8.1 Estimation Using Normal

# Sampling distribution 6 for n 25 38 x 3 x find z

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ample 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 find z-scores for 2.5% and 97.5%: z = ±1.96 2.5% find corresponding sample means: − z = xσ µ x x lower tail : upper tail : Interval Estimation −1.96 = x −µ 3 −5.88 = x − µ 1.96 = x −µ 3 5.88 = x − µ sampling error—difference between sample mean and population mean xL 95% xU -1.96 μ 1.96 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: x ± 5.88 − z = xσ µ x x lower tail : upper tail : Interval Estimation −1.96 = x −µ 3 −5.88 = x − µ If sample mean is known, use to find a range for the population mean: x ± 5.88 1.96 = x −µ 3 5.88 = x − µ sampling error—difference between sample mean and population mean xL 95% xU -1.96 μ 1.96 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 lower tail : upper tail : Interval Estimation −1.96 = x −µ 3 −5.88 = x − µ If sample mean is known, use to find a range for the population mean: x ± 5.88 xL 95% xU 2.5% -1.96 μ 1.96 1.96 = x −µ 3 5.88 = x − µ sampling error—difference between sample mean and population mean 5.88 is the margin of error 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 lower tail : upper tail : Interval Estimation −1.96 = x −µ 3 −5.88 = x...
View Full Document

## This document was uploaded on 03/05/2014.

Ask a homework question - tutors are online