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Hypothesis Testing
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• Unlike prop.test and t.test, we needed to specify that we wanted a confidence interval computed.
• For this data, the confidence interval is enormous as the size of the sample is small and the range is huge.
• We couldn’t have used a t-test as the data isn’t even close to normal.
Problems
9.1 Create 15 random numbers that are normally distributed with mean 10 and s.d. 5. Find a 1-sample z-test at
the 95% level. Did it get it right?
9.2 Do the above 100 times. Compute what percentage is in a 95% confidence interval. Hint: The following might
prove useful
> f=function () mean(rnorm(15,mean=10,sd=5))
> SE = 5/sqrt(15)
> xbar = simple.sim(100,f)
> alpha = 0.1;zstar = qnorm(1-alpha/2);sum(abs(xbar-10) < zstar*SE)
[1] 87
> alpha = 0.05;zstar = qnorm(1-alpha/2);sum(abs(xbar-10) < zstar*SE)
[1] 92
> alpha = 0.01;zstar = qnorm(1-alpha/2);sum(abs(xbar-10) < zstar*SE)
[1] 98
9.3 The t-test is just as easy to do. Do a t-test on the same data. Is it correct now? Comment on the relationship
between the confidence intervals.
9.4 Find an 80% and 95% confidence interval for the median for the exec.pay dataset.
9.5 For the Simple data set rat do a t-test for mean if the data suggests it is appropriate. If not, say why not. (This
records survival times for rats.)
9.6 Repeat the previous for the Simple data set puerto (weekly incomes of Puerto Ricans in Miami.).
9.7 The median may be the appropriate measure of center. If so, you might want to have a confidence interval for it
too. Find a 90% confidence interval for the median for the Simple data set malpract (on the size of malpractice
awards). Comment why this distribution doesn’t lend itself to the z-test or t-test.
9.8 The t-statistic has the t-distribution if the Xi ’s are normally distributed. What if they are not? Investigate the
distribution of the t-statistic if the Xi ’s have different distributions. Try short-tailed ones (uniform), long-tailed
ones (t-distributed to begin with), Uniform (exponential or log-normal).
(For example, If the Xi are nearly normal, but there is a chance of some errors introducing outliers. This can
be modeled with
Xi = ζ(µ + σZ) + (1 − ζ)Y
where ζ is 1 with high probability and 0 otherwise and Y is of a different distribution. For concreteness, suppose
µ = 0, σ = 1 and Y is normal with mean 0, but standard deviation 10 and P (ζ = 1) = .9. Here is some R
code to simulate and investigate. (Please note, the simulations for the suggested distributions should be much
simpler.)
>
+
+
+
>
>
>
>
f =
function(n=10,p=0.95) {
y = rnorm(n,mean=0,sd=1+9*rbinom(n,1,1-p))
t = (mean(y) - 0) / (sqrt(var(y))/sqrt(n))
}
sample = simple.sim(100,f)
qqplot(sample,rt(100,df=9),main="sample vs. t");qqline(sample)
qqnorm(sample,main="sample vs. normal");qqline(sample)
hist(sample)
The resulting graphs are shown. First, the graph shows the sample against the t-quantiles. A bad, fit. The
normal plot is better but we still see a skew in the histogram due to a single large outlier.)
Hypothesis Testing
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−2
0
2
−2
sample
0
2
Theoretical Quantiles
0 5 10 15 20 25 30
Histogram of sample
Frequency
Sample Quantiles
−4
−4 −3 −2 −1 0 1 2 3
sample vs. normal
−3 −2 −1 0 1 2 3
rt(100, df = 9)
sample vs. t
−4
0
2
sample
Figure 47: t-statistic for contaminated normal data
Section 10: Hypothesis Testing
Hypothesis testing is mathematically related to the problem of finding confidence intervals. However, the approach
is different. For one, you use the data to tell you where the unknown parameters should lie, for hypothesis testing, you
make a hypothesis about the value of the unknown parameter and then calculate how likely it is that you observed
the data or worse.
However, with R you will not notice much difference as the same functions are used for both. The way you use
them is slightly different though.
Testing a population parameter
Consider a simple survey. You ask 100 people (randomly chosen) and 42 say “yes” to your question. Does this
support the hypothesis that the true proportion is 50%?
To answer this, we set up a test of hypothesis. The null hypothesis, denoted H 0 is that p = .5, the alternative
hypothesis, denoted HA , in this example would be p 6= 0.5. This is a so called “two-sided” alternative. To test the
assumptions, we use the function prop.test as with the confidence interval calculation. Here are the commands
> prop.test(42,100,p=.5)
1-sample proportions test with continuity correction
data: 42 out of 100, null probability 0.5
X-squared = 2.25, df = 1, p-value = 0.1336
alternative hypothesis: true p is not equal to 0.5
95 percent confidence interval:
0.3233236 0.5228954
sample estimates:
p
0.42
Note the p-value of 0.1336. The p-value reports how likely we are to see this data or worse assuming the null
hypothesis. The notion of worse, is implied by the alternative hypothesis. In this example, the alternative is twosided as too small a value or too large a value or the test statistic is consistent with H A . In particular, the p-value is
the probability of 42 or fewer or 58 or more answer “yes” when the chance a person will answer “yes” is fifty-fifty.
Now, the p-value is not so small as to make an observation of 42 seem unreasonable in 100 samples assuming the
null hypothesis. Thus, one would “accept” the null hypothesis.
Next, we repeat, only suppose we ask 1000 people and 420 say yes. Does this still support the null hypothesis
that p = 0.5?
> prop.test(420,1000,p=.5)
1-sample proportions test with continuity correction