Lecture 5 12.2 Inference for a population mean when the stdev is unknown; one more example 12.3 Testing a population variance 12.4 Testing a population.
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1 Lecture 512.2 Inference for a population mean when the stdev is unknown; one more example12.3 Testing a population variance12.4 Testing a population proportion
2 Announcements Answer to Problem 12.77: Sample size of 752. Extra office hour this week: Wednesday, 9-10Homework – due Thursday, see web page for correction on last problemType II error calculation from last lecture – solution on web page
3 Hypothesis Testing – Basic Steps Set up alternative and null hypothesesChoose appropriate test statistic and values of test statistic that will be considered evidence in favor of H1, e.g., for testing , reject for large values of z-scoreFind critical values and compare the observed test statistic to critical value (rejection region method) or find p-value (p-value method)Make substantive conclusions.
4 Estimating m when s is unknown Example 12.2An investor is trying to estimate the return on investment in companies that won quality awards last year.A random sample of 83 such companies is selected, and the return on investment is calculated had he invested in them.Construct a 95% confidence interval for the mean return.Is there evidence that the returns are greater than 10%?
5 Estimating m when s is unknown Solution:The problem objective is to describe the population of annual returns from buying shares of quality award-winners.Given: x-bar=15.02, s=8.31, n=83Data: Xm12-02There is evidence that the returns are >10% at the 2.5% significance level. (Why?)t.025,80
6 12.3 Inference About a Population Variance Sometimes we are interested in making inference about the variability of processes.Examples:The consistency of a production process for quality control purposes.Investors use variance as a measure of risk.To draw inference about variability, the parameter of interest is s2.
7 12.3 Inference About a Population Variance The sample variance s2 is an unbiased, consistent and efficient point estimator for s2.The statistic has a distribution called Chi-squared, if the population is normally distributed.d.f. = 5d.f. = 10
8 Confidence Interval for Population Variance From the following probability statement P(c21-a/2 < c2 < c2a/2) = 1-a we have (by substituting c2 = [(n - 1)s2]/s2.)
9 Testing the Population Variance Example 12.3 (operation management application)A container-filling machine is believed to fill 1 liter containers so consistently, that the variance of the filling will be less than 1 cc (.001 liter).To test this belief a random sample of 25 1-liter fills was taken, and the results recorded (Xm12-03). s2=Do these data support the belief that the variance is less than 1cc at 5% significance level?Find a 99% confidence interval for the variance of fills.
12 12.4 Inference About a Population Proportion When the population consists of nominal data (e.g., does the customer prefer Pepsi or Coke), the only inference we can make is about the proportion of occurrence of a certain value.When there are two categories (success and failure), the parameter p describes the proportion of successes in the population. The probability of obtaining X successes in a random sample of size n from a large population can be calculated using the binomial distribution.
13 12.4 Inference About a Population Proportion Statistic and sampling distributionthe statistic used when making inference about p is:Under certain conditions, [np > 5 and n(1-p) > 5], is approximately normally distributed, with m = p and s2 = p(1 - p)/n.
14 Testing and Estimating a Proportion Test statistic for pInterval estimator for p (1-a confidence level)
15 Why are Proportions Different? The true variance of a proportion is determined by the true proportion:The CI of a proportion is NOT derived from the z-test:The denominator of the z-statistic is NOT estimated, but the width of the CI is estimated.=> “CI test” and z-test can differ sometimes.
16 Testing the Proportion Example 12.5 (Predicting the winner in election day)Voters are asked by a certain network to participate in an exit poll in order to predict the winner on election day.The exit poll consists of 765 voters say that they voted for the Republican candidate.The polls close at 8:00. Should the network announce at 8:01 that the Republican candidate will win?
18 Selecting the Sample Size to Estimate the Proportion Recall: The confidence interval for the proportion isThus, to estimate the proportion to within W, we can writeThe required sample size is:
19 Sample Size to Estimate the Proportion ExampleSuppose we want to estimate the proportion of customers who prefer our company’s brand to within .03 with 95% confidence.Find the sample size neededSolutionW = .03; 1 - a = .95,therefore a/2 = .025,so z.025 = 1.96Since the sample has not yetbeen taken, the sample proportionis still unknown.We proceed using either one of thefollowing two methods:
20 Sample Size to Estimate the Proportion Method 1:There is no knowledge about the value ofLet This results in the largest possible n needed for a a confidence interval of the formIf the sample proportion does not equal .5, the actual W will be narrower than .03 with the n obtained by the formula below.Method 2:There is some idea about the value ofUse the value of to calculate the sample size
21 Chapter 12: Introduction Variety of techniques are presented whose objective is to compare two populations.We are interested in:The difference between two means.The ratio of two variances.The difference between two proportions.
22 Inference about the Difference between Two Means Example 13.1Do people who eat high-fiber cereal for breakfast consume, on average, fewer calories for lunch than people who do not eat high-fiber cereal for breakfast?A sample of 150 people was randomly drawn. Each person was identified as an eater or non-eater of high fiber cereal.For each person the number of calories consumed at lunch was recorded. There were 43 high-fiber eaters who had a mean of calories for lunch with s= There were 107 non-eaters who had a mean of calories for lunch with s=
23 13.2 Inference about the Difference between Two Means: Independent Samples Two random samples are drawn from the two populations of interest.Because we compare two population means, we use the statistic
24 The Sampling Distribution of is normally distributed if the (original) population distributions are normal .is approximately normally distributed if the (original) population is not normal, but the samples’ size is sufficiently large (greater than 30).The expected value of is m1 - m2The variance of is s12/n1 + s22/n2
25 Making an inference about m1 – m2 If the sampling distribution of is normal or approximately normal we can write:Z can be used to build a test statistic or a confidence interval for m1 - m2
26 Making an inference about m1 – m2 Practically, the “Z” statistic is hardly used, because the population variances are not known.??Instead, we construct a t statistic using thesample “variances” (s12 and s22) to estimate
27 Making an inference about m1 – m2 Two cases are considered when producing the t-statistic:The two unknown population variances are equal.The two unknown population variances are not equal.
29 Inference about m1 – m2: Equal variances Construct the t-statistic as follows:Perform a hypothesis testH0: m1 - m2 = 0H1: m1 - m2 > 0Build a confidence intervalor < 0or 0
30 Example 13.1Assuming that the variances are equal, test the scientist’s claim that people who eat high-fiber cereal for breakfast consume, on average, fewer calories for lunch than people who do not eat high-fiber cereal for breakfast at the 5% significance level.There were 43 high-fiber eaters who had a mean of calories for lunch with s= There were 107 non-eaters who had a mean of calories for lunch with s=
33 Inference about m1 – m2: Unequal variances Conduct a hypothesis testas needed, or,build a confidence interval
34 Which case to use: Equal variance or unequal variance? Whenever there is insufficient evidence that the variances are unequal, it is preferable to perform the equal variances t-test.This is so, because for any two given samplesThe number of degreesof freedom for the equalvariances caseThe number of degreesof freedom for the unequalvariances case
35 Example 13.1 continuedTest the scientist’s claim about high-fiber cereal eaters consuming less calories than non-high fiber cereal eaters assuming unequal variances at the 5% significance level.There were 43 high-fiber eaters who had a mean of calories for lunch with s= There were 107 non-eaters who had a mean of calories for lunch with s=