Lecture 17c: SPSS output for One Sample Hypothesis Test of Means (or t-tests)

Transcription

1 17c_SPSS.pdf Michael Hallstone, Ph.D. Lecture 17c: SPSS output for One Sample Hypothesis Test of Means (or t-tests) The purpose of this lecture is to illustrate the SPSS output for one-sample hypothesis test of means (or one sample t-tests). One sample hypothesis test of means or t-test of mean number of complaints per month Here I use the last example from (17_one_sample_t_test.pdf). lecture 17: one sample hypothesis testing Please see the lecture, but in this example we pretend there is a Big Boss in the City and County of Honolulu who wants examine the mean number of complaints coming from a department she supervises. The hypothesized population mean number of complaints per month in step 1 is 16. You can test any of the following using the SPSS output below: H o : µ = 16 H 1 : µ 16 H o : µ 16 H 1 : µ < 16 H o : µ 16 H 1 : µ > 16 1 of 10

2 Doing the test using SPSS To have SPSS perform this operation from the menus choose: Analyze Compare Means > One Sample T-Test... of 10

3 A window will pop up and you highlight and move your continuous variable over to the Test Variables: box with the arrow. Then below there is a box that says Test Value. You put the number of the hypothesized population mean in the Test Value box! You put the number that appears in Step 1! You put the number in the H o and H 1! Above in our step 1 we use H o : µ = 16 H 1 : µ 16 so, we put 16 in the Test Value box. Do not confuse this with alpha in step which is typically.05! Do not put.05 in this box! You put the number that appears in your null and alternative hypothesis! Push OK and you will see the following 3 of 10

4 One-Sample Statistics N Mean Std. Deviation Std. Error Mean # of complaints One-Sample Test t df Sig. (-tailed) Test Value = 16 Mean Difference 95% Confidence Interval of the Difference Lower Upper # of complaints Test Ratio (TR) value in step 6 Look in the lower box under t and you will see the test ratio statistical test from step 6. I have highlighted it in green. x µ σ Ho In this case TR= =.05 / 10 = = x For our take home tests you do NOT have to show me the math by hand like above. Simply say SPSS says TR= of 10

5 p-values The p-value is found in the box Sig. (-tailed). This the actual p-value for a two-tailed test, but computing the p-value for one-tailed tests is more complicated. P value for a two tailed test The output below can be used for the following two-tailed test: H o : µ = 16 H 1 : µ 16 One-Sample Statistics N Mean Std. Deviation Std. Error Mean # of complaints One-Sample Test t df Sig. (-tailed) Test Value = 16 Mean Difference 95% Confidence Interval of the Difference Lower Upper # of complaints Look a the bottom box. SPSS provides the two tailed p value in the box that says Sig. (- tailed). I have highlighted in yellow. In this case p=.013 (or 1.3%) That means you if you reject the null you have to accept a 1.3% chance of being wrong. Or there is a 1.3% chance that the null hypothesis is correct. p-value in plain English In this case p=.013. That means you if you reject the null and conclude the mean number of complaints per month is not equal to 16, you would have to accept a 1.3% chance of being wrong. Whether or not you would actually reject null or fail to reject null would depend upon your amount of error in step of the seven steps. If your error or alpha was.05 or 5% you would reject null. If you error or alpha was.01 or 1% you would fail to reject the null. See the practice problem answers to this lecture for examples of plain English language when you reject null and fail to reject null.) 5 of 10

6 P value for a one tailed test TR value falls in correct tail Divide the p-value by two and that is the one-tailed p-value if the mathematical sign of the t- value is in the correct tail! If t is positive it lands to the right of the mean, negative t-values are to the left. The t-value must land in the appropriate tail. If testing the following: H o : µ 16 H 1 : µ > 16 The t value needs to be positive and our step 4/5 would look like In this case the p-value is the total area under the curve from the TR all the way out through the right or positive side of the curve regardless of whether or not the TR is negative or positive! H 0 : µ 16 H 1 : µ > 16 If the TR value is in the correct tail the p value is half of the Sig. (-tailed) value or.013 divided by..013 =.0065 so p=.0065 or 0.65%. p value in plain English Recall we tested H 0 : µ 16 complaints per month H 1 : µ > 16 complaints per month 6 of 10

7 In plain English, the Big Boss can conclude that the mean number of complaints per month is greater than 16, but there is a 0.65% chance that his conclusion is wrong or in error. 7 of 10

8 t value falls in the wrong tail If testing the following: H o : µ 16 H 1 : µ < 16 In this case the p-value is the total area under the curve from the TR all the way out through the left or negative side of the curve regardless of whether or not the TR is negative or positive! The TR value needs to be NEGATIVE and ours is POSITIVE (TR= ). Thus p-value = 1 - Sig.( tailed) using the output above p= = = so p=.9935 or 99.35%. p value in plain English If you were to reject the null and conclude the mean number of complaints a month was less than 16, you would have to accept a 95.95% chance of being wrong. (This amount of error is too high. Therefore you would fail to reject the theory in the null hypothesis.) 8 of 10

9 p-values when Sig. (-tailed). =.000 The p-value is found in the box Sig. (-tailed). Above it is p=.000, but p is NOT really equal to zero. p-value for two tailed tests The output above can be used for the following two-tailed test: H o : µ =.08 H 1 : µ.08. In this case p=.000, but that does not mean p=0. It is just smaller than.000 so p< For One-Tailed Tests: 1. Divide the p-value by two and that is the one-tailed p-value if the mathematical sign of the t-value is in the correct tail! If t is positive it lands to the right of the mean, negative t- values are to the left. The t-value must land in the appropriate tail. t value falls in correct tail If testing the following: H o : µ.08 H 1 : µ >.08 The t value needs to be positive and ours is (t=19.154). Thus p value is half of the Sig. (- tailed) value or p<.0001 divided by = p< or p<.005% So you could reject the null and conclude the mean Blood Alcohol Content was greater than.08 (µ >.08), but you would have to accept LESS than a.005% chance of being wrong. (That is very small amount of error, so you would reject the theory in the null and conclude the alternative!) 9 of 10

10 t value falls in the wrong tail If testing the following: H o : µ.08 H 1 : µ <.08 The t value needs to be NEGATIVE and ours is POSITIVE (t=19.154). Thus p-value > 1 - Sig.( tailed) NOTE that P is GREATER THAN: 1 - Sig.( tailed) Using the output above p> = or % If you were to reject the null and conclude mean Blood Alcohol Content was less than.08 (µ <.08) you would have to accept MORE THAN a % chance of being wrong. 10 of 10