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Social Science Statistics Module I
Gwilym Pryce
Lecture 6
Hypothesis Tests for Proportions & Two
Populations
Slides available from Statistics & SPSS page of www.gpryce.com
1
Notices:
• Register
• Class Reps and Staff Student committee.
Aims & Objectives
• Aim
• the aim of this lecture is to continue with our introduction of the
method of hypothesis testing and to demonstrate a number of
applications
• Objectives
– by the end of this lecture students should be able to carry out
hypothesis tests on:
• two population means
• one population proportion
• two population proportions
Plan:
• 1. Review of Significance
• 2. Review of one sample tests on the mean
• 3. Hypothesis tests about Two population means
• Homogenous variances
• Heterogeneous variances
• 4. Deciding on whether variances are equal
• 5. Hypothesis tests about proportions
– One population
– Two populations
Confidence
Intervals (CI)
Macro
commands:
Macro
command
CI_L1M
Large sample CI for one mean
Macro
Command
H_L1M
CI_S1M
Small sample CI for one mean
H_S1M
CI_S2MP
Small independent samples CI for
difference between 2 means
(pooled variance)
Small independent samples CI for
difference between 2 means
(different variances)
Large sample CI for one
proportion (presents output for
both Traditional and Wilson
methods of calculation)
Large sample CI for comparing
two proportions (presents output
for both Traditional and Wilson
methods of calculation)
H_S2MP
CI_S2MD
CI_L1P
CI_L2P
N_L1M
Definition
Sample size for desired margin or
error for the mean
H_S2MD
H_L1P
Hypothesis tests
Definition
Large sample significance test on
one mean
Small sample significance test on
one mean
Small
independent
samples
significance test for equality of 2
means (pooled variance)
Small
independent
samples
significance test for equality of 2
means (different variances)
Large sample significance test on
one proportion
H_L2P
Large samples significance test on
two proportions
H_S2VF
Simple small sample F-test on
equality of two variances (see also
Levene’s test in the SPSS help
menu for more sophisticated test
of homogenous variances).
1.
Review of Significance
• P = significance level = chances of our
observed sample mean occurring given that our
assumption about the population (denoted by
“H0”) is true.
• So if we find that this probability is small, it
might lead us to question our assumption about
the population mean.
– I.e. if our sample mean is a long way from our
assumed population mean then it is:
• either a freak sample
• or our assumption about the population mean is wrong.
• If we draw the conclusion that it is our assumption re
m that is wrong and reject H0 then we have to bear in
mind that there is a chance that H0 was in fact true.
– In other words:
• when P = 0.05, for every twenty times we reject H0, then on one
of those occasions we would have rejected H0 when it was in
fact true.
2. Review of one sample tests on the mean
• We introduced a common framework for hypothesis
testing:
4 Steps of Hypothesis testing:
Step (1) state H0 and H1
Step (2) state a and formula
Step (3) state decision rule
Step (4) compute P & decide
We also looked at 2 specific tests:
• Large sample sig. Test on one mean:
• Formula:
xm
z
s/ n
• Macro syntax:
H_L1M
n=(?) x_bar=(?)
m=(?)
s=(?).
• Small sample sig. Test on one mean:
• Formula:
xm
ti 
, df  n  1
s/ n
• Macro syntax:
H_S1M
n=(?) x_bar=(?)
m=(?)
s=(?).
3. Hypothesis tests about two population means
• In SPSS: this is called the “Independent Sample t-test”
• go to Analyse, Compare Means...
• Two different formulas for computing t:
Equal Variances
Unequal Variances
(formula has an exact t-distribution)
(does not have an exact t-distribution)
tc 
( x1  x2 )  ( m1  m 2 )
1 1
sp

n1 n2
df  n1  n2  1
tc 
( x1  x2 )  ( m1  m 2 )
2
2
s1
s2

n1 n2
df  min[n1  1, n2  1]
Example where variances are different:
As part of your PhD, you want to test whether the new
“Fun Phonics” reading method is better than the
“Letterland” method. You examine the reading power of 6
year old children from two similar schools.
– The first used the FP method and you found that this produced an average
reading proficiency score of 53.7 (based on a sample of 22 children; s.d. =
11.5).
– The second school used the Letterland method and you found that this
produced an average reading proficiency score of 42.51 (sample = 24; s.d. =
16.9).
Test whether the FP method produces higher results at the 1% significance
level.
xFP  53.7, nFP  22, sFP  11.5
xL  42.51, nL  24, sL  16.9
xFP  53.7, nFP  22, sFP  11.5
xL  42.51, nL  24, sL  16.9
• Use the 4 steps and the following formula to test whether the FP
method produces higher results at the 1% significance level.
4 Steps of Hypothesis testing:
Step (1) state H0 and H1
Step (2) state a and formula
Step (3) state decision rule
tc 
( x1  x2 )  ( m1  m 2 )
2
2
s1
s
 2
n1 n2
df  min[n1  1, n2  1]
Step (4) compute P & decide
• Can you use the canned SPSS procedure to do this problem?
xFP  53.7, nFP  22, sFP  11.5
xL  42.51, nL  24, sL  16.9
• (1) H0: mFP = mL
H1: mFP > mL
(means are equal)
(upper tail test)
• (2) a = 0.01 (implies critical t value of 2.528),
tc 
x1  x2
2
2
s1 s2

n1 n2

53.7  42.51
 2.644
132.25 285.61

22
24
df  min[n1  1, n2  1]  21
• (3) Reject H0 iff P < a, I.e. if P < 0.01
• (4) P = Prob(t > 2.644) = 0.0076, so reject H0
Doing the calculation in SPSS:
• You cannot use the canned SPSS procedure unless you
have the original data.
• But you can use the following macro commands:
– Homogenous variances:
• H_S2Mp
n1=(?) n2=(?) x_bar1=(?) x_bar2=(?) s1=(?) s2=(?).
– Heterogeneous variances:
• H_S2Md
n1=(?) n2=(?) x_bar1=(?) x_bar2=(?) s1=(?) s2=(?).
For the Letterland/FP example we would use the diff.
Variances syntax:
H_S2Md
n1=(22) n2=(24) x_bar1=(53.7) x_bar2=(42.51) s1=(11.5) s2=(16.9).
• The upper tail sig. = 0.007588
• I.e. less than 1% chance of false rejection,
therefore reject H0 of equal means in favour of the
alternative hypothesis that Fun Phonics results in
higher reading scores on average than Letterland.
4. How do we decide on whether the variances are similar?
• Where variances are hugely different or
exactly the same, the decision is simple.
• When there is any ambiguity, we can use one
of two tests to help us:
• Simple Ratio of Variances Test
• Levene’s Test
Simple Ratio of Variances test:
• If we divide the ratio of variances of samples from two
independent populations we find that that ratio has an F
distribution in repeated samples:
F = s 12 / s 2 2
• where the denominator degrees of freedom calculated as
n1–1 and the numerator degrees of freedom calculated as
n2–1.
• NB Because the critical values for the F distribution are only
calculated for the upper tail, if the F value you are have calculated is
less than one, you need to invert it
– i.e. swap round the numerator and denominator.
• This is the formula behind the following command:
H8_S2VF
n1=(?) n2=(?) s1(?) s2(?)
• E.g. For the Letterland/FP example:
H8_S2VF n1=(22) n2=(24) s1=(11.5) s2=(16.9).
•
Which tells us that there is less than a 5% chance of false rejection if
we reject the null of equal variances. So reject the null
• I.e. we can be sure that the population variances are indeed different.
The Levene’s test
• If we have the original data (rather than just the summary
statistics) we can use Levene’s test which is a canned routine in
SPSS.
• The Levene’s test is more sophisticated & robust than the
simple ratio of variances test:
– If P (I.e. “sig.”) from the Levene’s test is small reject the H0 of equal
variances & use the 1st t-formula.
– If P from the Levene’s test is large, accept H1 & use the 2nd tformula to compute the test statistic.
SPSS Output from test equal purchase prices between Cumberland and Durham (Nationwide)
S
t
E
H
N
e
e
i
a
f
P
7
1
3
8
9
3
0
7
S
s
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f
a
V
o
n
a
e
r
S
.e
i
g
E
S
e
o
e
p
F
d
a
i
t
r
w
g
r
p
f
P
E
0
4
6
4
7
8
0
0
5
a
E
4
3
8
8
0
7
2
n
Two tails from one:
• Along with the Levene’s test results, SPSS automatically supplies
t-test results for both the equal and different variances formulas.
• One problem with the SPSS t-test, however, is that it only gives
the 2 tail sig., but you can work out the one tail sigs as follows:
• The two tailed significance is twice that of the smallest one tailed
significance:
2 tailed sig. = 2  min[lower tail sig., upper tail sig.]
• But it can be a bit confusing working out which one tail significance level
is the one you want (see notes).
Testing for 2 means summary:
• If you’ve got the original data,
• First do the Levene’s test in SPSS
Analyze, Compare Means, Independent Samples
• Then do the appropriate macro t-test to avoid confusion.
H_S2Mp for equal variances or H_S2Md for different variances
• If you don’t have the original data,
• First do the ratio of variances test
H8_S2VF
• Then do the appropriate macro t-test
H_S2Mp for equal variances or H_S2Md for different variances
5.1 Hypothesis tests on proportions: 1 population (large samples only)
• So far looked at:
– how to make inferences about the population mean from our
sample mean.
• But sometimes the variable of interest is categorical
• household has or has not insurance;
• person is homosexual or not homosexual;
• a person has Aids or does not have Aids
• In such cases, what we are interested in is
the proportion of cases that fall into a
particular category:
• the proportion of households with insurance;
• the proportion of people who are homosexual;
• the proportion of people with Aids
• Calculating the sample proportion:
p=x/n
– where:
• x = cases with the attribute of interest
e.g. the number of households with insurance
• n = sample size
CLT and Proportions:
• Q/ Does the Central Limit Theorem apply to sample
proportions?
• A/ Yes.
– Proportions from repeated random samples will be normally
distributed around the population proportion p.
– We can then translate any sample proportion onto the standard
normal curve by calculating its z score:
p p
zi 
p (1  p )
n
Example:
E.g. 1 As a historian, you want to find the proportion
of citizens in medieval Scotland that contracted
the plague. From a sample of 400 parish records,
you find that 22 died of the plague. The
assumption in the literature has been that 10% of
the population had died. Test whether this
assumption is valid using both 2 and 1 tailed tests.
Summary of data:
– (1)
n = 400
x = 22
p0 = 0.1
H0: p = 10%
H1: p  10% (2-tailed test)
– (2) a = 0.02, for example.
p p
zi 
p (1  p )
n
p
p
x

=
=
n
=
the population proportion
the sample proportion = x/n
the no. of items in sample
with the attribute of interest
the sample size.
– (3) Reject H0 iff P < a, I.e. if P < 0.02
(this will happen if zc < - 2.33 or if zc > 2.33, where 2.33 is the z value
associated with a = 0.02. Since zc = -3.948, we know we can safely reject
H0).
– (4) Calculate z:
n = 400
x = 22
p0 = 0.1
P
zi 
0.055 0.1
 .045  .045
p p
zc 


 3.00
p (1  p )
0.1(1  0.1)
0.09 0.015
n
400
400
= 2x(Prob(z < -3.00))
= 2x 0.0013 = 0.0026
since P < 0.02 (I.e. less than one in 50 chance of type I error) we can
reject H0.
In fact, the chances of incorrect rejection of H0 are less than one in
3,000.
I.e. the chances of observing p (our sample proportion) assuming H0 (p = 10%) to
be true are so small that we are forced to question this assumption about p
One tailed test:
– (1)
H0: p = 10%
H1: p < 10% (lower tail test)
– (2) a = 0.02
– (3) Reject H0 iff P < a, I.e. if P < 0.02
– (4) Lower tail sig. = P = Prob(z < -3.00) = .001350
since P < 0.02 we can reject H0 knowing that the chances of incorrect
rejection of H0 are less than one in 740
» our cut-off rule for rejecting H0 was no more than a one in 50 chance
» one in 740 is a lot less than one in 50 so we can reject H0 with
confidence.
• The macro syntax for one proportion tests is
as follows:
• H6_L1P n=(400) x=(22) pi=(0.1).
Which comes to the same result.
5.2 Hypothesis tests about Two population proportions
• To test the hypothesis that the population proportions are equal:
H0: p1 = p2
compute the z statistic:
( p1  p2 )
z
SEDp
where:
SEDp is the pooled standard
error =
and x  x
p
1
2
n1  n2
SEDp 
1 1 
p(1  p)  
 n1 n2 
Example:
• Two surveys of mortgage payment protection insurance (MPPI)
are carried out, one on single parents with 1 child and one on
single parents with 3 children. Amongst the first group, 67 out of
a sample of 300 were found to have taken out MPPI, compared
with 15 out of a sample of 101 in the second group. Is take-up
significantly lower amongst the HHs with three children?
• p1 = 67/300
= 0.2233;
• p2 = 15/101
= 0.1485;
• p = (300 + 101)/(67+15)
= 0.2045;
H 0: p1 = p2
H1: p1 > p2
– (2) a = 0.01 (z* = 2.33)
– (1)
1 
 1
SEDp  0.205(1  0.205)

  0.0464
 300 100
z
( p1  p2 ) (0.2233 0.1485)

 1.6125
SEDp
0.0466
– (3) Reject H0: if P < 0.01
– (4) P = 0.053.
Take-up is not significantly lower amongst HHs with 3 children at
the1% sig. level; or even at 5% significance level.
• I.e. we cannot say that the difference in proportions is anything more
than the effect of sampling variation.
H7_L2P
n1=(300)
n2=(101) x1=(67) x2=(15) .
Summary:
• 1. Review of Significance
• 2. Review of one sample tests on the mean
• 3. Hypothesis tests about Two population means
• Homogenous variances
• Heterogeneous variances
• 4. Deciding on whether variances are equal
• 5. Hypothesis tests about proportions
– One population
– Two populations