Presentation is loading. Please wait.

Presentation is loading. Please wait.

Biostatistics Case Studies 2005 Peter D. Christenson Biostatistician Session 6: “Number Needed to Treat” to Prevent One Case.

Similar presentations


Presentation on theme: "Biostatistics Case Studies 2005 Peter D. Christenson Biostatistician Session 6: “Number Needed to Treat” to Prevent One Case."— Presentation transcript:

1 Biostatistics Case Studies 2005 Peter D. Christenson Biostatistician http://gcrc.humc.edu/Biostat Session 6: “Number Needed to Treat” to Prevent One Case

2 Case Study

3 Results

4 Main Figures for Results Studies in Glaucoma Subjects: Studies in Ocular Hypertensive Subjects:

5 Goals of this Session 1.Define “Number Needed to Treat” (NNT). 2.Show how to calculate NNT for published studies analyzed with methods for equal follow up for all subjects. 3.Show how to calculate NNT for published studies analyzed with survival (time-to-event) methods. 4.Reproduce NNT values for the case study. 5.Disadvantages of NNT.

6 NNT: Subjects Followed Equally NNT = the number of subjects who need to be treated in order to expect to prevent one case, relative to subjects not treated. Let p t = proportion of treated subjects who have the event in the fixed time period, and p c the proportion of control subjects with the event. The absolute risk reduction, ARR, is p c -p t. Since ARR is the proportion of treated subjects in whom events were prevented, NNT=1/ARR. 95% CI for NNT is 1/A U to 1/A L, where (A L,A U ) is the 95% CI for ARR.

7 Examples: Subjects Followed Equally NPcPc PtPt ScSc StSt ARRNNTALAL AUAU NNT 95%CI 1000.20.10.80.90.10100.0020.1985.1 to 500 1000.40.20.60.80.2050.0760.3233.1 to 13.2 1000.80.40.20.60.402.50.2760.5231.9 to 3.6 10000.20.10.80.90.10100.0690.1307.6 to 14.5 10000.40.20.60.80.2050.1600.2394.2 to 6.2 10000.80.40.20.60.402.50.3600.4392.3 to 2.8 S t = treated proportion surviving, i.e., w/o event = 1-P t. A L = ARR – 1.96*Standard Error (ARR) = ARR – 1.96*square root [P c *S c /N + P t *S t /N].

8 Comments: Subjects Followed Equally 1.NNT (and ARR) refer only to the fixed follow up time for all subjects. 2.NNT is based on risk difference, not risk ratio (RR). Note that RR = 2.0 for all of the previous examples. 3.CIs can be obtained from reported Ps (or Ss) and Ns using formula on previous slide. 4.Cannot obtain NNT (or ARR) from RR only. Need risk or survival (one of P c, P t, S c, or S t ). Usually underlying risk Pc for target population is used.

9 Generalization to Unequal Subject Follow Up 1.NNT is still specific to a particular specified time of follow up. 2.NNT is still 1/ARR = 1/(S t -S c ), but S t and S c are found from survival methods: either Kaplan-Meier or Cox regression. 3.To find NNT from a published paper, we need either: S t and S c, and for a CI on NNT, either their standard errors (which could be found from their CIs) or the numbers of subjects at risk at the F/U time (which are often in graphs). The hazard ratio (HR) and Sc (or St), and for a CI on NNT, the standard error of HR (which could be found from its CI). This is the typical way that results are reported.

10 NNT Results for the Case Study

11 Reproduce NNT for the Case Study Studies in Ocular Hypertensive Subjects: Hazard Ratio HR = 0.56 (95% CI: 0.39 to 0.81). S c is assumed to be 0.80. The key to obtaining ARR, and thus NNT, is that the Cox regression model assumes that S t = [S c ] HR. Here, S t = [0.80] 0.56 = 0.8825. Thus, ARR = 0.8825-0.80 = 0.0825. NNT = 1/ARR = 1/0.0825 = 12.12, reported as 12. A U = [0.80]0.39 – 0.80 = 0.9167-0.80 = 0.1167. A L = [0.80]0.81 – 0.80 = 0.8346-0.80 = 0.0346. 95% CI for NNT is 1/A U to 1/A L = 1/0.1167 to 1/0.0346 = 8.6 to 28.9, reported as 9 to 29.

12 Reproduce NNT for the Case Study Studies in Glaucoma Subjects: Hazard Ratio HR = 0.65 (95% CI: 0.49 to 0.87). S c is assumed to be 0.40. The key to obtaining ARR, and thus NNT, is that the Cox regression model assumes that S t = [S c ] HR. Here, S t = [0.40] 0.65 = 0.5512. Thus, ARR = 0.5512-0.40 = 0.1512. NNT = 1/ARR = 1/0.1512 = 6.61, reported as 7. A U = [0.40]0.49 – 0.40 = 0.6383-0.40 = 0.2383. A L = [0.40]0.87 – 0.40 = 0.4506-0.40 = 0.0506. 95% CI for NNT is 1/A U to 1/A L = 1/0.2383 to 1/0.0506 = 4.2 to 19.8, reported as 4 to 20.

13 Disadvantages of NNT: Heterogeneity Scaling: NNT differs for subpopulations with different underlying risk, but equal RR. Questionable use for meta analysis summary overall measure since underlying risk may differ among studies. Could find NNT for studies separately and give range of NNTs. In a single study, subgroups may also have differing underlying risk, so again separate NNTs may be more useful.

14 Disadvantages of NNT: Non-Significant Treatment Effect If p<0.05 for treatment effect, then the 95% CI for NNT contains negative numbers; e.g., need to treat between -18 and 8 subjects from the Kamal study! Negative values refer to NNT for harm. Positive values refer to NNT for benefit. Some have suggested an interpretation using the fact that treatment effect zero corresponds to ∞ subjects. In my opinion, NNT should not be used here: if the CI is narrow and contains 0, we are sure there is no effect, so NNT is irrelevant, and if the CI is wide, NNT is not useful anyway.

15 Other References Original NNT suggestion: Cook et al. BMJ 1995; 310: 452-454. More detail on NNT in survival analyses: Altman et al. BMJ 1999; 319: 1492-1495. Negative NNT confidence interval issues: Altman et al. BMJ 1998; 317: 1309-1312.

16 Personal Conclusions NNT can be useful way to express the effort needed (many treated subjects) for treatments with moderate relative risks and small underlying risk. Limit NNT to groups that are homogeneous for underlying risk, and to treatments that show significant effects.


Download ppt "Biostatistics Case Studies 2005 Peter D. Christenson Biostatistician Session 6: “Number Needed to Treat” to Prevent One Case."

Similar presentations


Ads by Google