In the preceding discussion we have been using s, the population standard deviation, to compute the standard error. However, we don't really know the population standard deviation, since we are working from samples. To get around this, we have been using the sample standard deviation [s] as an estimate. This is not a problem if the sample size is 30 or greater because of the central limit theorem. However, if the sample is small [ t.test[bmi]
The output would look like this:
One Sample t-test
data: bmi
t = 228.5395, df = 3231, p-value < 2.2e-16
alternative hypothesis: true mean is not equal to
0
95 percent confidence interval:
26.66357 27.12504
sample estimates:
mean of x
26.8943
R defaults to computing a 95% confidence interval, but you can specify the confidence interval as follows:
> t.test[bmi,conf.level=.90]
This would compute a 90% confidence interval.
Lozoff and colleagues compared developmental outcomes in children who had been anemic in infancy to those in children who had not been anemic. Some of the data are shown in the table below.
Mean + SD | Anemia in Infancy [n=30] | Non-anemic in Infancy [n=133] |
Gross Motor Score | 52.4+14.3 | 58.7+12.5 |
Verbal IQ | 101.4+13.2` | 102.9+12.4 |
Source: Lozoff et al.: Long-term Developmental Outcome of Infants with Iron Deficiency, NEJM, 1991
Compute the 95% confidence interval for verbal IQ using the t-distribution
Link to the Answer in a Word file
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