Stats Tutorial

Is My Coin Fair?

• Assume it is no different from others (null hypothesis)

• When will you no longer accept this assumption?

Beware Multiple Comparisons!

If you give patients a placebo and test

for 6 different things, what’s the chance

that one thing will be found to be“statistically significant (p< 0.05)”

bychance alone?

Answer

.95x.95x.95x.95x.95x.95=0.74

74% chance that you won’t cross the magic 0.05 threshold

OR26% chance that you will conclude there’s a real difference when there isn’t

Confidence Intervals

• RR =1 no difference

• RR < 1 treatment helps

• RR > 1 treatment harms

Treatment Event Rate

Control Event Rate

RR Estimate 95% CI

¼ 2/4 0.50 ?

5/20 10/20 0.5 ?

10/40 20/40 0.5 ?

25/100 50/100 0.5 ?

250/1000 500/1000 0.5 ?

Confidence Intervals

• RR =1 no difference

• RR < 1 treatment helps

• RR > 1 treatment harms

Treatment Event Rate

Control Event Rate

RR Actual 95% CI

¼ 2/4 0.50 0.07-3.5

5/20 10/20 0.5 0.21-1.2

10/40 20/40 0.5 0.26-0.9

25/100 50/100 0.5 0.34-0.74

250/1000 500/1000 0.5 0.44-0.57

Sample Questions

Question

• In a sample of 100 neonates, the mean total WBC is 7500 cells/mm3 with SD 1500 cells/mm3. If total WBC is normally distributed, then approximately 50% of individuals will have a value

a) between 6000 and 9000b) above 9000c) below 6000 or above 9000d) below 7500

Normal Distribution

mean = 7500

68%

95%

Answer

• Below 7500• In a normal distribution, the mean and

the median are the same. If the median is 7500, then half are above and below.

a) 6000-9000 is +/- 1 SD, this encompasses 68% of the population

b) Above 9000 is > 1 SD and that’s (50% - 34%) = 16%

c) Below 6000 or above 9000 is < 1 SD or > 1 SD or 16% +16%=32%

Question

• Again, assuming a normal distribution of WBC, a randomly selected neonate would have a WBC > 10,500:

a) 1% of the timeb) 2.5% of the timec) 5% of the timed) 16% of the time

Answer

• 2.5% of the time• You are 2 SD from the mean;

95% of the distribution lies within 2 SD of the mean.

• 5% of the distribution lies outside the 95% area. Half (2.5%) above and half (2.5%) below.

Question

A case-control study compares treatment in insured and Medicaid patients with preterm labor. The 95% CI for the odds ratio for Medicaid patients being more likely to be under-treated than insured patients was 1.1 to 2.5. Which is true?

a) 95% of the time Medicaid patient are more likely than insured patients to be under-treated

b) The results are not statistically significant (p> 0.05)c) The probability is 95% that odds ratios in similar

studies would fall within these limitsd) Since the observed odds ratio falls in the center of

these limits, the probability is 95% that it is the correct value

Answer

• The probability is 95% that odds ratios in similar studies would fall within these limits Two of the 3 are pure-nonsense

distractors. The 95% CI for the OR does not

cross 1, so there is a statistically significant difference, p < 0.05.

Question

• An evaluation of prophylactic vancomycin to prevent line-associated sepsis enrolls 500 neonates. Subjects are randomly assigned to either vancomycin (n=250) or placebo (n=250). A 20% reduction in risk is considered clinically significant and a = 0.05.

Question

• Line-sepsis occurs in 65 of the placebo group and 50 in the vancomycin group. How many neonates would need to receive prophylactic vancomycin to prevent one from developing line-sepsis?

Answer

• NNT =17

Sepsis

No Sepsis

Vanco 50 200 250

Placebo 65 185 250

Totals 115 385 500

EER = 50/250=.2

CER = 65/250=0.26

RR= 0.2/0.26= 0.77

ARR=0.26-0.20=0.06

NNT=1/0.06=17

Question

• The authors report RR = 0.77 (95% CI 0.5435 to 1.089) You conclude that:

a) There is a statistically significant decrease in line-sepsis in the vancomycin treated group

b) The proportion with line-sepsis is the same between groups

c) The study has proven that prophylactic vancomycin is not effective

d) A larger study is indicated

Answer

• A larger study is indicated The 95% CI of the RR crosses 1 so

there is not a statistically significant difference.

The 95% CI includes a clinically significant effect.

Question

• The results of 5 screening tests for identifying fetal anemia are presented below. Test Sensitivity Specificity

Peak systolic velocity

.69 .89

SD ratio .56 .94

Sinusoidal FHR

.22 .98

Serial growth .55 .91

MSAFP .47 .82

Question

1. The test, if negative, that best helps to rule OUT fetal anemia is:

2. The test, if positive, that best helps to rule IN fetal anemia is:

Answer

1. Peak systolic velocity• It has the highest SENSITIVITY. • SnOUT

2. Sinusoidal FHR• It has the highest SPECIFICITY• SpIN

Question

• A study evaluates the use of physical exam (PE) to identify PDA. The “gold standard” is echo. Eighty patients with PDA by echo were evaluated with PE as were 50 patients without PDA. The PE was positive in 56 of the neonates with PDA and in 10 of the neonates without PDA.

Question

• The sensitivity of PE for PDA in this study is

a) 10/50 = 20%b) 24/80 = 30%c) 56/80 = 70%d) 40/50 = 80%e) 56/66 = 85%

Question

• The prevalence of PDA in this study sample was

a) 130/yearb) 56/130 = 43% c) 80/130 = 62%d) 96/130 = 74%

Answer

__________ __________

____________________ __________

____________________ __________

Answer

• Sensitivity = PID Disease = 80 Positive in

disease = 56 Sensitivity =

56/80 or 70%

• Prevalence is disease in population at risk = 80/130 = 62%

PDA No PDA

PE + 56 10 66

PE - 24 40 64

80 50 130

Question

• PE is used in a patient that you estimate has a 50% chance of having PDA based on risk factors. What are the chances of having a PDA if the PE is positive?

a) 35/100 or 35%b) 50/100 or 50%c) 35/50 or 70%d) 35/45 or 78%

Answer

• 35/45 = 78%

PDA No PDA

PE + 35 (10) (45)

PE - (15) 40 (55)

50 50 100

1. Assume 100 patients

2. Set desired prevalence (50%)

3. You already know that Sens = 0.7 and Spec = 0.8

4. Fill in TP=50 x 0.7 = 35

5. Fill in TN=50 x 0.8 = 40

6. Subtract to get the other 2 boxes

7. Calculate the PPV

Another Way

• Use Likelihood Ratio

• LR (+) =(A/A+C)/(B/B+D)

• LR (-) =(C/A+C)/(D/B+D)

True (+)A

False (+)B

False (-)C

True (-)D

Another Way

• Fagan Nomogram• Plot pre-test

probability estimate (prevalence, best guess)

• Draw line through the LR

• Read the post-test probability

Test Names to KnowNominal Ordinal Interval or Ratio

Difference in proportions

Chi-square or Fisher’s exact (small frequencies)

One or 2 means Student’s t-testWilcoxon signed rank test (not normally distributed)

More than 2 means

ANOVA

Association Relative Risk Spearman rho Pearson r

Predict one variable from another

Logistic regression

Linear regression

Sample Question

• Because of my superior intellect, good study habits, and confident attitude, I will: A. Approach this test with a smile

on my face and a song in my heart. B. Miss a few questions, but hey

don’t we all? C. Kick butt on this exam D. All of the above