Online Participation Topic: Week 5 For week 5, the online participation topic is on confidence intervals. We basically want to find the mean of a population (µ) by taking a sample and inferring a range of µ from the sample’s properties. Let N = the population size n = the sample size µ = the population mean = the sample mean σ = the population standard deviation s = the sample standard deviation The sample’s properties (n, , s) are known or can be computed. The population’s properties (N, µ, σ) are the unknowns. However, n < n="" (usually).="" sometimes="" σ="" is="" known.="" if="" this="" is="" the="" case,="" we="" can="" compute="" a="" z-score:="" z="=" if="" σ="" is="" unknown,="" we="" then="" compute="" a="" t-score="" t="=" k="" is="" a="" measure="" of="" the="" number="" of="" standard="" deviations="" from="" the="" mean.="" by="" choosing="" an="" appropriate="" value="" of="" z="" or="" t,="" we="" can="" construct="" a="" “confidence="" interval”="" about="" the="" mean="" (="" –="" k)="">< µ="">< ( + k) if we let z = 1.645 so that 90% of a normal distribution would fall between +/-1.645σ of the mean, then the above inequality would be a 90% confidence interval. this means that 90% of the time, the population µ will be inside the confidence interval constructed using the sample data. for week 5: 1. download the excel spreadsheet onlineweek2.xlsx, or onlineweek2data again from week 2. 2. reserve one the of the data set columns (a-v) different from the one you did for week 2. 3. compute , s using excel, and df (degrees of freedom) = n – 1. 4. find the 90% t-score using the t-distribution applet. we are not using the z-distribution because we do not know σ, the standard deviation of the population (stdev.p). thus, we are using the t-distribution because we know only the standard deviation of the sample (stdev.s). 5. in the t-distribution applet, input df, two tails (because a confidence interval is + and -), and p = .10 to get the 90% t-value. 6. k = t * s / 7. compute the 90% “confidence interval”. 8. write in words the meaning of your confidence interval. “there is a 90% chance that the population mean lies inside the confidence interval.” 9. post your answer to “week 5 discussion group inputs”. (="" +="" k)="" if="" we="" let="" z="1.645" so="" that="" 90%="" of="" a="" normal="" distribution="" would="" fall="" between="" +/-1.645σ="" of="" the="" mean,="" then="" the="" above="" inequality="" would="" be="" a="" 90%="" confidence="" interval.="" this="" means="" that="" 90%="" of="" the="" time,="" the="" population="" µ="" will="" be="" inside="" the="" confidence="" interval="" constructed="" using="" the="" sample="" data.="" for="" week="" 5:="" 1.="" download="" the="" excel="" spreadsheet="" onlineweek2.xlsx,="" or="" onlineweek2data="" again="" from="" week="" 2.="" 2.="" reserve="" one="" the="" of="" the="" data="" set="" columns="" (a-v)="" different="" from="" the="" one="" you="" did="" for="" week="" 2.="" 3.="" compute="" ,="" s="" using="" excel,="" and="" df="" (degrees="" of="" freedom)="n" –="" 1.="" 4.="" find="" the="" 90%="" t-score="" using="" the="" t-distribution="" applet.="" we="" are="" not="" using="" the="" z-distribution="" because="" we="" do="" not="" know="" σ,="" the="" standard="" deviation="" of="" the="" population="" (stdev.p).="" thus,="" we="" are="" using="" the="" t-distribution="" because="" we="" know="" only="" the="" standard="" deviation="" of="" the="" sample="" (stdev.s).="" 5.="" in="" the="" t-distribution="" applet,="" input="" df,="" two="" tails="" (because="" a="" confidence="" interval="" is="" +="" and="" -),="" and="" p=".10" to="" get="" the="" 90%="" t-value.="" 6.="" k="t" *="" s="" 7.="" compute="" the="" 90%="" “confidence="" interval”.="" 8.="" write="" in="" words="" the="" meaning="" of="" your="" confidence="" interval.="" “there="" is="" a="" 90%="" chance="" that="" the="" population="" mean="" lies="" inside="" the="" confidence="" interval.”="" 9.="" post="" your="" answer="" to="" “week="" 5="" discussion="" group="">