Homogeneity of Proportions Worksheet - Page 2

Homogeneity of Proportions Worksheet
• Page 2
11.
A college administration reported that more than 60% of the student body is in favour of candidate John being elected as the student council officer. From a random sample of 100 students it was found that 62 voted in favour of John. Find the $p$-value. Is there enough evidence to support the college administration's report on student council officer at α = 0.01?
 a. 0.3409; yes b. 0.3409; no c. 0.1704; no d. 0.1704; yes

Solution:

H0: p ≤ 0.60 and H1: p > 0.60
[Null and Alternate Hypotheses.]

Sample proportion, pˆ = 62 / 100 = 0.62

The test value, z = pˆ-pp(1-p)n, where p is the population proportion and n the sample size.

z = 0.62-0.60(0.60)(0.40)100 = 0.41
[Substitute the values and simplify.]

Area under the standard normal distribution curve for z = 0.41 is 0.1591. That is P(0 < z < 0.41) = 0.1591
[Use the standard normal distribution table.]

= 0.5000 - 0.1591= 0.3409
P-value for a right tailed test = P(z > 0.41)

The P-value 0.3409 > 0.01, do not reject the null hypothesis.

So, there is not enough evidence to support the college administration's report on student council officer that more than 60% of the student body is in favour of candidate John.

12.
A survey suggests that 90% of the chief executives of the largest companies in the United States are of above average height. From a random sample of 140 executives it was observed that 119 are of above average height. Find the $p$-value for the test. Is there enough evidence to reject the survey report at α = 0.01?
 a. 0.0244; no b. 0.0488; no c. 0.0488; yes d. 0.0244; yes

Solution:

H0: p = 0.90 and H1: p ≠ 0.90
[Null and Alternate Hypotheses.]

Sample proportion, pˆ = 119 / 140 = 0.85

The test value, z = pˆ-pp(1-p)n, where p is the population proportion and n the sample size.

z = 0.85-0.90(0.90)(0.10)140 = - 1.97
[Substitute the values and simplify.]

Area under the standard normal distribution curve for z = - 1.97 is 0.4756. That is P(- 1.97 < z < 0)
[Use the standard normal distribution table.]

= 2(0.5000 - 0.4756) = 0.0488
P-value for a two tailed test = 2 × P(z < - 1.97)

The P-value 0.0488 > 0.01, do not reject the null hypothesis.

So, there is not enough evidence to reject the survey report that 90% of the chief executives of the largest companies in the United States are of above average height.

13.
According to the U.S. Department of Education, 26% of adults completed a four year degree after the age of 25. A researcher enquired 140 adults and found that 42 of them completed the degree course after the age of 25. Find the p-value for the test. Is the education department's report at α = 0.05 correct?
 a. 0.2802; no b. 0.2802; yes c. 0.1401; no d. 0.1401; yes

Solution:

H0: p = 0.26 and H1: p ≠ 0.26
[Null and Alternate Hypotheses.]

Sample proportion, pˆ = 42 / 140 = 0.30

The test value, z = pˆ-pp(1-p)n, where p is the population proportion and n the sample size.

z = 0.30-0.26(0.26)(0.74)140 = 1.08
[Substitute the values and simplify.]

Area under the standard normal distribution curve for z = 1.08 is 0.3599. That is P(0 < z < 1.08) = 0.3599
[Use the standard normal distribution table.]

= 2(0.5000 - 0.3599) = 0.2802
P-value for a two tailed test = 2 × P(z > 1.08)

The P-value 0.2802 > 0.05, do not reject the null hypothesis.

So, there is not enough evidence to reject the education department report that 26% of adults completed their four year degree after the age of 25.

14.
The population survey in 2001 stated that 37% of urban workers held multiple jobs in order to earn extra money. From a random sample of 200 urban workers taken, it was found that 70 of them held multiple jobs. Find the $p$-value for the test. Does the sample proportion differ from the population survey report at α = 0.01?
 a. 0.562; yes b. 0.281; yes c. 0.281; no d. 0.562; no

Solution:

H0: p = 0.37 and H1: p ≠ 0.37
[Null and Alternate Hypotheses.]

Sample proportion, pˆ = 70 / 200 = 0.35

The test value, z = pˆ-pp(1-p)n, where p is the population proportion and n the sample size.

z = 0.35-0.37(0.37)(0.63)200 = - 0.58
[Substitute the values and simplify.]

Area under the standard normal distribution curve for z = - 0.58 is 0.2190. That is P(- 0.58 < z < 0) = 0.2190
[Use the standard normal distribution table.]

= 2(0.5000 - 0.2190) = 0.562
P-value for a two tailed test = 2 × P(z < - 0.58)

The P-value 0.562 > 0.01, do not reject the null hypothesis.

There is not enough evidence to support the claim that the sample proportion differ from the population proportion 0.37.

15.
In a statistical hypothesis test, $p$ is the population proportion, $\stackrel{ˆ}{p}$ the sample proportion and $n$ the sample size then the formula for the $z$ test is

 a. I only b. IV only c. III only d. II only

Solution:

Formula for z test for proportions is pˆ-pp(1-p)n, where p is the population proportion, pˆ the sample proportion and n the sample size.