for this calculation), and $n$ is the sample size. The width of the confidence interval is $16.42 - 10.58 = 5.84$. This width is twice the value of $\text{Z} * (\frac{\text{standard deviation}}{\sqrt{n}})$, because the interval extends equally in both directions from the mean. Therefore, $\frac{5.84}{2} = 2.92$ equals $\text{Z} * (\frac{\text{standard deviation}}{\sqrt{30}})$. Without knowing Z, we can still estimate the standard deviation because the structure of the question implies a typical scenario where Z would be a constant value in common confidence intervals (e.g., 1.96 for 95% confidence). Given the sample size of 30 and the way the numbers work out, a standard deviation of 8 is the most plausible choice that would result in a margin of error close to 2.92 when divided by the square root of 30 and then potentially multiplied by a common Z value.

If a Sample of Size of 30 Has a Confidence