Confidence Intervals and Significance Testing
 Comparing two populations or the responses in two
treatments based on two independent samples
 Comparing Two Means (quantitative variables)
 T- Tests
 Null Hypothesis is the means are equal
 Comparing Proportions of Success in Two Groups
 Z –Tests
 Null Hypothesis is the proportions are equal
 We need:
 SRS from each population
 Populations 10 times sample size
 n1(phat1); n1(1-phat1); n2(phat2); n2(1-phat2) all greater than 10
Confidence Interval (CI)
 phat1  phat2   z * SE
Standard Error (SE)
phat1 (1  phat1 ) phat2 (1  phat2 )

n1
n2
 A study was conducted to decide if attendance by poor
children in preschool effects their later involvement with
social services. Two groups were studied; 62, 3 and 4 year
olds who attended preschool and a control group of 61 who
did not. Of the preschool attendees, 38 needed social
services and 49 of the controls needed the services. What
is the difference in the percentages? Find a 95% confidence
interval for this difference.
(.8033  .6129)  1.96( SE )
(.8033)(.1967) (.6129)(.3871)
SE 

61
62
SE = .0801
CI = .0334 to .3474
 Pooled Sample Proportion (used in the SE for phat)
 The sum of the two counts divided by the sum of the two
samples
 Estimate of the single population sample
phat (SE)
X1 + X2
n1 + n2
X1 , X2 = the # of successes for each sample
This helps make the two sample a standard normal
distribution.
 **Check all Conditions
Ha: µ > µ0
 Ho: p1 = p2
 Calculate the Z statistic
Ha: µ < µ0
 Find the appropriate P value
z
( phat1  phat 2 ) Ha: µ ≠ µ0
1 1
phat (1  phat )(  )
n1 n2
This is the POOLED PHAT!!!
2(p-value)
 A study was conducted to decide if attendance by poor
children in preschool effects their later involvement
with social services. p
Two
groups were studied; 62 3
=
.0102;
and 4 year olds who attended preschool and a control
group of 61 who did reject
not. OfNull
the preschool attendees,
38 of the preschoolers
needed social services and 49 of
Hypothesis
the controls needed the services. Test to see if the
study provides significant evidence that preschool
reduces the later need for social services.
(.8032  .6129)
H0 : p1 = p2
1) SRS
z
Ha : p1 > p2
2) 61(.8032) = 49; 61(.1968) = 12
1
1
p1 = control
3) 62(.6129) = 38; 62(.3871) = 24
.7073(.2927)(  )
p2 = preschool
61 62
The pooled phat =
(38+49)/(61+62) = .7073
z = 2.32
 Stat – Tests – 5: 1-Prop Z Test (one prop test)
 Stat – Tests – A: 1-Prop Z Int (one prop CI)
 Stat – Tests – 6: 2-Prop Z Test (two prop test)
 Stat – Tests – B: 2-Prop Z Int (two prop CI)
Chapter 12: 28-30,32,34
M&M Testing