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Every confidence interval we've talked about has the form "statistics plus or minus a quantile times a standard deviation". Suppose the population variance is unknown and the sample standard deviation is 1.88. Compute the following for a 80% confidence interval with a sample of size 6 (we can assume the sampling distribution of the sample average is normal): (Use 3 decimal places)

(a) The "quantile" part for the confidence interval.

(b) The "standard deviation" part for the confidence interval.

User Paulm
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(a) The "quantile part" for the confidence interval is:

χ²(α/2) = 7.779 and χ²(1 - α/2) = 0.773.

(b) The "standard deviation" part for the confidence interval is: √49,954.62 ≈ 223.80 for the lower bound and √4,690.26 ≈ 68.49 for the upper bound.

Confidence Interval for the Standard Deviation:

We're trying to estimate the population standard deviation σ, with a sample standard deviation s = 188, a sample size n = 5, and a confidence level of 80%. Since the population variance is unknown and the sample size is small, we'll use the chi-square distribution to construct the confidence interval.

Degrees of freedom: Calculate the degrees of freedom

ν = n - 1 = 5 - 1 = 4.

Chi-square quantiles:

Find the chi-square quantiles χ²(α/2) and χ²(1 - α/2) for the desired confidence level α = 1 - 0.8 = 0.2.

Using a chi-square table or calculator, we get χ²(α/2) = 7.779 and χ²(1 - α/2) = 0.773.

Confidence interval: Calculate the lower and upper bounds of the confidence interval:

Lower bound: s² / χ²(1 - α/2) = 188² / 0.773 ≈ 49,954.62

Upper bound: s² / χ²(α/2) = 188² / 7.779 ≈ 4,690.26

Take the square root of both bounds to get the interval in terms of standard deviation:

Lower bound: √49,954.62 ≈ 223.80

Upper bound: √4,690.26 ≈ 68.49

User Zhang TianYu
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