8-5 Estimating mean differences

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 Dependent samples are completely unrelated to each other. (Drawing two random samples such as a drug trial with placebo) 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 μ 1 – μ 2 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 = μ 1 – μ 2 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 = μ 1 – μ 2 3. the standard deviation will equal σ 1 2 n 1  σ 2 n 2 2 Have you seen this before?

As before

If x 1 and x 2 applies. 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

Confidence Interval

For μ 1 – μ 2 The confidence interval will be (  1 –  2 ) – E < μ 1 Where – μ 2 < (  1 –  2 ) + E

Confidence Interval

For μ 1 – μ 2 The confidence interval will be (  1 –  2 ) – E < μ 1 Where – μ 2 < (  1 –  2 ) + E c σ 1 2 n 1



σ 2 2 n 2

And logic dictates that

If σ 1 – σ 2 are unknown, the student T distribution applies. The degrees of freedom idea still applies, and you choose the smaller of n 1 –1 and n 2 – 1. The formula looks a little different

Confidence Interval

For μ 1 – μ 2 The confidence interval will be (  1 –  2 ) – E < μ 1 Where – μ 2 < (  1 –  2 ) + E

Confidence Interval

For μ 1 – μ 2 The confidence interval will be (  1 –  2 ) – E < μ 1 Where – μ 2 < (  1 –  2 ) + E c s 1 2 n 1



s 2 2 n 2 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 x 1 245 262 255 251 244 276 240 265 257 252 282 256 250 264 270 275 245 275 253 265 270 Weight in lbs of pro Basketball players x 2 205 200 220 210 191 215 221 216 228 207 225 208 195 191 207 196 181 193 201

1. Use your calculator to verify that x 1  259.6 x 2  205.8 s 1  12.1 s 2  12.9 2. Let μ let μ 2 1 be the population mean for be the population mean for   2 1 and . Find a 99% confidence interval for μ professional basketball players?

1 – μ 2 .

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

Interpretation of confidence Intervals

If c% contains only negative values, then one is c% confident that μ 1 < μ 2 If c% contains only positive values, then one is c% confident that μ to be made.

1 > μ 2 If c% contains both positive and negative then no conclusion can be made, but reducing c may allow either of the above 2 conclusions

Confidence Interval for p

1

– p

2 Theorem 8-2 will likely look confusing. Essentially you must look for two samples with size n 1 and n 2 a point estimate for each ( , probability for success (and failure) for each trial, success and failure) then a ê 1 and confidence interval can be found.

ê 2 (sample size times point estimate for ) As long as four quantities are above five

Confidence Interval for p

1

– p

2 Theorem 8-2 will likely look confusing. Essentially you must look for two samples with size n 1 and n 2 a point estimate for each ( , probability for success (and failure) for each trial, ê 1 and ê 2 ) As long as four quantities are above thirty (sample size times point estimate for success and failure) then a confidence interval can be found.

2

ˆ

2 2

ˆ

2

Confidence Interval for p

1

– p

2 For μ 1 – μ 2 The confidence interval will be ( ê 1 – ê 2 ) – E < p 1 Where – p 2 < ( ê 1 – ê 2 ) + E

Confidence Interval for p

1

– p

2 For μ 1 – μ 2 The confidence interval will be ( ê 1 – ê 2 ) – E < p 1 Where – p 2 < ( ê 1 – ê 2 ) + E z c n 1



n 2 2

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 p 1 be the population proportion of all married couples who have three preferences in common and let p 2 be the population proportion of all married couples who have two personality preferences in common.

Find a 90% confidence interval for p 1 – p 2 .

Interpretation of confidence Intervals

If c% contains only negative values, then one is c% confident that p 1 < p 2 If c% contains only positive values, then one is c% confident that p 1 > p 2 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.