1

Section 8.3Testing a claim about a Proportion

Objective

For a population with proportion p, use a sample (with a sample proportion) to test a claim about the proportion.

Testing a proportion uses the standard normal distribution (z-distribution)

2

Notation

3

(1) The sample used is a a simple random sample (i.e. selected at random, no biases)

(2) Satisfies conditions for a Binomial distribution

(3) n p0 ≥ 5 and n q0 ≥ 5

Requirements

Note: p0 is the assumed proportion, not the sample proportion

Note: 2 and 3 satisfy conditions for the normal approximation to the binomial distribution

4

Test StatisticDenoted z (as in z-score) since the test uses the z-distribution.

5

If the test statistic falls within the critical region, reject H0.

If the test statistic does not fall within the critical region, fail to reject H0 (i.e. accept H0).

Traditional method:

6

Types of Hypothesis Tests:Two-tailed, Left-tailed, Right-tailed

The tails in a distribution are the extreme regions where values of the test statistic agree with the alternative hypothesis

7

Left-tailed Test “<”

H0: p = 0.5

H1: p < 0.5 significance level

Area =

-z

(Negative)

8

Right-tailed Test “>”

H0: p = 0.5

H1: p > 0.5 significance level

Area =

z

(Positive)

9

Two-tailed Test “≠”

H0: p = 0.5

H1: p ≠ 0.5 significance level

z

Area = Area =

-z

10

The XSORT method of gender selection is believed to increases the likelihood of birthing a girl.

14 couples used the XSORT method and resulted in the birth of 13 girls and 1 boy.

Using a 0.05 significance level, test the claim that the XSORT method increases the birth rate of girls.

(Assume the normal birthrate of girls is 0.5)

What we know: p0 = 0.5 n = 14 x = 13 p = 0.9286

Claim: p > 0.5 using α = 0.05

Example 1

n p0 = 14*0.5 = 7 n q0 = 14*0.5 = 7

Since n p0 > 5 and n q0 > 5, we can perform a hypothesis test.

11

H0 : p = 0.5

H1 : p > 0.5

Example 1

Right-tailed

What we know: p0 = 0.5 n = 14 x = 13 p = 0.9286

Claim: p > 0.5 using α = 0.01

z in critical region

z = 3.207zα = 1.645

Test statistic:

Critical value:

Initial Conclusion: Since z is in the critical region, reject H0

Final Conclusion: We Accept the claim that the XSORT method increases the birth rate of girls

12

P-Value

The P-value is the probability of getting a value of the test statistic that is at least as extreme as the one representing the sample data, assuming that the null hypothesis is true.

z Test statistic

zα Critical value

z zαP-value = P(Z > z)

p-value(area)

Example

13

P-Value

Critical region in the left tail:

Critical region in the right tail:

Critical region in two tails:

P-value = area to the left of the test statistic

P-value = area to the right of the test statistic

P-value = twice the area in the tail beyond the test statistic

14

P-Value method:

If the P is low, the null must go.If the P is high, the null will fly.

If P-value , reject H0.

If P-value > , fail to reject H0.

15

Caution

Don’t confuse a P-value with a proportion p.Know this distinction:

P-value = probability of getting a test statistic at least as extreme as the one representing sample data

p = population proportion

16

Calculating P-value for a Proportion

Stat → Proportions → One sample → with summary

17

Calculating P-value for a Proportion

Enter the number of successes (x) and the number of observations (n)

18

Calculating P-value for a Proportion

Enter the Null proportion (p0) and select the alternative hypothesis (≠, <, or >)

Then hit Calculate

19

Calculating P-value for a Proportion

The resulting table shows both the test statistic (z) and the P-value

Test statistic

P-value

P-value = 0.0007

20

Using P-value

Initial Conclusion: Since p-value < α (α = 0.05), reject H0

Final Conclusion: We Accept the claim that the XSORT method increases the birth rate of girls

P-value = 0.0007

Stat → Proportions→ One sample → With summary

Null: proportion=

Alternative

Number of successes:

Number of observations:

H0 : p = 0.5

H1 : p > 0.5

Example 1What we know: p0 = 0.5 n = 14 x = 13 p = 0.9286

Claim: p > 0.5 using α = 0.01

13

14

0.5

>

● Hypothesis Test

21

Do we prove a claim?

A statistical test cannot definitely prove a hypothesis or a claim.

Our conclusion can be only stated like this:

The available evidence is not strong enough to warrant rejection of a hypothesis or a claim

We can say we are 95% confident it holds.

“The only definite is that there are no definites” -Unknown

22

Mendel’s Genetics Experiments

When Gregor Mendel conducted his famous hybridization experiments with peas, one such experiment resulted in 580 offspring peas, with 26.2% of them having yellow pods. According to Mendel’s theory, ¼ of the offspring peas should have yellow pods. Use a 0.05 significance level to test the claim that the proportion of peas with yellow pods is equal to ¼.

What we know: p0 = 0.25 n = 580 p = 0.262

Claim: p = 0.25 using α = 0.05

Example 2

n p0 = 580*0.25 = 145 n q0 = 580*0.75 = 435

Since n p0 > 5 and n q0 > 5, we can perform a hypothesis test.

Problem 32, pg 424

23

H0 : p = 0.25

H1 : p ≠ 0.25

Example 2

Two-tailed

z not in critical region

z = 0.667

zα = -1.960

Test statistic:

Critical value:

Initial Conclusion: Since z is not in the critical region, accept H0

Final Conclusion: We Accept the claim that the proportion of peas with yellow pods is equal to

¼

What we know: p0 = 0.25 n = 580 p = 0.262

Claim: p = 0.25 using α = 0.05

zα = 1.960

24

Using P-valueExample 2

Initial Conclusion: Since P-value > α, accept H0

Final Conclusion: We Accept the claim that the proportion of peas with yellow pods is equal to

¼

H0 : p = 0.25

H1 : p ≠ 0.25

What we know: p0 = 0.25 n = 580 p = 0.262

Claim: p = 0.25 using α = 0.05

x = np = 580*0.262 ≈ 152 P-value = 0.5021

Stat → Proportions→ One sample → With summary

Null: proportion=

Alternative

Number of successes:

Number of observations:

152

580

0.25

≠

● Hypothesis Test