Healthcare Statistics
Chapter 6
Confidence Interval
Estimates
Learning Objectives (1 of 2)
- Define point estimate, standard error, confidence level, and margin of error
- Compare and contrast standard error and margin of error
- Compute and interpret confidence intervals for means and proportions
- Differentiate independent and matched or paired samples
Learning Objectives (1 of 2)
- Compute confidence intervals for the difference in means and proportions in independent samples and for the mean difference in paired samples
- Identify the appropriate confidence interval formula based on type of outcome variable and number of samples
Statistical Inference (1 of 2)
- There are two broad areas of statistical inference: estimation and hypothesis testing.
- Estimation—the population parameter is unknown, and sample statistics are used to generate estimates of the unknown parameter.
Statistical Inference (2 of 2)
- Hypothesis testing—an explicit statement or hypothesis is generated about the population parameter; sample statistics are analyzed and determined to either support or reject the hypothesis about the parameter.
- In both estimation and hypothesis testing, it is assumed that the sample drawn from the population is a random sample.
Estimation (1 of 2)
- Process of determining likely values for unknown population parameter
- Point estimate is best single-valued estimate for parameter.
- Confidence interval is range of values for parameter.
point estimate ± margin of error
Estimation (2 of 2)
- A point estimate for a population parameter is the “best” single number estimate of that parameter.
- A confidence interval estimate is a range of values for the population parameter with a level of confidence attached (e.g., 95% confidence that the range or interval contains the parameter).
Confidence Interval Estimates
point estimate ± margin of error
point estimate ± Z SE (point estimate)
where Z = value from standard normal distribution for desired confidence level and SE (point estimate) = standard error of the point estimate
Confidence Intervals for m
- Continuous outcome
- One sample
n ≥ 30 (Find Z in Table 1B)
n < 30 (Find t in Table 2 [next slide], df = n – 1)
Table 2. Critical Values of
the t Distribution
- Table entries represent values from t distribution with upper tail area equal to a.
Confidence level 80% 90% 95% 98% 99%
Two-sided test a .20 .10 .05 .02 .01
One-sided test a .10 .05 .025 .01 .005
df
1 3.078 6.314 12.71 31.82 63.66
2 1.886 2.920 4.303 6.965 9.925
3 1.638 2.353 3.182 4.541 5.841
4 1.533 2.132 2.776 3.747 4.604
5 1.476 2.015 2.571 3.365 4.032
6 1.440 1.943 2.447 3.143 3.707
7 1.415 1.895 2.365 2.998 3.499
8 1.397 1.860 2.306 2.896 3.355
9 1.383 1.833 2.262 2.821 3.250
10 1.372 1.812 2.228 2.764 3.169
Example 6.1.
Confidence Interval for m (1 of 2)
- In the Framingham Offspring Study (n = 3534), the mean systolic blood pressure (SBP) was 127.3 with a standard deviation of 19.0.
- Generate a 95% confidence interval for the true mean SBP.
127.3 ± 0.63
(126.7, 127.9)
Example 6.2.
Confidence Interval for m (2 of 2)
- In a subset of n = 10 participants attending the Framingham Offspring Study, the mean SBP was 121.2 with a standard deviation of 11.1.
- Generate a 95% confidence interval for the true mean SBP.
121.2 ± 7.94
(113.3, 129.1)
df = n – 1 = 9, t = 2.262
New Scenario
- Outcome is dichotomous (p = population proportion)
- Result of surgery (success, failure)
- Cancer remission (yes/no)
- One study sample
- Data
- On each participant, measure outcome (yes/no)
- n, x = number of positive responses
Confidence Intervals for p
- Dichotomous outcome
- One sample
(Find Z in Table 1B)
Example 6.3.
Confidence Interval for p
- In the Framingham Offspring Study (n = 3532), 1219 patients were on antihypertensive medications.
- Generate a 95% confidence interval for the true proportion on antihypertensive medication.
0.345 ± 0.016
(0.329, 0.361)
New Scenario
- Outcome is continuous
- SBP, weight, cholesterol
- Two independent study samples
- Data
- On each participant, identify group and measure outcome
Two Independent Samples (1 of 2)
RCT: Set of Subjects Who Meet
Study Eligibility Criteria
Randomize
Treatment 1 Treatment 2
Mean Trt 1 Mean Trt 2
Two Independent Samples (2 of 2)
Cohort Study: Set of Subjects Who
Meet Study Inclusion Criteria
Group 1 Group 2
Mean Group 1 Mean Group 2
Confidence Intervals for (m1 - m2)
- Continuous outcome
- Two independent samples
n1 ≥ 30
and n2 ≥ 30 (Find Z in
Table 1B)
n1 < 30
or n2 < 30 (Find t in
Table 2,
df = n1 + n2 – 2)
Pooled Estimate of Common Standard Deviation, Sp
- Previous formulas assume equal variances (s12 = s22)
- If 0.5 ≤ s12/s22 ≤ 2, assumption is reasonable
Example 6.5.
Confidence Interval for (m1 - m2)
- Using data collected in the Framingham Offspring Study, generate a 95% confidence interval for the difference in mean SBP between men and women.
n Mean Std Dev
Men 1623 128.2 17.5
Women 1911 126.5 20.1
Assess Equality of Variances
- Ratio of sample variances: 17.52/20.12 = 0.76
Confidence Intervals for (m1 - m2)
1.7 ± 1.26
(0.44, 2.96)
New Scenario
- Outcome is continuous
- SBP, weight, cholesterol
- Two matched study samples
- Data
- On each participant, measure outcome under each experimental condition.
- Compute differences (D = X1 – X2)
Two Dependent/Matched Samples
Subject ID Measure 1 Measure 2
1 55 70
2 42 60
.
.
Measures taken serially in time or under different experimental conditions
Crossover Trial
Treatment Treatment
Eligible R
Participants
Placebo Placebo
Each participant measured on treatment and placebo
Confidence Intervals for md
- Continuous outcome
- Two matched/paired samples
n ≥ 30 (Find Z in Table 1B)
n < 30 (Find t in Table 2,
df = n – 1)
Example 6.8.
Confidence Interval for md (1 of 3)
- In a crossover trial to evaluate a new medication for depressive symptoms, patients’ depressive symptoms were measured after taking new drug and after taking placebo.
- Depressive symptoms were measured on a scale of 0 to100 with higher scores indicative of more symptoms.
Example 6.8.
Confidence Interval for md (2 of 3)
- Construct a 95% confidence interval for the mean difference in depressive symptoms between drug and placebo.
- The mean difference in the sample (n = 100) is –12.7 with a standard deviation of 8.9.
Example 6.8.
Confidence Interval for md (3 of 3)
–12.7 ± 1.74
(–14.1, –10.7)
New Scenario
- Outcome is dichotomous
- Result of surgery (success, failure)
- Cancer remission (yes/no)
- Two independent study samples
- Data
- On each participant, identify group and measure outcome (yes/no)
Confidence Intervals for (p1 - p2)
- Dichotomous outcome
- Two independent samples
(Find Z in Table 1B)
Example 6.10.
Confidence Interval for (p1 – p2) (1 of 3)
- A clinical trial compares a new pain reliever to that considered standard care in patients undergoing joint replacement surgery; the outcome of interest is reduction in pain by 3+ scale points.
- Construct a 95% confidence interval for the difference in proportions of patients reporting a reduction between treatments.
Example 6.10.
Confidence Interval for (p1 – p2) (2 of 3)
Reduction of 3+ Points
Treatment n Number Proportion
New 50 23 0.46
Standard 50 11 0.22
Example 6.10.
Confidence Interval for (p1 – p2) (3 of 3)
0.24 ± 0.18
(0.06, 0.42)
Confidence Intervals for Relative Risk (RR)
- Dichotomous outcome
- Two independent samples
exp(lower limit), exp(upper limit)
(Find Z in Table 1B)
Example 6.12.
Confidence Interval for RR (1 of 2)
Reduction of 3+ Points
Treatment n Number Proportion
New 50 23 0.46
Standard 50 11 0.22
Construct a 95% CI for the relative risk.
Example 6.12.
Confidence Interval for RR (2 of 2)
0.737 ± 0.602 exp(0.135), exp(1.339)
(0.135, 1.339) (1.14, 3.82)
Confidence Intervals for Odds Ratio (OR)
- Dichotomous outcome
- Two independent samples
exp(lower limit), exp(upper limit)
(Find Z in Table 1B)
Example 6.14.
Confidence Interval for OR (1 of 2)
Reduction of 3+ Points
Treatment n Number Proportion
New 50 23 0.46
Standard 50 11 0.22
Construct a 95% CI for the odds ratio.
Example 6.14.
Confidence Interval for OR (2 of 2)
1.105 ± 0.870 exp(0.235), exp(1.975)
(0.235, 1.975) (1.26, 7.21)
n
s
Z
X
±
n
s
t
X
±
3534
19.0
96
.
1
127.3
±
n
s
t
X
±
10
11.1
262
.
2
121.2
±
n
x
p
ˆ
=
5
)]
p
ˆ
n(1
,
p
ˆ
min[n
³
-
p̂ ± Z p̂(1 – p̂) n
ˆ
p±Z
ˆ
p(1 –
ˆ
p)
n
p̂ ± Z p̂(1 – p̂) n
ˆ
p±Z
ˆ
p(1 –
ˆ
p)
n
0.345
3532
1219
p
ˆ
=
=
0.345 ±1.96 0.345(1 – 0.345) 3532
0.345±1.96
0.345(1 – 0.345)
3532
)
s
(or
s
,
X
,
n
),
s
(or
s
,
X
,
n
2
2
2
2
2
1
2
1
1
1
(X1 – X2 )± ZSp 1 n1 +
1 n2
(X
1
– X
2
)± ZSp
1
n
1
+
1
n
2
(X1 – X2 )± tSp 1 n1 +
1 n2
(X
1
– X
2
)± tSp
1
n
1
+
1
n
2
2
n
n
1)s
(n
1)s
(n
Sp
2
1
2
2
2
2
1
1
-
+
-
+
-
=
0
.
19
12
.
359
2
1911
1623
1)20.1
(1911
1)17.5
(1623
Sp
2
2
=
=
-
+
-
+
-
=
(128.2 – 126.5) ± 1.96 (19.0) 1 1623
+ 1
1911
(128.2 – 126.5) ± 1.96 (19.0)
1
1623
+
1
1911
(X1 – X2 )± ZSp 1 n1 +
1 n2
(X
1
– X
2
)± ZSp
1
n
1
+
1
n
2
d
d
s
,
X
n,
n
s
Z
X
d
d
±
n
s
t
X
d
d
±
n
s
Z
X
d
d
±
–12.7±1.96 8.9 100
–12.7±1.96
8.9
100
2
2
1
1
p
ˆ
,
n
,
p
ˆ
,
n
5
)]
p
ˆ
(1
n
,
p
ˆ
n
),
p
ˆ
(1
n
,
p
ˆ
min[n
2
2
2
2
1
1
1
1
³
-
-
(p̂1 – p̂2 ) ± Z p̂1(1 – p̂1 )
n1 +
p̂2 (1 – p̂2 ) n2
(
ˆ
p
1
–
ˆ
p
2
)±Z
ˆ
p
1
(1 –
ˆ
p
1
)
n
1
+
ˆ
p
2
(1–
ˆ
p
2
)
n
2
(p̂1 – p̂2 ) ± Z p̂1(1 – p̂1 )
n1 +
p̂2 (1 – p̂2 ) n2
(
ˆ
p
1
–
ˆ
p
2
)±Z
ˆ
p
1
(1 –
ˆ
p
1
)
n
1
+
ˆ
p
2
(1–
ˆ
p
2
)
n
2
(0.46 – 0.22) ±1.96 0.46(1 – 0.46) 50
+ 0.22(1 – 0.22)
50
(0.46 – 0.22)±1.96
0.46(1 – 0.46)
50
+
0.22(1–0.22)
50
ln(R̂R)± Z (n1 – x1 )/x1 n1
+ (n2 – x2 )/x2
n2
ln(
ˆ
RR)±Z
(n
1
– x
1
)/x
1
n
1
+
(n
2
– x
2
)/x
2
n
2
2.09
0.22
0.46
p
ˆ
p
ˆ
R
R
ˆ
2
1
=
=
=
50
39/11
50
27/23
1.96
ln(2.09)
+
±
)
x
(n
1
n
1
)
x
(n
1
x
1
Z
R)
O
ˆ
ln(
2
2
2
1
1
1
-
+
+
-
+
±
ÔR = x1 /(n1 – x1 ) x2 /(n2 – x2 )
= 23/27 11/39
= 3.02
ˆ
OR=
x
1
/(n
1
– x
1
)
x
2
/(n
2
– x
2
)
=
23/27
11/39
=3.02
39
1
11
1
27
1
23
1
1.96
ln(3.02)
+
+
+
±