8-5 Estimating mean differences

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## 8-5 Estimating mean differences

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**Comparing two populations**Are two populations different? Really? Just a little? What if I wanted to compare the mean test grades of two classes. How different might they be? Independent samples are completely unrelated to each other. (Drawing two random samples such as a drug trial with placebo) Dependent samples can be paired based on correspondence. (Before and After – double measuring)**Two samples**Given populations 1 and 2 Take samples (not necessarily of the same size) and find the mean of the samples. Do this over and over and you will see the result, stated as Theorem 8-1. The distribution of mean differences will be normal, the mean 1 – 2 will equal μ1 – μ2 and the standard deviation will equal**Two samples**Given populations 1 and 2 Take samples (not necessarily of the same size) and find the mean of the samples. Do this over and over and you will see the result, stated as Theorem 8-1. 1.The distribution of mean differences will be normal 2. the mean 1 – 2 = μ1 – μ2 3. the standard deviation will equal**Two samples**Given populations 1 and 2 Take samples (not necessarily of the same size) and find the mean of the samples. Do this over and over and you will see the result, stated as Theorem 8-1. 1.The distribution of mean differences will be normal 2. the mean 1 – 2 = μ1 – μ2 3. the standard deviation will equal Have you seen this before?**As before**If x1 and x2 have normal distributions then the difference will be a normal distribution. If not, as long both n subsets are 30 or larger, then the CLT applies.**Confidence Interval**For μ1 – μ2 The confidence interval will be (1 – 2) – E < μ1 – μ2 < (1 – 2) + E Where**Confidence Interval**For μ1 – μ2 The confidence interval will be (1 – 2) – E < μ1 – μ2 < (1 – 2) + E Where**And logic dictates that**If σ1 – σ2 are unknown, the student T distribution applies. The degrees of freedom idea still applies, and you choose the smaller of n1 –1 and n2 – 1. The formula looks a little different**Confidence Interval**For μ1 – μ2 The confidence interval will be (1 – 2) – E < μ1 – μ2 < (1 – 2) + E Where**Confidence Interval**For μ1 – μ2 The confidence interval will be (1 – 2) – E < μ1 – μ2 < (1 – 2) + E Where Warning: The calculator will give a slightly different d.f.**Example**Independent random samples of professional football and basketball players gave the following information Sports Encyclopedia of Pro Football; Official NBA Basketball Encyclopeia)**Weight in lbs of pro football players x1**245 262 255 251 244 276 240 265 257 252 282 256 250 264 270 275 245 275 253 265 270 Weight in lbs of pro Basketball players x2 205 200 220 210 191 215 221 216 228 207 225 208 195 191 207 196 181 193 201**1. Use your calculator to verify that**2. Let μ1 be the population mean for 1 and let μ2 be the population mean for 2 . Find a 99% confidence interval for μ1 – μ2. 3. Examine this confidence interval and explain what it means in the context of the problem. At the 99% level of confidence, do professional football players tend to have a higher population mean weight than professional basketball players?**Interpretation of confidence Intervals**If c% contains only negative values, then one is c% confident that μ1< μ2 If c% contains only positive values, then one is c% confident that μ1> μ2 If c% contains both positive and negative then no conclusion can be made, but reducing c may allow either of the above 2 conclusions to be made.**Confidence Interval for p1 – p2**Theorem 8-2 will likely look confusing. Essentially you must look for two samples with size n1and n2, probability for success (and failure) for each trial, a point estimate for each (ê1 and ê2) As long as four quantities are above five (sample size times point estimate for success and failure) then a confidence interval can be found.**Confidence Interval for p1 – p2**Theorem 8-2 will likely look confusing. Essentially you must look for two samples with size n1and n2, probability for success (and failure) for each trial, a point estimate for each (ê1 and ê2) As long as four quantities are above thirty (sample size times point estimate for success and failure) then a confidence interval can be found.**Confidence Interval for p1 – p2**For μ1 – μ2 The confidence interval will be (ê1 – ê2) – E < p1 – p2 < (ê1 – ê2) + E Where**Confidence Interval for p1 – p2**For μ1 – μ2 The confidence interval will be (ê1 – ê2) – E < p1 – p2 < (ê1 – ê2) + E Where**Most married couples have two or three personality**preferences in common. Myers used a random sample of 375 couples and found that 132 had three preferences in common. Another random sample of 571 couples showed that 217 had two personality preferences in common. Let p1 be the population proportion of all married couples who have three preferences in common and let p2 be the population proportion of all married couples who have two personality preferences in common. Find a 90% confidence interval for p1 – p2.**Interpretation of confidence Intervals**If c% contains only negative values, then one is c% confident that p1< p2 If c% contains only positive values, then one is c% confident that p1> p2 If c% contains both positive and negative then no conclusion can be made, but reducing c may allow either of the above 2 conclusions to be made.**Calculator**Page 457 outlines the way to use the calculator (under STATS: TESTS) Choose 2 sample z interval, 2 sample t interval or 2 proportion z interval.