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Student's t-test

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Short notes on how to answer t-test question for the coming exam

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Student's t-test

  1. 1. Dr Azmi Mohd Tamil Normally distributed data Comparison between 2 groups Student’s T-Test 1 drtamil@gmail.com 2015
  2. 2. Concept drtamil@gmail.com 20152  Basically we are comparing the mean of the outcome (as measured in continuous data) among the treated against the mean of the outcome among the control.  If the treatment is effective, we would expect those treated to have a significantly higher (“causative”) or lower (“protective”) mean of the outcome compared to those in the control group. n Mean HDL S.D. Fish Oil + 30 1.4767 0.45549 Control - 30 1.2593 0.44488
  3. 3. Research Question  Will those treated with fish oil supplement have a higher level of HDL compared to control group?  If it is true that fish oil is good for you, then we expect a significantly higher level of HDL among those treated with fish oil group (1.48+0.46) compared to the control group (1.26+0.44).  Student’s t-test will test whether the HDL mean among treated group (1.48) is significantly higher than the HDL mean of control group (1.26). 3 drtamil@gmail.com 2015 n Mean HDL S.D. Fish Oil + 30 1.4767 0.45549 Control - 30 1.2593 0.44488
  4. 4. Null hypothesis  Null hypothesis assumes that there is no association between fish oil usage and HDL level. As though taking fish oil has no effect on the HDL level.  If taking fish oil has no effect on HDL, then there should be no difference of mean of HDL level between treatment and control groups.  The statistical test conducted is to decide whether or not to reject the null hypothesis. So if the mean HDL level of treatment group is significantly different than the mean HDL of control group, null hypothesis is rejected. 4 drtamil@gmail.com 2015
  5. 5. Significance  We expect better HDL level among the treated group, compared to the control group. The better the improvement, the larger is the mean difference between the two groups.  The larger the difference, the more significant is the p value.  The bigger the mean difference, the bigger is the t value, therefore the smaller is the p value, therefore more likely to be significant.  The smaller the variance, the smaller is the p value.
  6. 6. Significance  The larger the difference, the more significant is the p value. But here the difference is only 0.2174, which is even smaller than the standard deviation (0.45549 & 0.44488)  The small mean difference, and the large standard deviation/variance will lead to a smaller t value, therefore larger p value, therefore more likely to be insignificant. n Mean HDL S.D. Fish Oil + 30 1.4767 0.45549 Control - 30 1.2593 0.44488
  7. 7. Example:  t = (1.4767 – 1.2593)_____ (0.455492/30 + 0.444882/30)0.5  t = 0.2174/0.1162=1.870  df = (30-1)+(30-1) = 58  Critical value for df = 60 for p=0.05 is 2.00,  Critical value for df = 40 for p=0.05 is 2.02,  The calculated t is smaller than the critical value, therefore the null hypothesis is NOT REJECTED.  Conclusion: Although the treated group HDL level is slightly higher (1.48+0.46) compared to the control group (1.26+0.44), the difference is not statistically significant (p>0.05)  Hint: Memorise critical value of 1.96 for df=infinity.7 drtamil@gmail.com 2015
  8. 8. Refer to Table A3. We don’t have df=58, so we use df=40 instead. Calculated t = 1.870 < 2.02 If t=2.02, p=0.05 Therefore if t=1.870, p>0.05.
  9. 9. As analysed by SPSS drtamil@gmail.com 20159
  10. 10. Summary drtamil@gmail.com 201510  Hypothesis – those treated with fish oil supplement have a higher level of HDL compared to control group.  Null hypothesis  No difference of mean of HDL level between treatment and control groups, or;  There is no association between fish oil treatment and HDL level.  Suitable statistical test – Student’s t-test (compare mean between 2 groups, n=60-assume normally distributed data).
  11. 11. Summary drtamil@gmail.com 201511  Calculation;  t = (1.4767 – 1.2593)_____ (0.455492/30 + 0.444882/30)0.5 = 0.2174/0.1162=1.870  df = (30-1)+(30-1) = 58  p > 0.05  Null hypothesis NOT REJECTED because p > 0.05  Conclusion:  Although the treated group HDL level is slightly higher (1.48+0.46) compared to the control group (1.26+0.44), the difference is not statistically significant (p>0.05); or  There is no association between fish oil usage and HDL level.

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