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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

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Chapter 12

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  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

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