1 / 29

Chapter 12

Chapter 12. Tests of a Single Mean When σ is Unknown. A Research Question. Children ’ s growth is stunted by a number of chemicals (lead, arsenic, mercury) The tap water in the local community contains a bit of each of these chemicals

dwight
Download Presentation

Chapter 12

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 12 Tests of a Single Mean When σ is Unknown

  2. A Research Question • Children’s growth is stunted by a number of chemicals (lead, arsenic, mercury) • The tap water in the local community contains a bit of each of these chemicals • Are children in this town smaller than other children their age?

  3. A Research Project • 16 (n = 16) 6-yr old children are randomly selected from around town • Each child’s height is measured • In the US the average height of 6-yr olds is 42”(μ = 42) • The variance of 6-yr-old’s height, however is not known

  4. The Data • The 16 kid’s heights were: 44, 38, 42, 37, 35, 41, 46, 39, 40, 42, 34, 39, 41, 42, 45, 35

  5. Hypothesis Test 1. State and Check Assumptions Heights normally distributed ? - probably (n = 16 large enough) Interval level data Random Sample Population variance unknown

  6. Hypothesis Test • Null and Alternative Hypotheses HO : μ = 42 (6-yr old’s height is 42”) HA : μ < 42 (6yr-old’s height is less than 42”)

  7. Hypothesis Test 3. Choose Test Statistic Parameter of interest - μ Number of Groups - 1 Independent Sample Normally distributed - probably Variance - unknown

  8. What do we do? • z-test requires that we know the population standard deviation (σ) • Can we use s as a substitute for σ? • Not with a z statistic, but… • We can use s with a t statistic (Student’s t) and a t sampling distribution

  9. Single Sample t statistic

  10. Back to the Hypothesis Test 3. Choose the test statistic Parameter of interest - μ Number of Groups - 1 Independent Samples Normally distributed - probably Variance - unknown One Sample t-test

  11. Hypothesis Test 4. Set significance level α = .05 critical value is found in table C What’s a df?

  12. Degrees of Freedom (df) • Degrees of Freedom (df) - the number of components in a statistic’s calculation that are free to vary

  13. df Explained • If you have a M = 10 obtained from 5 scores, what are the scores? • Let’s say the first four are 15, 10, 11, and 5 • in this case the last score has to be 9, in order to have a mean of 10 • As a second example, let’s say the first four are 8, 14, 3, and 11 • the last score has to be 14 in order to have a mean of 10

  14. df Explained • Therefore, the first 4 scores can vary, the fifth score is not free to vary - it must take on some value (in order to maintain the mean of 10) • In our example, there are 4 degrees of freedom • The first four scores can take on any value (they are free to vary), but that last one is fixed in order to maintain the mean

  15. One Sample t test • In a one sample t test the degrees of freedom are always equal to n - 1 • df = n -1

  16. Back to the Hypothesis test 4. Set significance level and make decision rule α= .05 df = n -1 = 16 - 1 = 15 critical value at .05 of t(15) = 1.753 (read: “critical value at .05 of a t test with 15 degrees of freedom is 1.753”) But, since we have a directional hypothesis (< 42), then the critical value is -1.753 Thus, if our computed t ≤ -1.753, we reject HO

  17. Or… • If we compute the p-value associated with our t, with 15 df, we can state the decision rule as: • If p ≤ α, Reject the HO

  18. Hypothesis Test 5. Compute test statistic

  19. Hypothesis Test 6. Draw conclusions Since our obtained t (-2.236) is less than the critical t (-1.753) we, Reject HO, and conclude That our town’s 6-yr olds are smaller, on average, than 6-yr olds in the US

  20. Careful…a warning • We have rejected the HOand concluded that our town’s 6-yr-olds are smaller, on average, than 6-yr-olds in the US • But, we are not allowed, in this case, to conclude that it is because of chemicals in the water, or any other cause

  21. Alternative Explanations • There are likely many causes for children’s small stature, not limited to: • Genetics • Diet • Environmental contaminants • Chemicals in ground water • Etc. • The hypothesis test allows us to conclude that these children are smaller, on average, but does not allow us to say why

  22. Before we move on… • Although we already rejected the null hypothesis, • We can determine the actual probability of our results if the null hypothesis were true (p-value) • We know that it is less than .05, but how much less?

  23. Ugghh!!!

  24. Excel recognizes only positive values for a t distribution, but because the t is symmetrical, use the absoute value function (ABS) to find the p-value

More Related