Hypothesis testing for parametric data (1)
- 1. Hypothesis : a statement about one or more populations The null hypothesis, 𝐻0, is the hypothesis that is to be tested. The alternative hypothesis, 𝐻1is the hypothesis that in some sense contradicts the null hypothesis.
- 2. Variable :characteristic that we would like to measure on individuals. Data :measurements recorded on individuals Dependent: outcome Independent : influence on the outcome
- 4. Steps in hypothesis testing 1.State the null &alternative hypothesis 2.Decide one or two sides test 3.Select the test statistic 4.Select the confidence level, establish cut-off level and draw the acceptance and rejection region 5.Compute the test statistic 6.Plot the test statistic in the curve and make the conclusions
- 5. Using P-value to make a decision about whether to reject, or not reject, your null hypothesis
- 6. Example: Interpreting a p-value – blood pressure before and after exercise In the example of examining change in blood pressure before and after exercise in 16 men the p-value was less than 0.001. What does p < 0.001 mean? Your results are unlikely when the null hypothesis is true. Is this result statistically significant? The result is statistically significant because the p- value is less than the significance level α set at 5% or 0.05. You decide? That there is sufficient evidence to reject the null hypothesis and accept the alternative hypothesis that there is a difference (a rise) in the mean blood pressure of middle-aged men before and after exercise.
- 7. Preparing for statistical testing 1) Identify type of data 2) Identify number of groups studied
- 9. One Group: Comparison of one sample mean with mean of reference population Z= 𝑿−𝝁 𝝈/ 𝒏 T= 𝑿−𝝁 𝑺/ 𝒏 X: sampe mean s:estimated standard deviation n: sample size μ:hypothesized mean Z or T test are basically the same T test: - If a sample is small (n< 30) - If we do know the population standard deviation
- 10. Boys of a certain age have a mean weight of 8.5 Kg. An observation was made that in a city neighborhood, children were underfed. As evidence, all 25 boys in the neighborhood of that age were weighed and found to have a mean 𝑥 of 8.9 and a standard deviation s of 1.16 Kg . An application of the procedure above yields SE(𝑥)= 𝑆 𝑛 = 1.16 25 =0.232 t= 8.9−8.5 0.232 =1.724
- 11. The underfeeding complaint corresponds to the one-sided alternative 𝐻𝐴: μ< 8.5so that we would reject the null hypothesis if t ≤ tabulated value. From Appendix C and with 24 degrees of freedom (n-1) , we find that tabulated value= 1.71 under the column corresponding to a 0.05 upper tail area; the null hypothesis is rejected at the 0.05 level. In other words, there is enough evidence to support the underfeeding complaint.
- 12. Comparing means two independent groups
- 13. This test is referred to as a two-sample t test and its rejection region is determined using t distribution at 𝑛1 + 𝑛2 − 2 degrees of freedom: For a one-tailed test, use the column corresponding to an upper tail area of 0.05 and 𝐻0 is rejected if t≤ tatabulated value for 𝐻𝐴:μ1<μ2ort ≥tabulated value for 𝐻𝐴:μ1 > μ2 For a two-tailed test or 𝐻𝐴 :μ1#μ2 , use the column corresponding to an upper tail area of 0.025 and 𝐻0 is rejected if t ≤ tabulated value
- 14. In an attempt to assess the physical condition of joggers, a sample of 𝑛1 =25 joggers was selected and their maximum volume of oxygen uptake (V𝑂2) was measured with the following results: 𝑥1=47.5 mL/kg 𝑠1= 4.8 mL/kg Results for a sample of 𝑛2 =26 nonjoggers were 𝑥2= 37.5 mL/kg 𝑠2= 5.1 mL/kg To proceed with the two-tailed, two-sample t test, we have
- 15. indicating a significant difference between joggers and nonjoggers (at 49 degrees of freedom , the tabulated t value, with an upper tail area of 0.025, is about 2.0).
- 16. comparing means two dependent groups It is applied to cases where each subject or member of a group is observed twice (e.g., before and after certain interventions),or matched pairs are measured for the same continuous characteristic. What we really want to do is to compare the means, before versus after or cases versus controls, and use of the sample of differences ( 𝑑𝑖 , one for each subject, helps to achieve that.
- 17. The systolic blood pressures of n=12 women between the ages of 20 and 35 were measured before and after administration of a newly developed oral contraceptive: SubjectBefore After Differences (𝑑𝑖) 𝑑𝑖 2 1 122 127 5 25 2 126 128 2 4 3 132 140 8 64 4 120 119 -1 1 5 142 145 3 9 6 130 130 0 0 7 142 148 6 36 8 137 135 -2 4 9 128 129 1 1 10 132 137 5 25 11 128 128 0 0 12 129 133 4 16 31 185
- 18. 𝑑= 31 12 =2.58 mm Hg = 𝑑2 − 𝑑 2/𝑛 𝑛−1 = 185− 31 2/12 11 =9.54 𝒔𝒅=3.09 SE(𝑑)= 3.09 12 = 0.89 t= 2.58 0.89 =2.90 Degree of freedom=n-1 =12-1=11 Using the column corresponding to the upper tail area of 0.05 in Appendix C, we have a tabulated value of 1.796 for 11 df. Since t= 2.90 >1.796 we conclude that the null hypothesis of no blood pressure change should be rejected at the 0.05 level; there is enough evidence to support the hypothesis of increased systolic blood pressure (one-sided alternative).
- 20. Exercise : Vision, especially visual acuity, depends on a number of factors. A study was undertaken in to determine the effect of one of these factors: racial variation. Visual acuity of recognition as assessed in clinical practice has a defined normal value of 20/20 (or zero on the log scale). The following summarize the data on monocular visual acuity (expressed on a log scale).