This document discusses statistical methods for comparing two independent sample means and two independent sample proportions. It provides steps and examples for conducting significance tests to compare population means and proportions. For means, it describes using a z-test where the test statistic is the difference between sample means divided by the pooled standard error. For proportions, it describes using a z-test where the test statistic is the difference between sample proportions divided by the pooled standard error. Examples provided show conducting these tests to analyze differences in housework hours and attitudes between years.

1. By:
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2. Sociology 601 Class 8: September 24, 2009
 6.6: Small-sample inference for a proportion
 7.1: Large sample comparisons for two
independent sample means.
 7.2: Difference between two large sample
proportions.
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3. 7.1 Large sample comparisons for two independent
means
 So far, we have been making estimates and
inferences about a single sample statistic
 Now, we will begin making estimates and inferences
for two sample statistics at once.
 many real-life problems involve such comparisons
 two-group problems often serve as a starting point for
more involved statistics, as we shall see in this class.
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4. Independent and dependent samples
 Two independent random samples:
 Two subsamples, each with a mean score for some other
variable
 example: Comparisons of work hours by race or sex
 example: Comparison of earnings by marital status
 Two dependent random samples:
 Two observations are being compared for each “unit” in
the sample
 example: before-and-after measurements of the same
person at two time points
 example: earnings before and after marriage
 husband-wife differences
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5. Comparison of two large-sample means
for independent groups
Hypothesis testing as we have done it so far:
 Test statistic: z = (Ybar - o) / (s /SQRT(n))
 What can we do when we make inferences about a
difference between population means (2 - 1)?
 Treat one sample mean as if it were o ?
 (NO: too much type I error)
 Calculate a confidence interval for each sample mean
and see if they overlap?
 (NO: too much type II error)
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6. Figuring out a test statistic
for a comparison of two means
Is Y2 –Y1an appropriate way to evaluate 2 - 1?
• Answer: Yes. We can appropriately define (2 - 1) as a
parameter of interest and estimate it in an unbiased way
with (Y2 – Y1) just as we would estimate  with Y.
• This line of argument may seem trivial, but it becomes
important when we work with variance and standard
deviations.
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7. Figuring out a standard error for a comparison of two
means
Comparing standard errors:
 A&F 213: formula without derivation
 Is s2
Ybar2 - s2
Ybar1an appropriate way to estimate
2
(Ybar2-Ybar1)?
 No!
 2
(Ybar2-Ybar1)= 2
(Ybar2) - 2(Ybar2,Ybar1) + 2
(Ybar1)
 Where 2(Ybar2,Ybar1) reflects how much the observations
for the two groups are dependent.
 For independent groups, 2(Ybar2,Ybar1) = 0,
so 2
(Ybar2-Ybar1)= 2
(Ybar2) + 2
(Ybar1)
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8. Step 1: Significance test for 2 - 1
 The parameter of interest is 2 - 1
 Assumptions:
 the sample is drawn from a random sample of some sort,
 the parameter of interest is a variable with an interval
scale,
 the sample size is large enough that the sampling
distribution of Ybar2 – Ybar1 is approximately normal.
 The two samples are drawn independently
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9. Step 2: Significance test for 2 - 1
 The null hypothesis will be that there is no
difference between the population means. This
means that any difference we observe is due to
random chance.
 Ho: 2 - 1 = 0
 (We can specify an alpha level now if we want)
 Q: Would it matter if we used
Ho: 1 - 2 = 0 ?
Ho: 1 = 2 ?
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10. Step 3: Significance test for 2 - 1
 The test statistic has a standard form:
 z = (estimate of parameter – Ho value of parameter)
standard error of parameter
Y Y
(  ) 
0
2 1 2
2
2
1
s
s
 Q: If the null hypothesis is that the means are the
same, why do we estimate two different standard
deviations?
10
2
1
n
n
z


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11. Step 4: Significance test for 2 - 1
P-value of calculated z:
• Table A
• Stata: display 2 * (1 – normal(z) )
• Stata: testi (no data, just parameters)
• Stata: ttest (if data file in memory)
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12. Step 5: Significance test for 2 - 1
Step 5: Conclusion.
 Compare the p-value from step 4 to the alpha level
in step 1.
If p < α, reject H0 If p ≥ α, do not reject H0
 State a conclusion about the statistical significance
of the test.
 Briefly discuss the substantive importance of your
findings.
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13. Significance test for 2 - 1: Example
 Do women spend more time on housework than
men?
 Data from the 1988 National Survey of Families and
Households:
 sex sample size mean hours s.d
 men 4252 18.1 12.9
 women 6764 32.6 18.2
 The parameter of interest is 2 - 1
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14. Significance test for 2 - 1: Example
1. Assumptions: random sample, interval-scale variable,
sample size large enough that the sampling distribution of
2 - 1is approximately normal, independent groups
2. Hypothesis: Ho: 2 - 1= 0
3. Test statistic:
z = ((32.6 – 18.1) – 0) / SQRT((12.9)2/4252 + (18.2)2/6764) = 48.8
4. p-value: p<.001
5. conclusion:
a. reject H0: these sample differences are very unlikely to occur if men
and women do the same number of hours of housework.
b. furthermore, the observed difference of 14.5 hours per week is a
substantively important difference in the amount of housework.
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15. Confidence interval for 2 - 1:
2
2
2
1
s
s
c i  Y Y  z 
 housework example with 99% interval:
 c.i….
= (32.6 – 18.1) +/- 2.58*( √((12.9)2/4252 + (18.2)2/6764))
= 14.5 +/- 2.58*.30
= 14.5 +/- .8, or (13.7,15.3)
 By this analysis, the 99% confidence interval for the
difference in housework is 13.7 to 15.3 hours.
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 
2
1
2 1 . .
n
n
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16. Stata: Large sample significance test for
2 - 1
 Immediate (no data, just parameters)
 ttesti 4252 18.1 12.9 6764 32.6 18.2, unequal
• Q: why ttesti with large samples?
 For the immediate command, you need the following:
 sample size for group 1 (n = 4252)
 mean for group 1
 standard deviation for group 1
 sample size for group 2
 mean for group 2
 standard deviation for group 2
 instructions to not assume equal variance (, unequal)
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17. Stata: Large sample significance test for
2 - 1, an example
. ttesti 4252 18.1 12.9 6764 32.6 18.2, unequal
Two-sample t test with unequal variances
------------------------------------------------------------------------------
| Obs Mean Std. Err. Std. Dev. [95% Conf. Interval]
---------+--------------------------------------------------------------------
x | 4252 18.1 .1978304 12.9 17.71215 18.48785
y | 6764 32.6 .221294 18.2 32.16619 33.03381
---------+--------------------------------------------------------------------
combined | 11016 27.00323 .1697512 17.8166 26.67049 27.33597
---------+--------------------------------------------------------------------
diff | -14.5 .2968297 -15.08184 -13.91816
------------------------------------------------------------------------------
Satterthwaite's degrees of freedom: 10858.6
Ho: mean(x) - mean(y) = diff = 0
Ha: diff < 0 Ha: diff != 0 Ha: diff > 0
t = -48.8496 t = -48.8496 t = -48.8496
P < t = 0.0000 P > |t| = 0.0000 P > t = 1.0000
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18. Large sample significance test for 2 - 1: command for a
data set (#1)
. ttest YEARSJOB, by(nonstandard) unequal
Two-sample t test with unequal variances
------------------------------------------------------------------------------
Group | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval]
---------+--------------------------------------------------------------------
0 | 980 9.430612 .2788544 8.729523 8.883391 9.977833
1 | 379 7.907652 .3880947 7.555398 7.144557 8.670747
---------+--------------------------------------------------------------------
combined | 1359 9.005887 .2290413 8.443521 8.556573 9.4552
---------+--------------------------------------------------------------------
diff | 1.522961 .4778884 .5848756 2.461045
------------------------------------------------------------------------------
diff = mean(0) - mean(1) t = 3.1869
Ho: diff = 0 Satterthwaite's degrees of freedom = 787.963
Ha: diff < 0 Ha: diff != 0 Ha: diff > 0
Pr(T < t) = 0.9993 Pr(|T| > |t|) = 0.0015 Pr(T > t) = 0.0007
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19. Large sample significance test for 2 - 1: command for a
data set (#2)
. ttest conrinc if wrkstat==1, by(wrkslf) unequal
Two-sample t test with unequal variances
------------------------------------------------------------------------------
Group | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval]
---------+--------------------------------------------------------------------
self-emp | 190 48514.62 2406.263 33168.05 43768.03 53261.2
someone | 1263 34417.11 636.9954 22638 33167.43 35666.8
---------+--------------------------------------------------------------------
combined | 1453 36260.56 648.5844 24722.9 34988.3 37532.82
---------+--------------------------------------------------------------------
diff | 14097.5 2489.15 9191.402 19003.6
------------------------------------------------------------------------------
diff = mean(self-emp) - mean(someone) t = 5.6636
Ho: diff = 0 Satterthwaite's degrees of freedom = 216.259
Ha: diff < 0 Ha: diff != 0 Ha: diff > 0
Pr(T < t) = 1.0000 Pr(|T| > |t|) = 0.0000 Pr(T > t) = 0.0000
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20. 7.2: Comparisons of two independent
population proportions
 In 1982 and 1994, respondents in the General Social Survey
were asked: “Do you agree or disagree with this statement?
‘Women should take care of running their homes and leave
running the country up to men.’”
 Year Agree Disagree Total
 1982 122 223 345
 1994 268 1632 1900
 Total 390 1855 2245
 Do a formal test to decide whether opinions differed in the
two years.
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21. Step 1: Significance test for π2 - π1
 The parameter of interest is π2 - π1
 Assumptions:
 the sample is drawn from a random sample of some sort,
 the parameter of interest is a variable with an interval
scale,
 the sample size is large enough that the sampling
distribution of Pihat2 – Pihat1 is approximately normal.
 The two samples are drawn independently
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22. Step 2: Significance test for π2 - π1
The null hypothesis will be that there is no
difference between the population proportions. This
means that any difference we observe is due to
random chance.
Ho: π2 - π1 = 0
(State an alpha here if you want to.)
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23. Step 3: Significance test for π2 - π1
The test statistic has a standard form:
 z = (estimate of parameter – Ho value of parameter)
standard error of parameter

(  ˆ 

ˆ )
2 1

1 1
 
 Where pihat is the overall weighted average
 This means we are assuming equal variance in the two
populations.
 Q: why do we use an assumption of equal variance to
estimate the standard error for the t-test?
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   

 


1 2
ˆ 1 ˆ
n n
z
 
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24. Step 4: Significance test for π2 - π1
P-value of calculated z:
• Table A, or
• Stata: display 2 * (1 – normal(z) ), or
• Stata: testi (no data, just parameters)
• Stata: ttest (if data file in memory)
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25. Step 5: Significance test for π2 - π1
Conclusion:
 Compare the p-value from step 4 to the alpha level
in step 1.
If p < α, reject H0 If p ≥ α, do not reject H0
 State a conclusion about the statistical significance
of the test.
 Briefly discuss the substantive importance of your
findings.
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26. Significance test for π2 - π1: Example
1. Assumptions: random sample, interval-scale variable,
sample size large enough that the sampling distribution of
2 - 1is approximately normal, independent groups
2. Hypothesis: Ho: π2 - π1= 0
3. Test statistic:
z = (122/345 – 268/1900) /
SQRT[(390/2245)*(1 - 390/2245)*(1/345 + 1/1900)]
= 9.59
4. p-value: p<<.001
5. conclusion:
a. reject H0: attitudes were clearly different in 1994 than in 1982.
b. furthermore, the observed difference of .21 is a substantively
important change in attitudes.
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27. Comparisons of two independent population proportions:
Confidence Interval
 confidence interval:
P P
2 2
P P
1 1
 Notice that there is no overall weighted average
Pihat, as there is in a significance test for
proportions.
 Instead, we estimate two separate variances from the
separate proportions.
 Why?
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 
2
1
2 1
(1 ) (1 )
. .
n
n
c i P P z



  
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28. STATA: Significance test for π2 - π1:
immediate command
. prtesti 345 .3536 1900 .1411
 STATA needs the following information:
 sample size for group 1 (n = 345)
 proportion for group 1 (p = 122/345)
 sample size for group 2 (n = 1900)
 proportion for group 2 (p = 268/1900)
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29. STATA: Significance test for π2 - π1:
immediate command
. prtesti 345 .3536 1900 .1411
Two-sample test of proportion x: Number of obs = 345
y: Number of obs = 1900
------------------------------------------------------------------------------
Variable | Mean Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
x | .3536 .0257393 .3031518 .4040482
y | .1411 .0079865 .1254467 .1567533
-------------+----------------------------------------------------------------
diff | .2125 .0269499 .1596791 .2653209
| under Ho: .0221741 9.58 0.000
------------------------------------------------------------------------------
Ho: proportion(x) - proportion(y) = diff = 0
Ha: diff < 0 Ha: diff != 0 Ha: diff > 0
z = 9.583 z = 9.583 z = 9.583
P < z = 1.0000 P > |z| = 0.0000 P > z = 0.0000
Note the use of one standard error (unequal variance) for the
confidence interval, School.and another edhole.(equal com
variance) for the
significance test. 29

30. STATA command for a data set (#1)
. prtest nonstandard if (RACECEN1==1 | RACECEN1==2), by(RACECEN1)
Two-sample test of proportion 1: Number of obs = 1389
2: Number of obs = 260
------------------------------------------------------------------------------
Variable | Mean Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
1 | .2800576 .0120482 .2564436 .3036716
2 | .3538462 .0296544 .2957247 .4119676
-------------+----------------------------------------------------------------
diff | -.0737886 .0320084 -.1365239 -.0110532
| under Ho: .0307147 -2.40 0.016
------------------------------------------------------------------------------
diff = prop(1) - prop(2) z = -2.4024
Ho: diff = 0
Ha: diff < 0 Ha: diff != 0 Ha: diff > 0
Pr(Z < z) = 0.0081 Pr(|Z| < |z|) = 0.0163 Pr(Z > z) = 0.9919
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31. STATA command for a data set (#1)
. gen byte wrkslf0=wrkslf-1
(152 missing values generated)
. prtest wrkslf0 if wrkstat==1, by(sex)
Two-sample test of proportion male: Number of obs = 874
female: Number of obs = 743
------------------------------------------------------------------------------
Variable | Mean Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
male | .8272311 .0127876 .8021678 .8522944
female | .9044415 .0107853 .8833027 .9255802
-------------+----------------------------------------------------------------
diff | -.0772103 .0167286 -.1099978 -.0444229
| under Ho: .0171735 -4.50 0.000
------------------------------------------------------------------------------
diff = prop(male) - prop(female) z = -4.4959
Ho: diff = 0
Ha: diff < 0 Ha: diff != 0 Ha: diff > 0
Pr(Z < z) = 0.0000 Pr(|Z| < |z|) = 0.0000 Pr(Z > z) = 1.0000
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