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on 1 = proportion of items in the success category.
f items in the failure category.
e of n
1
items, let X
1
be the number of items from the success category.
= ,
on 2= proportion of items in the success category.
f items in the failure category.
e of n
2
items, let X
2
be the number of items from the success category. = ,
STAT 2103 Class Topics
Chapter 6, Section 6.6
Testing of Hypotheses about the Binomial Proportion p
The methodology requires a formulation of the null and alternative Hypotheses. In any hypothesis testing
problem, only one of the following formulations will be suitable for testing.
H
0
: p
p
≤
0
against
H
a
: p > p
0
(
onesided)
Or
H
0
: p
p
≥
0
against
H
a
: p < p
0
(
onesided)
Or
H
0
: p = p
0
against
H
a
: p
p
≠
0
(
twosided).
In the hypotheses,
p
0
is a specified value for
p.
For either formulation, the test statistic is:
,
For one sided alternative,
Pvalue = P(zdistribution
>
the absolute value of zstatistic).
For two sided alternative,
Pvalue = Twice the Pvalue for onesided alternative.
Decision rule:
Reject H
0
if Pvalue <
, level of significance.
α
Hippocrates magazine (May/June, 1989) reported that 30% of American children skip breakfast.
Suppose
a foreign social scientist believes that the rate is higher for children in her country.
What can we
conclude in this regard if a sample of 500 children revealed that 172 skip breakfast?
Test at the .05
significance level.
Using StatCrunch and following the same steps as above until you will select hypothesis test.
1
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View Full Document Click on next and enter the value for the Null: prop=.30, and select “>” for the Alternative.
2
Click on Calculate
Hypothesis test results:
p : proportion of successes for population
H
0
: p = 0.3
H
A
: p > 0.3
Proportio
n
Coun
t
Tota
l
Sample
Prop.
Std. Err.
ZStat
Pvalue
p
172
500
0.344
0.020493902
2.14698
0.0159
Since the pvalue of 0.0159 is lower than alpha of .05 the social scientist can conclude that the rate of
children who skip breakfast in her country is higher than 30%.
Chapter 7
Confidence Interval for Estimating the Difference Between Two Means
μ
A (1 –
)100% degree of confidence interval for estimating
α
μ(1) μ(2)
is:
Margin of Error, ME is to the right of,

X1 X2

±
X1 X2 ME
Comparing Means of Two Populations
Comparison of two populations’ means is required when seeking answers to questions of the following
types: Is it true that those who eat highfiber cereal for breakfast consume on average fewer calories at
3
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View Full Document lunch than those who don’t eat highfiber cereal?
The mean caloric intakes may be denoted by μ
1
for
those who eat highfiber cereal and by μ
2
for those who don’t eat highfiber cereal. The question then can
be formulated as a hypothesis testing:
H
a
: (μ
1
 μ
2
) < 0
against
the null
H
0
: (μ
1
 μ
2
)
0.
≥
In general let, μ
1
and μ
2
denote the means of two populations. One of the following formulations will be
formulated for testing.
H
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This note was uploaded on 05/02/2011 for the course STATISTICS 2103 taught by Professor Zhao during the Spring '11 term at Temple.
 Spring '11
 Zhao
 Statistics

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