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Hypothesis Testing

Learn about hypothesis testing, null hypothesis, types of errors, decision-making in statistics, and practical examples. Understand the statistical decision-making framework and interpretation of results.

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Hypothesis Testing

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  1. Hypothesis Testing Introduction to Inductive Statistics

  2. Background Website • http://www.intuitor.com/statistics/T1T2Errors.html

  3. Terms • Descriptive Statistics: what we’ve done so far… • Inductive Statistics: making decisions on the basis of statistical evidence • Hypothesis: the relationship or proposition you wish to test, stated to affirm the relationship or proposition • Null Hypothesis: the negative form of the proposition to be tested

  4. Some propositions… • Statistics is making decisions on the basis of incomplete or imperfect information. • A hypothesis or proposition can be refuted by one observation but not proved by many. • We thus proceed by determining the likelihood that the null hypothesis can be rejected.

  5. The Dilemma… There is always the possibility of making the wrong decision: • Rejecting a true hypothesis: Type 1 Error • Failing to reject a false hypothesis: Type 2 Error

  6. An Example of the Issue: A Jury’s Decision Making

  7. An Example of the Issue: A Jury’s Decision Making

  8. The Statistical Decision Making Framework

  9. The Statistical Decision Making Framework

  10. The Jury and the Researcher Compared

  11. Steps for Making a Decision • Specify a hypothesis and the null hypothesis • Specify a level of probability which you will use to decide whether to reject the null hypothesis. • Specify the test statistic and the sampling distribution you will use to make a decision. • Calculate the statistics and compare to the theoretical probability distribution, for example, the t distribution. • Interpret the results.

  12. General Form of the Test for a Mean • Z tests: • (Sample mean – population mean) / SE of sample mean, or • (Sample mean – population mean)/ (s / √n) • T tests: • (Sample mean – population mean) / SE of sample mean, or • (Sample mean – population mean)/ (s / √n-1)

  13. General Form of a T Test • t = sample estimate – null hypothesis/ SE • Which simplifies to: • t = sample estimate/SE • When the null hypothesis is that the sample statistic is 0.

  14. T Distribution

  15. Example 1 • Hypothesis: There is a difference in the average number of persons per household in the 18th and the 14th wards. • Null Hypothesis: There is no difference in the average number of persons per household in the 18th and the 14th wards, or more specifically, any difference we measure is a matter of the particular sample we have.

  16. Example, cont. • Level of probability: 95% confidence level, so that only 1 in 20 times would the results be different. • Test statistic: Means and a T-Test of the difference of two groups. • t = (mean1 – mean2)/ (SE of the difference of mean1-mean2) • Calculate the statistics….

  17. Results Two-sample t test on PERSONS grouped by WARD Group N Mean SD 14 316 8.15 3.70 18 120 5.25 2.55 Separate Variance t = 9.29 df = 310.5 Prob = 0.00 Difference in Means = 2.90 95.00% CI = 2.29 to 3.52 Pooled Variance t = 7.90 df = 434 Prob = 0.00 Difference in Means = 2.90 95.00% CI = 2.18 to 3.62

  18. Results, Graphically Displayed

  19. Interpret the Results • Let’s look at the t distribution again. • We can reject the null hypothesis that the two means are the same in the underlying population (the unknown truth). • We say that there is a statistically significant difference between the average number of persons in the two wards.

  20. Example 2 • Hypothesis: There is a difference in the average number of persons per household in the 18th and the 20th wards. • Null Hypothesis: There is no difference in the average number of persons per household in the 18th and the 20th wards, or more specifically, the difference is a matter of the particular sample we have.

  21. Results… TEST PERSONS * WARD Data for the following results were selected according to: (WARD<> 14) AND (ward<> 22) Two-sample t test on PERSONS grouped by WARD Group N Mean SD 18 120 5.25 2.55 20 342 5.72 2.62 Separate Variance t = -1.73 df = 213.5 Prob = 0.08 Difference in Means = -0.47 95.00% CI = -1.01 to 0.07 Pooled Variance t = -1.71 df = 460 Prob = 0.09 Difference in Means = -0.47 95.00% CI = -1.02 to 0.07

  22. Results, Graphically Displayed…

  23. Interpret the Results • We cannot reject the null hypothesis that the two means are the same in the underlying population (the unknown truth). • We say that there is not a statistically significant difference between the average number of persons in the two wards.

  24. An example of the issue: A jury’s decision making

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