Outline

• Hypothesis tests – population proportion• Example 1• Example 2

Lecture 10

Hypothesis Tests – Population proportion

• Inferences about population proportions are often made in the context of the probability, p, of success for a binomial distribution.

– For example, how many people plan to vote ‘Liberal’ in the next election?

– We would define Success = vote Liberal, Failure = vote non-Liberal.

Lecture 10

Hypothesis Tests – Population proportion

• The sample mean proportion, , (“p-hat”) is obtained by dividing # of successes by sample size.

• For large samples, is approximately normally distributed. By C.L.T., we can use Z to test a hypothesis about p.

– We use to test a hypothesis about p.

p̂

p̂

p̂

Lecture 10

Hypothesis Tests – Population Proportion

• Before using the Z test for p, we need to know 2 things:

1. That n is large enough. n is large enough if:

0 < p0 ± 3 < 1

Alternative: n is large enough if np > 5 and nq > 5 (where q = 1-p)

p̂

Lecture 10

Hypothesis Tests – Population Proportion

• Before using the Z test for p, we need to know 2 things:

2. The standard error of the proportion,

= √(pq/n) √(p0q0/n)

Given these, we can now do a Z test…

p̂

p̂

The values in the null hypothesis

Lecture 10

Hypothesis Tests – Population Proportion

•H0: p = p0

•HA: p < p0 HA: p ≠ p0 •or HA: p > p0

•(One-tailed test) (Two-tailed test)

•Test Statistic: Z = - p0

•

p̂

p̂

Lecture 10

Hypothesis Tests – Population proportion

Rejection region:

Z < -Zα │Z│< Zα/2

or Z > Zα

p̂

Lecture 10

Population Proportion – Example 1

• Prior to the use of fertility drugs, one in 80 pregnancies in Canada resulted in multiple births (twins, triplets, quadruplets, etc.). Since the use of fertility drugs has become more common, it is suspected that a higher proportion of pregnancies are now resulting in multiple births. A 2004 survey of 1000 women who gave birth at randomly selected hospitals across Canada found that 20 of these were multiple births.

Lecture 10

Population Proportion – Example 1

• a. Do the data from the 2004 survey support the notion that more multiple births are occurring these days? (α = .05)

• b. Use the 2004 survey data to form a 92% C.I. to estimate the true current proportion of multiple births.

Population Proportion – Example 1a

• Test: σp = √p0q0/n = √(.0125 *.9875)/1000 = .003

• np = 1000 * .0125 = 12.5• nq = 1000 * .9875 = 987.5

• So, we can use the normal approximation…

Lecture 10

^

Lecture 10

Population Proportion – Example 1a

• H0: p = 1/80 = .0125

• HA: p > .0125

• Test Statistic: Z = p – p0

•

• Rejection region: Z > 1.645

p̂

^

Lecture 10

Population Proportion – Example 1a

• Zobt = (20/1000) - .0125• √(.0125)(.9875)/1000

• = 2.135

• Decision: Reject H0 – There is evidence that more multiple births are occurring nowadays.

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Population Proportion – Example 1b

• For 92% C.I., use Z.46 = 1.75.

• p = .02 ± 1.75 (√(.02) (.98)/1000• p = .02 ± .0077

• (.01225 ≤ p ≤ .02775)

Lecture 10

Population proportion – Example 2

• A film company has been very successful producing action films. Recently, the company has started to worry that the population is aging, meaning that the average age of people in the population is going up. Since younger people are more likely to go to action movies, an aging population could mean trouble for the company.

Lecture 10

Population proportion – Example 2

• a. The film company has a researcher sample 334 people, asking their age. 124 of these people are aged 25 or less. Use this result to form a 99% C.I. for the true proportion of people in the population who are currently in the age group 25 or less.

• b. At the 1990 census, 40% of the population was in the age group 25 or less. Based on the sample of 334 people, has this significantly decreased since then? (α ≤ .05)

Lecture 10

Population Proportion – Example 2a

• a. C.I. = ± Zα/2 ( ) ± Zα/2 (√(pq/n)

• p = x• n• p = 124 = .3713• 334

• q = (1 – p) = .6287

p̂p̂

^^

Lecture 10

Population Proportion – Example 2a

• = √(.3713)(.6287)/334 = √.0006989 = .0264

• For 99% C.I., use α/2 = .01/2 = .005. Z.005 = 2.575

• C.I. = .3713 ± 2.575 (.0264)• = .3713 ± .0681

• (.3032 ≤ p ≤ .4394)

p̂

Note “p,” not p-hat

Lecture 10

Population Proportion – Example 2b

• H0: p = .40

• HA: p < .40

• Test Statistic: Z = p - p0

•

• Rejection region: Zobt < -Z.45 = -1.645

p̂

^

Lecture 10

Population Proportion – Example 2b

• Zobt = .3713 – .40 = -.0287• √(.40)(.60)/334 .0268

• Zobt = -1.071

• Decision: Do not reject H0. There is not sufficient evidence that proportion of the population aged 25 or less has changed.