96 x 3 588 x x 368 588 36212 25 xl 95 xu 196

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Unformatted text preview: µ x x low tail: er upper tail : Interval Estimation −1.96 = x −µ 3 −5.88 = x − µ x = 368 −5.88 = 362.12 2.5% xL 95% xU -1.96 μ 1.96 1.96 = x −µ 3 5.88 = x −µ x = 368+5.88 = 373.88 2.5% σ Known Example Find an interval symmetrically distributed around the population mean that includes 95% of the sample means. Sampling distribution: µ= 6; σ = for n = 25 38 x 3 x find z-scores for 2.5% and 97.5%: z = ±1.96 find corresponding sample means: − z = xσ µ x x low tail: er upper tail : Interval Estimation −1.96 = x −µ 3 −5.88 = x − µ x = 368 −5.88 = 362.12 2.5% xL 95% xU -1.96 μ 1.96 1.96 = x −µ 3 5.88 = x −µ x = 368+5.88 = 373.88 Using MS Excel Lower tail 2.5%: =NORM.INV(.025,368,3)=362.120108 Upper tail 97.5%: =NORM.INV(.975,368,3)=373.879892 2.5% σ Known Example Find an interval symmetrically distributed around the population mean that includes 95% of the sample means. Sampling distribution: µ= 6; σ = for n = 25 38 x 3 x find z-scores for 2.5% and 97.5%: z = ±1.96 2.5% find corresponding sample means: − z = xσ µ x x low tail: er upper tail : Interval Estimation −1.96 = x −µ 3 −5.88 = x − µ x = 368 −5.88 = 362.12 xL 95% xU 2.5% -1.96 μ 1.96 1.96 = x −µ 3 5.88 = x −µ x = 368+5.88 = 373.88 3 2 ≤. 6 ≤3 8 2 x7 . 1 3 8 Therefore, if the population mean is 368 grams, then 95% of sample means would be between 362.12 and 373.88 grams. σ Known Example Find an interval symmetrically distributed around the population mean that includes 95% of the sample means. Sampling distribution: µ= 6; σ = for n = 25 38 x 3 x Interval Estimation Instead of solving for sample mean, solve for populatio n mean given a sample mean. find z-scores for 2.5% and 97.5%: z = ±1.96 2.5% find corresponding sample means: − z = xσ µ x x low tail: er upper tail : −1.96 = x −µ 3 −5.88 = x − µ x = 368 −5.88 = 362.12 xL 95% xU 2.5% -1.96 μ 1.96 1.96 = x −µ 3 5.88 = x −µ x = 368+5.88 = 373.88 3 2 ≤. 6 ≤3 8 2 x7 . 1 3 8 Therefore, if the population mean is 368 grams, then 95% of s...
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