Download Section 9.3 ~ Hypothesis Tests for Population Proportions

Section 9.3 ~
Hypothesis Tests for Population Proportions
Introduction to Probability and Statistics
SPRING 2016
Sec. 9.3
Objective

After this section you will understand and
interpret hypothesis tests for claims made
about population proportions.
Sec. 9.3
Hypothesis Tests for Population Proportions

The four step process for hypothesis testing is the same when
dealing with population proportions as it is when dealing with
population means; the only difference is the method for
calculating the standard deviation

The standard deviation of a distribution of sample proportions is:
p(1  p)
n

The z-score is then found by comparing the sample statistic (
the null hypothesis (p) and dividing by the standard deviation:
z
pˆ  p
p(1  p) / n
p̂ ) to
Sec. 9.3
Hypothesis Tests for Population Proportions

Example 1 ~

Suppose the national unemployment rate is 3.5%. In a survey of n = 450
people in a rural Wisconsin county, 22 people are found to be unemployed.
County officials apply for state aid based on the claim that the local
unemployment rate is higher than the national average. Test this claim at the
.05 significance level.

Step 1: Write the null and alternative hypotheses
H 0 : p  .035

H a : p  .035
Step 2: Draw a sample and come up with a sample statistic and the standard
deviation of that sample
 n = 450
 The sample proportion is:
pˆ 

22
 0.0489
450
The standard deviation is:
p(1  p)
n

.035(1  .035)
450
Sec. 9.3
Hypothesis Tests for Population Proportions

Example 1 Cont’d…

Step 3: Determine the P-value (or compare to critical values) and
determine the level of significance
z



pˆ  p
p(1  p) / n


z
.0489  .035
.00866
.035(1
 .035)
 z  1.6
450
A z-score of 1.6 corresponds to a probability of .9452, but since this is a righttailed test, the probability that the unemployment rate is higher than .0489 is
.0548 (1 - .9452)
 This is higher than .05, which means that it’s not significant at the .05
level
You could have also compared the z-score of 1.6 to the critical value for a
right-tailed test at the .05 level (1.645) to determine if it’s significant at the
.05 level
Step 4: Determine whether to reject or not reject the null hypothesis


Since an unemployment rate of .0489 is not significant at the .05 level, this
tells us that we should NOT reject the null hypothesis
In other words, there is not enough evidence to support the claim that this
particular county has an unemployment rate higher than the national average
Sec. 9.3
Hypothesis Tests for Population Proportions

Example 2 ~

A random sample of n = 750 people is selected, of whom 92 are left-handed.
Use these sample data to test the claim that 10% of the population is left
handed.

Step 1: Write the null and alternative hypotheses
H 0 : p  .10

H a : p  .10
Step 2: Draw a sample and come up with a sample statistic and the standard
deviation of that sample
 n = 750
 The sample proportion is:
pˆ 

92
 0.12267
750
The standard deviation is:
p(1  p)
n

.10(1  .10)
750
Sec. 9.3
Hypothesis Tests for Population Proportions

Example 2 Cont’d…

Step 3: Determine the P-value (or compare to critical values) and
determine the level of significance
z

pˆ  p
p(1  p) / n

z

.12267  .10
.01095
.10(1
 .10)
 z  2.07
750
The probability that corresponds with 2.07 is .9808, so the probability above
is .0192 (1 - .9808).
 Since 2*.0192 = .0384 is than .05, which means that it is
significant at the .05 level


You could have also just compared the z-score of 2.07 to the critical values of
a two-tailed test at the .05 level (1.96)
Step 4: Determine whether to reject or not reject the null hypothesis
 Since the sample provided data that was statistically significant, we
can reject the null hypothesis and support the claim that the
proportion of the population that is left-handed is not 10%
Time to Practice!
Pg 399 #10, 11, 17
HOMEWORK
Pg 399
#1, 4-6, 12-16all
Pg 391-393
#3, 53