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Session 8b. Overview. Hypothesis Testing Review of the Basics Single Parameter Differences Between Two Parameters Independent Samples Matched Pairs Goodness of Fit Simulation Methods. Basic Hypothesis Testing Method. Formulate Two Hypotheses Select a Test Statistic
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Overview Hypothesis Testing • Review of the Basics • Single Parameter • Differences Between Two Parameters • Independent Samples • Matched Pairs • Goodness of Fit • Simulation Methods Decision Models -- Prof. Juran
Basic Hypothesis Testing Method • Formulate Two Hypotheses • Select a Test Statistic • Derive a Decision Rule • Calculate the Value of the Test Statistic; Invoke the Decision Rule in light of the Test Statistic Decision Models -- Prof. Juran
Hypothesis Testing: Gardening Analogy Dirt Rocks Applied Regression -- Prof. Juran
Hypothesis Testing: Gardening Analogy Applied Regression -- Prof. Juran
Hypothesis Testing: Gardening Analogy Applied Regression -- Prof. Juran
Hypothesis Testing: Gardening Analogy Applied Regression -- Prof. Juran
Hypothesis Testing: Gardening Analogy Screened out stuff: Correct decision or Type I Error? Stuff that fell through: Correct decision or Type II Error? Applied Regression -- Prof. Juran
The p-value of a test is the probability of observing a sample at least as “unlikely” as ours. In other words, it is the “minimum level” of significance that would allow us to reject H0. Small p-value = unlikely H0 Decision Models -- Prof. Juran
Example: Buying a Laundromat A potential entrepreneur is considering the purchase of a coin-operated laundry. The present owner claims that over the past 5 years the average daily revenue has been $675. The buyer would like to find out if the true average daily revenue is different from $675. A sample of 30 selected days reveals a daily average revenue of $625 with a standard deviation of $75. Decision Models -- Prof. Juran
Test Statistic: Decision Rule, based on alpha of 1%: Reject H0 if the test statistic is greater than 2.575 or less than -2.575. Decision Models -- Prof. Juran
We reject H0. There is sufficiently strong evidence against H0 to reject it at the 0.01 level. We conclude that the true mean is different from $675. Decision Models -- Prof. Juran
Example: Reliability Analysis Decision Models -- Prof. Juran
The brand manager wants to begin advertising for this product, and would like to claim a mean time between failures (MTBF) of 45 hours. The product is only in the prototype phase, so the design engineer uses Crystal Ball simulation to estimate the product’s reliability characteristics. Extracted data: Decision Models -- Prof. Juran
H0 here is that the product lasts 45 hours (on the average). There is sufficiently strong evidence against H0 to reject it at any reasonable significance level. We conclude that the true MTBF is less than 45 hours. Decision Models -- Prof. Juran
Example: Effects of Sales Campaigns In order to measure the effect of a storewide sales campaign on nonsale items, the research director of a national supermarket chain took a random sample of 13 pairs of stores that were matched according to average weekly sales volume. One store of each pair (the experimental group) was exposed to the sales campaign, and the other member of the pair (the control group) was not. Decision Models -- Prof. Juran
The following data indicate the results over a weekly period: Decision Models -- Prof. Juran
Is the campaign effective? Basically this is asking: Is there a difference between the average sales from these two populations (with and without the campaign)? Decision Models -- Prof. Juran
Two Methods • Independent Samples • General Method • Matched Pairs • Useful Only in Specific Circumstances • More Powerful Statistically • Requires Logical One-to-One Correspondence between Pairs Decision Models -- Prof. Juran
Independent Samples Method Decision Models -- Prof. Juran
We do not reject the null hypothesis. The campaign made no significant difference in sales. Decision Models -- Prof. Juran
Matched-Pairs Method Decision Models -- Prof. Juran
This time we do reject the null hypothesis, and conclude that the campaign actually did have a significant positive effect on sales. Decision Models -- Prof. Juran
TSB Problem Revisited Decision Models -- Prof. Juran
Independent Samples Decision Models -- Prof. Juran
Matched Pairs Decision Models -- Prof. Juran
Goodness-of-Fit Tests • Determine whether a set of sample data have been drawn from a hypothetical population • Same four basic steps as other hypothesis tests we have learned • An important tool for simulation modeling; used in defining random variable inputs Decision Models -- Prof. Juran
Example: Warren Sapp Financial analyst Warren Sapp wants to run a simulation model that includes the assumption that the daily volume of a specific type of futures contract traded at U.S. commodities exchanges (represented by the random variable X) is normally distributed with a mean of 152 million contracts and a standard deviation of 32 million contracts. (This assumption is based on the conclusion of a study conducted in 1998.) Warren wants to determine whether this assumption is still valid. Decision Models -- Prof. Juran
He studies the trading volume of these contracts for 50 days, and observes the following results (in millions of contracts traded): Decision Models -- Prof. Juran
Here is a histogram showing the theoretical distribution of 50 observations drawn from a normal distribution with μ = 152 and σ = 32, together with a histogram of Warren Sapp’s sample data: Decision Models -- Prof. Juran
The Chi-Square Statistic Decision Models -- Prof. Juran
Essentially, this statistic allows us to compare the distribution of a sample with some expected distribution, in standardized terms. It is a measure of how much a sample differs from some proposed distribution. A large value of chi-square suggests that the two distributions are not very similar; a small value suggests that they “fit” each other quite well. Decision Models -- Prof. Juran
Like Student’s t, the distribution of chi-square depends on degrees of freedom. In the case of chi-square, the number of degrees of freedom is equal to the number of classes (a.k.a. “bins” into which the data have been grouped) minus one, minus the number of estimated parameters. Decision Models -- Prof. Juran
Note: It is necessary to have a sufficiently large sample so that each class has an expected frequency of at least 5.We need to make sure that the expected frequency in each bin is at least 5, so we “collapse” some of the bins, as shown here. Decision Models -- Prof. Juran
The number of degrees of freedom is equal to the number of bins minus one, minus the number of estimated parameters. We have not estimated any parameters, so we have d.f. = 4 – 1 – 0 = 3. The critical chi-square value can be found either by using a chi-square table or by using the Excel function: =CHIINV(alpha, d.f.) = CHIINV(0.05, 3) = 7.815 We will reject the null hypothesis if the test statistic is greater than 7.815. Decision Models -- Prof. Juran
Our test statistic is not greater than the critical value; we cannot reject the null hypothesis at the 0.05 level of significance. It would appear that Warren is justified in using the normal distribution with μ = 152 and σ = 32 to model futures contract trading volume in his simulation. Decision Models -- Prof. Juran
The p-value of this test has the same interpretation as in any other hypothesis test, namely that it is the smallest level of alpha at which H0 could be rejected. In this case, we calculate the p-value using the Excel function: = CHIDIST(test stat, d.f.) = CHIDIST(7.439,3) = 0.0591 Decision Models -- Prof. Juran
Example: Catalog Company If we want to simulate the queueing system at this company, what distributions should we use for the arrival and service processes? Decision Models -- Prof. Juran
