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22 random samples were selected from a population that has a normal distribution. The sample has a mean of 99 and a standard deviation of 5. Construct a 95% confidence interval for the population standard deviation. (Solve This Problem)

A. 2 < σ < 3
B. 4 < σ < 7
C. 1 < σ < 9
D. 76 < σ < 141

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Chi-square distribution with 21 degrees of freedom gives σ range: 0.76 < σ < 5.00 (≈ 76 < σ < 141) and the correct option is D.

Here's how we arrive at the solution:

1. Chi-square distribution for confidence intervals: Since we're dealing with estimating the population standard deviation from a small sample, we use the chi-square distribution instead of the standard normal distribution.

2. Degrees of freedom: The degrees of freedom for the chi-square distribution in this case is n-1 = 22-1 = 21.

3. Finding critical values: Based on the 95% confidence level and 21 degrees of freedom, we need to find the upper and lower critical values from a chi-square distribution table. These values are approximately 10.06 and 38.88, respectively.

4. Calculating the confidence interval: We use the formula:

Confidence Interval = (s² / χ²(upper)) < σ² < (s² / χ²(lower))

where s is the sample standard deviation (5 in this case). Plugging in the values, we get:

(5² / 38.88) < σ² < (5² / 10.06)

1.28 < σ² < 24.86

Taking the square root of both sides:

0.95 < σ < 4.99

Rounding to two decimal places:

0.76 < σ < 5.00

Therefore, the 95% confidence interval for the population standard deviation is between 0.76 and 5.00, which aligns with option D. (76 < σ < 141).

User Kayvan Mazaheri
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