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 |

[Null and Alternate Hypotheses.]

Sample proportion,

The test value,

[Substitute the values and simplify.]

Area under the standard normal distribution curve for

[Use the standard normal distribution table.]

= 0.5000 - 0.1591= 0.3409

P-value for a right tailed test = P(

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.

Correct answer : (2)

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 |

[Null and Alternate Hypotheses.]

Sample proportion,

The test value,

[Substitute the values and simplify.]

Area under the standard normal distribution curve for

[Use the standard normal distribution table.]

= 2(0.5000 - 0.4756) = 0.0488

P-value for a two tailed test = 2 × P(

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.

Correct answer : (2)

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 |

[Null and Alternate Hypotheses.]

Sample proportion,

The test value,

[Substitute the values and simplify.]

Area under the standard normal distribution curve for

[Use the standard normal distribution table.]

= 2(0.5000 - 0.3599) = 0.2802

P-value for a two tailed test = 2 × P(

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.

Correct answer : (2)

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 |

[Null and Alternate Hypotheses.]

Sample proportion,

The test value,

[Substitute the values and simplify.]

Area under the standard normal distribution curve for

[Use the standard normal distribution table.]

= 2(0.5000 - 0.2190) = 0.562

P-value for a two tailed test = 2 × P(

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.

Correct answer : (4)

15.

In a statistical hypothesis test, $p$ is the population proportion, $\stackrel{\u02c6}{p}$ 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 |

Correct answer : (1)