Using confidence function population population

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ce interval estimate of the mean 12-month weight gain for all quitters. Using margin of error: Population: Population: std dev σ: 6 Interval Estimation Sample: std dev σ: 6 Sample: mean: 13.15384615 sample size n: 13 mean: =AVERAGE(A2:A16) sample size n: =COUNT(A2:A16) Sampling Distribution: std error: 1.664100589 Sampling Distribution: see Smoking.xlsx Example 2 One of the few negative side effects of quitting smoking is weight gain. Suppose that the weight gain in the 12 months following a cessation in smoking is normally distributed with a standard deviation of 6 kilograms. To estimate the mean weight gain, a random sample of 13 quitters was drawn their weight gains recorded and listed here. Determine the 90% confidence interval estimate of the mean 12-month weight gain for all quitters. Using CONFIDENCE function: Population: Population: Interval Estimation std dev σ: 6 6 mean: std dev σ: =AVERAGE(A2:A14) Sample: Sample: mean: 13.15384615 Interval Estimate: see Smoking.xlsx lower: Interval Estimate: =E6-CONFIDENCE.NORM(0.1,E3,COUNT(A2:A14)) upper: =E6+CONFIDENCE.NORM(0.1,E3,COUNT(A2:A14)) To Try… 1. 2. Suppose that shopping times for customers at a local grocery store are normally distributed. A random sample of 16 shoppers in the local grocery store had a mean time of 25 minutes. Assume σ=6 minutes. Find the 95% confidence interval for the population mean. Interpret your results. A process produces bags of refined sugar. The weights of the contents of these bags are normally distributed with standard deviation 1.2 ounces. The contents of a random sample of 25 bags had a mean weight of 19.8 ounces. Find the upper and lower confidence limits of a 99% confidence interval for the true mean weight for all bags of sugar produced by the process. If the bags are intended to have 20 ounces of sugar, comment on the filling process. Interval Estimation To Try… Suppose that shopping times for customers at a local grocery store are normally distributed. A random sample of 16 shoppers in the local grocery store had a mean time of 25 minutes. Assume σ=6 minutes. Find the 95% confidence interval for the population mean. Interpret σ your results. Confidence interval: 1. x ±z 2 α n Population: std dev σ: 6 Interval Estimation confidence level = 95%, therefore, α=.05 zα/2 = z.05/2 = 1.960 confidence interval: Sample: σ x ± 1.960 size n: 16 n mean x­bar: 25 6 25 ± 1.960 16 25 ± 2.94 Interval Estimate: 22.06 ≤ µ ≤ 27.94 significance α: 0.05 Or, using MS Excel, CONFIDENCE.NORM(0.05,6,16)=2.939945977 confidence: 95% We are 95% confident that the mean shopping time for customers at the grocery store is between 22 and 28 minutes. Calulations: To Try… A process produces bags of refined sugar. The weights of the contents of these bags are normally distributed with standard deviation 1.2 ounces. The contents of a random sample of 25 bags had a mean weight of 19.8 ounces. Find the upper and lower confidence limits of a 99% confidence interval for the true mean weight for all bags of sugar produced by the process. If the bags are intended to have 20 ounces of sugar, comment on the filling σ Population: process. Confidence interval: x ±z 2. α2 n Interval Estimation confidence level = 99%, therefore, α=.01 zα/2 = z.01/2 = 2.576 1.2 confidence interval: 19.8 ± 2.576 25 19.8 ± 0.61824 19.18176 ≤ µ ≤ 20.41824 std dev σ: 1.2 size n: 25 mean x­bar: 19.8 Sample: Interval Estimate: Or, using MS Excel, CONFIDENCE.NORM (0.01,1.2,25)=0.618199033 significance α: 0.01 We are 99% confident that the mean weight of the bags of sugar is between confidence: 99% 19.2 and 20.4 ounces. Because 20 ounces is in the interval, we can assume the filling process is working correctly....
View Full Document

This document was uploaded on 03/05/2014.

Ask a homework question - tutors are online