Prof. Robert Martin Southeastern Louisiana University.

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Presentation transcript:

Prof. Robert Martin Southeastern Louisiana University

1. State null and research hypotheses. 2. Determine what probability distribution you want to use to test the hypothesis. 3. Determine what probability is associated with a “very unlikely” event (alpha). 4. Calculate probability of observing your data under your null hypothesis. 5. Decide whether to reject your null hypothesis.

 Are people who participate in a job training program more reliable workers than people in the larger community?  If we know how many days of work on average people in the whole community miss per year (and the standard deviation of this variable), we can take a random sample of people from the job training program to answer this question.

 Info we already know: Average number of missed days (community) is 7.2 days per year.  From a sample of 127 participants in the program, you learn that the average number of missed days per year is 6.8 days. ◦ The standard deviation of the sample is 1.43 days.

 Is this difference statistically significant? ◦ That is, does it suggest there’s a difference between EVERYONE in the job training program and everyone in the whole community?

 Null hypothesis (H 0 ) = No difference on average days missed of work between people in the community and people in the training program. ◦ H 0 : µ = 7.2 days  Research hypothesis (H 1 ) = There is a difference on average days missed of work between people in the community and people in the training program. ◦ H 1 : µ ≠ 7.2 days ◦ We test the null hypothesis to see how likely it is that it’s accurate.

 We often use the NORMAL DISTRIBUTION to help us test hypotheses. We do this if we have a sample of 100+ people. ◦ This is also known as the Z-distribution. If we assume the sampling distribution is “normal,” we calculate Z scores to test our hypothesis.  The point of picking a probability distribution is to determine what kind of statistic you need to calculate so you can draw a conclusion.

 We have to compare the average days missed in whole community (a population mean) with the average days missed from our sample.  We would reject our null hypothesis if it’s unlikely that the sample mean was drawn from a sampling distribution that has the same mean as the community (the population mean in the community). ◦ But how *unlikely* does this have to be? We pick.

 Probability of Unlikely Event = “Alpha level” or “critical region” - 95% of the area under the Z-distribution is between Z = and Z = These are “critical” Z scores. -95% confidence means you pick an alpha level of.05.

Popular Alpha Levels and critical Z-scores (2-tailed): α =.10, Z = ±1.65 α =.05; Z = ± 1.96 α =.01, Z = ± 2.58

 Alpha Level = Probability we’re willing to accept of making Type I Error  Type I Error = We incorrectly reject a TRUE null hypothesis. ◦ This is the error we least want to make. ◦ We could only find out if the null hypothesis is actually true by having information from the population of people from the program (which we don’t have.)  Type II Error = We fail to reject a FALSE null hypothesis

- If you wanted to be 95% confident that you can reject your null hypothesis, the “Z score” showing the difference between your sample mean was and what you ASSUME it would be if the null hypothesis were true has to be: -LESS THAN -1.96; or -GREATER THAN 1.96.

 The nice thing about a normal distribution is that we know “how far away” our sample statistic is from the mean… using standard deviations.  We calculate a “Z score” and see whether it falls into the critical region of the normal curve (which in this case is the area OUTSIDE the middle 95% of the normal curve).

 We use the information we have to calculate our test statistic, which is our “obtained Z score.”  That’s three pieces of information we got right from our sample, plus one missing piece (µ). ◦ But we got that by assuming our null hypothesis is true. Z obtained =

Our obtained Z score is

 Is the Z-score larger than the Z-score associated with our alpha level?  If it is, we REJECT our null hypothesis and can conclude with a certain amount of confidence that the research hypothesis is accurate.  If it isn’t, we FAIL TO REJECT our null hypothesis. ◦ What do we do here?

 We can do everything we just did to test hypotheses using data from one sample… AS LONG AS THE SAMPLE SIZE IS 100 OR MORE.  If not, we can’t pick the Z-distribution to use in Step 2. We have to use the t-distribution instead (and calculate t-scores). ◦ Critical t-scores can be found in Appendix B. The obtained t-scores are calculated with the same formula as the obtained Z-scores.

 Most hypothesis tests we do are called “two- tailed” tests (they involve both “tails” of the probability distribution). ◦ These are for testing “non-directional hypotheses.”  If you want to test a “directional hypothesis” (e.g., to see if recovering alcoholics from the program miss FEWER days of work than the community at large), you need to do a one- tailed test. ◦ See critical Z-scores on page 314; appropriate column in Appendix B for critical t-scores.

 We can use the same procedure to see whether proportions of people from a single sample are different from the proportions in the community they’re drawn from.  Differences are: we don’t have to assume that the variable is “interval-ratio,” and the test statistic is calculated differently.

From a random sample of 122 households in one neighborhood, a proportion of.43 say they plan to vote to re-elect the mayor. In the whole city,.39 of households say they plan to vote to re-elect the mayor. Is that neighborhood significantly different from the whole city?

p is the population proportion that you think exists. - Here, it’s the proportion of the WHOLE NEIGHBORHOOD that you think would vote to re-elect the mayor, IF the null hypothesis were right. is the sample proportion. This is referred to as “p – hat.”

The obtained Z-score is If we wanted to be 95% confident in rejecting our null hypothesis, we’d compare this obtained Z- score to the critical Z-score of ±1.96. What would we conclude?