Tests of Hypothesis
[Motivational Example]. It is claimed that the average grade of all 12 year old children in a
country in a particular aptitude test is 60%. A random sample of n= 49 students gives a mean
x = 55% with a standard deviation s = 2%. Is the sample finding consistent with the claim?
We regard the original claim as a null hypothesis (H0) which is tentatively accepted as TRUE:
H0 : m = 60.
n(0,1)
If the null hypothesis is true, the test statistic
n(0,1)
t= x-m
0.95
sn
-1.96
1.96
is a random variable with a N(0, 1) distribution.
Thus
55 - 60 = - 35 / 2 = - 17.5
2/  49
is a random value from n(0, 1).
rejection regions
But this lies outside the 95% confidence interval (falls in the rejection region), so either
(i) The null hypothesis is incorrect
or
(ii) An event with a probability of at most 0.05 has occurred.
Consequently, we reject the null hypothesis, knowing that there is a probability of 0.05 that we
are acting in error. In technical terms, we say that we are rejecting the null hypothesis at the
0.05 level of significance.
The alternative to rejecting H0, is to declare the test to be inconclusive. By this we mean that
there is some tentative evidence to support the view that H0 is approximately correct.
Modifications
Based on the properties of the Normal , student t and other distributions, we can generalise
these ideas. If the sample size n < 25, we should use a tn-1 distribution. We can also vary the
level of significance of the test and we can apply the tests to proportionate sampling
environments.
Example. 40% of a random sample of 1000 people in a country indicate satisfaction with
government policy. Test at the .01 level of significance if this consistent with the claim that
45% of the people support government policy?
Here,
H0: P = 0.45
p = 0.40,
n = 1000
so
 p (1-p) / n = 0.015 test statistic = (0.40 - 0.45) / 0.015 = - 3.33
99% critical value = 2.58
so H0 is rejected at the .01 level of significance.
One-Tailed Tests
If the null hypothesis is of the form H0 : P  0. then arbitrary large values of p are
acceptable, so that the rejection region for the test statistic lies in the left hand tail only.
Example. 40% of a random sample of 1000 people in a country indicate satisfaction with
government policy. Test at the .05 level of significance if this consistent with the claim that
at least 45% of the people support government policy?
N(0,1)
Here the critical value is -1.64, so the
0.95
the null hypothesis H0: P  0.
is rejected at the .05 level of
significance
-1.64
Rejection region
Testing Differences between Means
Suppose that
x 1 x2
…
xm is a random sample with mean x and standard deviation s1
drawn from a distribution with mean m1 and
y1 y2
…
yn is a random sample with mean y and standard deviation s1
drawn from a distribution with mean m2. Suppose that we wish to test the null hypothesis that
both samples are drawn from the same parent population (i.e.)
H0: m1 = m2.
The pooled estimate of the parent variance is
s 2 = { (m - 1) s12 + (n - 1) s22 } / ( m + n - 2)
and the variance of x - y, being the variance of the difference of two independent random
variables, is
s ’ 2 = s 2 / m + s 2 / n.
This allows us to construct the test statistic, which under H0 has a tm+n-2 distribution.
Example. A random sample of size m = 25 has mean x = 2.5 and standard deviation s1 = 2,
while a second sample of size n = 41 has mean y = 2.8 and standard deviation s2 = 1. Test at
the .05 level of significance if the means of the parent populations are identical.
Here
H0 : m1 = m2
x - y = - 0.3 and
s 2 = {24(4) + 38(1)} / 64 = 2.0938
so the test statistic is
- 0.3 /  .0    .0   = - .7
The .05 critical value for N(0, 1) is .6, so the test is inconclusive.
Paired Tests
If the sample values ( xi , yi ) are paired, such as the marks of students in two
examinations, then let
di = xi - yi be their differences and treat these
values as the elements of a sample to generate a test statistic for the hypothesis
H0: m1 = m2.
The test statistic
d / sd / n
has a tn-1 distribution if H0 is true.
Example. In a random sample of 100 students in a national examination their examination
mark in English is subtracted subtracted from their continuous assessment mark, giving a
mean of 5 and a standard deviation of 2. Test at the .01 level of significance if the true mean
mark for both components is the same.
Here
n = 100, d = 5,
sd / n = 2/10 = 0.2
so the test statistic is 5 / 0.2 = 10.
the 0.1 critical value for a N(0, 1) distribution is 2.58, so H0 is rejected at the .01 level of
significance.
Tests for the Variance.
For Normally distributed random variables, given
H0: s 2 = k, a constant, then
(n-1) s2 / k has a c 2n - 1 distribution.
Example. A random sample of size 30 drawn from a normal distribution has variance s2 = 5.
Test at the .05 level of significance if this is consistent with H0 : s 2 = 2 .
Test statistic = (29) 5 /2 = 72.5, while the .05 critical value for c 229 is 45.72,
so H0 is rejected at the .05 level of significance.