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T Test by SPSS

T Test by SPSS. by Aaed Al- Rabai. Two sample t. Standardized Difference. The case of equal variance. 2 sample t test (assuming equal variance). t-test for 2 independent samples. Null Hypothesis: No difference in mean blood PH levels between battery workers and control group

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T Test by SPSS

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  1. T Test by SPSS by Aaed Al- Rabai

  2. Two sample t

  3. Standardized Difference

  4. The case of equal variance

  5. 2 sample t test (assuming equal variance)

  6. t-test for 2 independent samples Null Hypothesis: No difference in mean blood PH levels between battery workers and control group i.e. Ho : mbattery = mcontrol Alternative Hypothesis: H1 : mbattery > mcontrol because battery workers are occupationally exposed. One-sided test Blood PH concentrations Battery workers Control (occupationally (not occupationally exposed) exposed) 0.082 0.040 0.080 0.035 0.079 0.036 0.069 0.039 0.085 0.040 0.090 0.046 0.086 0.040 Mean 0.08157 0.03943 Variance 0.00004495 0.00001262

  7. t-test for 2 independent samples From data: Difference=.08157-.03943=.04214 Sp=(6X.00004495+6X.00001262) 12 = 0.00002879 SE =Sp X sqrt(1/7+1/7) = 0.002868 T = standardized difference = .04214/.002868 = 14.7 with 12 df Blood PH concentrations Battery workers Control (occupationally (not occupationally exposed) exposed) 0.082 0.040 0.080 0.035 0.079 0.036 0.069 0.039 0.085 0.040 0.090 0.046 0.086 0.040 Mean 0.08157 0.03943 Variance 0.00004495 0.00001262 2

  8. t-test for 2 independent samples Question: If there were really no difference in the mean PH level of the 2 groups, what is the probability that the standardized difference between the 2 sample means will be 14.7 or more due to chance alone? Blood PH concentrations Battery workers Control (occupationally (not occupationally exposed) exposed) 0.082 0.040 0.080 0.035 0.079 0.036 0.069 0.039 0.085 0.040 0.090 0.046 0.086 0.040 Mean 0.08157 0.03943 Variance 0.00004495 0.00001262 Answer: 1-sided p-value = Pr(t12>=14.7) < 0.0005

  9. Upper percentiles of t-distributions Probability df .025 .01 .005 .0005 ------------------------------------------------------ 11 2.201 2.718 3.106 4.437 12 2.179 2.681 3.055 4.318 13 2.160 2.650 3.012 4.221 14 2.145 2.624 2.977 4.140 15 2.131 2.602 2.947 4.073 From our example: t=14.7 with 12 d.f. Value far exceeds 4.318, the upper 0.05-percentile of the t-distribution with 12 df i.e. Pr < 0.0005

  10. t-test for 2 independent samples Conclusion: Since p-value < 0.001, it is extremely unlikely that the observed difference is due to chance or sampling error alone.This suggests that there could be a real difference and there is some evidence that the battery workers may have a higher mean blood PH concentration. Blood PH concentrations Battery workers Control (occupationally (not occupationally exposed) exposed) 0.082 0.040 0.080 0.035 0.079 0.036 0.069 0.039 0.085 0.040 0.090 0.046 0.086 0.040 Mean 0.08157 0.03943 Variance 0.00004495 0.00001262

  11. Assumptions • PH levels normally distributed • Equal variance for the two populations • The two samples are independent The 2 sample t test with pooled variance is quite robust to non-normality and unequal variance when the sample sizes are equal If the sample sizes are quite different, the test will be affected by unequal variance and should be used with caution

  12. Topic 13: Paired T Test

  13. Blood samples from 11 individuals were collected before and after they smoked a cigarette and the % of blood platelet aggregation recorded

  14. Advantages of pairing • Since the same person acts as his/her own control, the observed difference is more likely to be due to treatment rather than by chance or other factors • Effect of confounding factors minimized or controlled for • Cut down extraneous source of variation. Difference between 2 measurements of the same individual is typically less variable than the difference in measurements between 2 individuals • Higher precision for estimating the mean difference, resulting in a more powerful t-test

  15. Why 2-sample t test not applicable to paired data? • The two observations within the same pair are likely to be positively correlated, violating the assumption of independence • SE for observed difference obtained assuming independence over-estimates the true SE, making the standardized difference smaller than it should be Simple Remedy: Take difference within each pair and apply 1-sample t test to the differences

  16. D

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