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Assume that the heights of men are normally distributed. A random sample of 19 men have a mean height of 65.5 inches and a standard deviation of 3.0 inches. Construct a 99% confidence interval for the population standard deviation,

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Lower bound = sqrt((n-1)*s^2 / χ^2(α/2,n-1))

Upper bound = sqrt((n-1)*s^2 / χ^2(1-α/2,n-1))

Where:

n = sample size = 19
s = sample standard deviation = 3.0 inches
α = significance level = 0.01 (since the confidence level is 99%, the significance level is 1%)
χ^2(α/2,n-1) = chi-squared value for α/2 and n-1 degrees of freedom
χ^2(1-α/2,n-1) = chi-squared value for 1-α/2 and n-1 degrees of freedom

Using a chi-squared distribution table with 18 degrees of freedom (since n-1 = 19-1 = 18), we find that:

χ^2(0.005,18) = 38.582
χ^2(0.995,18) = 7.962

Substituting the values into the formula, we get:

Lower bound = sqrt((19-1)*3^2 / 38.582) = 1.923
Upper bound = sqrt((19-1)*3^2 / 7.962) = 4.409

Therefore, the 99% confidence interval for the population standard deviation is (1.923, 4.409) inches.
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