# Homogeneity of Proportions Worksheet

Homogeneity of Proportions Worksheet
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1.
150 children of K - 12 grades whose parents (at least one or both) are employed are selected randomly. It was found that 18 of them participate in after school programs. Test the hypothesis that 14% of such K - 12 children participate in after school programs. [Use a 5% level of significance.]
 a. $z$ = 1.41, there is not enough evidence that 14% of the children participate in after school programs b. $z$ = - 0.71, there is enough evidence that 14% of the children participate in after school programs c. $z$ = - 0.71, there is not enough evidence that 14% of the children participate in after school programs d. $z$ = - 1.41, there is enough evidence that 14% of the children participate in after school programs

#### Solution:

H0: p = 0.14 and H1: p ≠ 0.14
[Null and Alternate Hypotheses.]

Sample proportion, pˆ = Xn = 18 / 150 = 0.12

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

z = 0.12-0.14(0.14)(0.86)150 = - 0.71
[Substitute the values and simplify.]

The critical values for a two tailed test at α = 0.05 are + 1.96 and - 1.96.
[Use the standard normal distribution table.]

The critical values, test value, the critical and noncritical regions are shown in the figure.

z = - 0.71 > - 1.96, we fail to reject the null hypothesis, since the test value falls in the non critical region as shown in the figure.

There is enough evidence that 14% of K -12 children whose both parents or single parent hold jobs, participate in after school programs.

2.
90 primary school children out of 100 randomly selected, view television on an average day. Test the hypothesis that more than 83% of the primary school children view television on an average day. Use α = 0.05 level of significance.
 a. $z$ = 1.86, there is enough evidence to support the hypothesis b. $z$ = - 1.32, there is not enough evidence to support the hypothesis c. $z$ = 1.32, there is enough evidence to support the hypothesis d. $z$ = - 1.86, there is not enough evidence to support the hypothesis

#### Solution:

H0: p ≤ 0.83 and H1: p > 0.83
[Null and Alternate Hypotheses.]

Sample proportion, pˆ = Xn = 90 / 100 = 0.90

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

z = 0.90-0.83(0.83)(0.17)100 = 1.86
[Substitute the values and simplify.]

The critical value for a right tailed test at α = 0.05 is 1.65.
[Use the standard normal distribution table.]

The critical value, test value, the critical and noncritical regions are shown in the figure.

z = 1.86 > 1.65, we reject the null hypothesis, since the test value falls in the critical region as shown in the figure.

There is enough evidence to support that more than 83% of the primary school children in a state view television on an average day.

3.
Internet services reported that 33% of the secondary school principals use internet email more than once a day either in home or school. From a random sample of 120 secondary school principals, it was found that 42 use internet email more than once a day. Is there enough evidence to reject the internet services report at α = 0.01? What is the test value?
 a. yes, 0.46 b. yes, - 0.46 c. no, - 0.46 d. no, 0.46

#### Solution:

H0: p = 0.33 and H1: p ≠ 0.33
[Null and Alternate Hypotheses.]

Sample proportion, pˆ = Xn = 42 / 120 = 0.35

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

z = 0.35-0.33(0.33)(0.67)120 = 0.46
[Substitute the values and simplify.]

The critical values at α = 0.01 for a two tailed test are + 2.58 and - 2.58.
[Use the standard normal distribution table.]

The critical values, test value, the critical and noncritical regions are shown in the figure.

z = 0.46 < 2.58, we fail to reject the null hypothesis, since the test value falls in the non critical region as shown in the figure.

There is not enough evidence to reject the internet services report that the secondary school principals use internet email more than once a day either in home or school.

4.
A survey reported that more than 60% of people who go to the movies on a weekend eat out either before or after the movie. A researcher interviewed 144 of them and found that 101 eat out before or after the movie. At α = 0.01, is there enough evidence to support the survey report that more than 60% of people who go the movies on a weekend eat out? Use the $p$-value method.
What is the $z$ value?
 a. no, $z$ = 2.45 b. no, $z$ = 1.73 c. yes, $z$ = 2.45 d. yes, $z$ = 1.73

#### Solution:

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

Sample proportion, pˆ = 101 / 144 = 0.70

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

z = 0.70-0.60(0.60)(0.40)144 = 2.45
[Substitute the values and simplify.]

Area under the standard normal distribution curve for z = 2.45 is 0.4929. That is P(0 < z < 2.45) = 0.4929
[Use the standard normal distribution table.]

= 0.5000 - 0.4929 = 0.0071
P-value for a right tailed test = Area in the right tail = P(z > 2.45)

The P-value 0.0071 < 0.01, reject the null hypothesis.

So, there is enough evidence to support the survey report that more than 60% of people who go the movies on a weekend eat out either before or after the movie.

5.
A survey reports that 40% of Americans planning for longer vacations travel by plane. 160 Americans who were planning for longer vacations were interviewed and it was found that 80 of them were planning to travel by plane. Find the $p$-value for the test. Is there enough evidence to reject the survey report at α = 0.02?
 a. 0.0049; yes b. 0.0049; no c. 0.0098; yes d. 0.0098; no

#### Solution:

H0: p = 0.40 and H1: p ≠ 0.40
[Null and Alternate Hypotheses.]

Sample proportion, pˆ = 80 / 160 = 0.50

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

z = 0.50-0.40(0.40)(0.60)160 = 2.58
[Substitute the values and simplify.]

Area under the standard normal distribution curve for z = 2.58 is 0.4951. That is P(0 < z < 2.58) = 0.4951
[Use the standard normal distribution table.]

= 2(0.5000 - 0.4951) = 0.0098
P-value for a two tailed test = 2 × P(z > 2.58)

The P - value 0.0098 < 0.02, reject the null hypothesis.

So, there is enough evidence to reject the survey report that 40% of Americans planning for longer vacations travel by plane.

6.
The manager of a theatre claims that 94% of people order drinks in the theatre. A sales person of drinks observes that 114 people out of 120 ordered drinks in the theatre. Find the $p$-value for the test. Use α = 0.01, is there enough evidence to reject the manager's claim?
 a. 0.6456; yes b. 0.3228; yes c. 0.3228; no d. 0.6456; no

#### Solution:

H0: p = 0.94 and H1: p ≠ 0.94
[Null and Alternate Hypotheses.]

Sample proportion, pˆ = 114 / 120 = 0.95

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

z = 0.95-0.94(0.94)(0.06)120 = 0.46
[Substitute the values and simplify.]

Area under the standard normal distribution curve for z = 0.46 is 0.1772. That is P(0 < z < 0.46) = 0.1772
[Use the standard normal distribution table.]

= 2(0.5000 - 0.1772) = 0.6456
P-value for a two tailed test = 2 × P(z > 0.46)

The P-value 0.6456 > 0.01, do not reject the null hypothesis.

So, there is not enough evidence to reject the manager's claim that 94% of people order drinks in the theatre.

7.
A survey reported that 38% of Americans go to the supermarket during noon to 5 p.m. In a random sample of 130 Americans, it was found that 47 responded the same. Find the $p$-value for the test. Is the survey report correct at α = 0.05?
 a. 0.3192; no b. 0.6384; no c. 0.6384; yes d. 0.3192; yes

#### Solution:

H0: p = 0.38 and H1: p ≠ 0.38
[Null and alternate hypotheses.]

Sample proportion, pˆ = 47 / 130 = 0.36

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

z = 0.36-0.38(0.38)(0.62)130 = - 0.47
[Substitute the values and simplify.]

Area under the standard normal distribution curve for z = - 0.47 is 0.1808. That is P(- 0.47 < z < 0) = 0.1808
[Use the standard normal distribution table.]

= 2(0.5000 - 0.1808) = 0.6384
P-value for a two tailed test = 2 × P(z < - 0.47)

The P-value 0.6384 > 0.05, do not reject the null hypothesis.

So, there is not enough evidence to reject the survey report that 38% of Americans go to the supermarket during noon to 5 p.m.

So, the survey report is correct at α = 0.05.

8.
In a random sample of 180 American home owners, it was found that 158 of them personaly attend to their yard or garden duties. Test the hypothesis that more than 80% of the American home owners personally work in their yard or garden at α = 0.01. Use the $p$-value method. What is the test value?
 a. there is enough evidence to support the hypothesis, $z$ = 2.68 b. there is not enough evidence to support the hypothesis, $z$ = 1.34 c. there is enough evidence to support the hypothesis, $z$ = 1.34 d. there is not enough evidence to support the hypothesis, $z$ = 2.68

#### Solution:

H0: p ≤ 0.80 and H1: p > 0.80
[Null and alternate hypotheses.]

Sample proportion, pˆ = 158 / 180 = 0.88

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

z = 0.88-0.80(0.80)(0.20)180 = 2.68
[Substitute the values and simplify.]

Area under the standard normal distribution curve for z = 2.68 is 0.4963. That is P(0 < z < 2.68) = 0.4963
[Use the standard normal distribution table.]

= 0.5000 - 0.4963 = 0.0037
P-value for a right tailed test = P(z > 2.68)

The P-value 0.0037 < 0.01, reject the null hypothesis.

So, there is enough evidence to support the hypothesis that more than 80% of the American home owners personally work in their yard or garden.

9.
A survey reports that 60% of the men around the age of 60 snore while sleeping. To test this report a researcher observed 120 men of this age group and found that 75 men snore while sleeping. Find the $p$-value for the test. Is there enough evidence to reject the survey report at α = 0.01?
 a. 0.5754; yes b. 0.5754; no c. 0.2877; yes d. 0.2877; no

#### Solution:

H0: p = 0.60 and H1: p ≠ 0.60
[Null and Alternate Hypotheses.]

Sample proportion, pˆ = 75 / 120 = 0.625

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

z = 0.625-0.60(0.6)(0.4)120 = 0.56
[Substitute the values and simplify.]

Area under the standard normal distribution curve for z = 0.56 is 0.2123. That is P(0 < z < 0.56) = 0.2123
[Use the standard normal distribution table.]

= 2(0.5000 - 0.2123) = 0.5754
P-value for a two tailed test = 2 × P(z > 0.56)

The P-value 0.5754 > 0.01, do not reject the null hypothesis.

So, there is not enough evidence to reject the survey report that 60% of the men around the age of 60 snore while sleeping.

10.
A manufacturer reported that 95% of electronic components manufactured by them are free from defects. From a random sample of 80 components, 5 had defects. Find the $p$-value for the test. Is there enough evidence to reject the manufacturer's report at α = 0.01?
 a. 0.305; yes b. 0.61; yes c. 0.305; no d. 0.61; no

#### Solution:

H0: p = 0.95 and H1: p ≠ 0.95
[Null and Alternate Hypotheses.]

Sample proportion, pˆ = 75 / 80 = 0.9375

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

z = 0.9375-0.95(0.95)(0.05)80 = - 0.51
[Substitute the values and simplify.]

Area under the standard normal distribution curve for z = - 0.51 is 0.1950. That is P(- 0.51 < z < 0) = 0.1950
[Use the standard normal distribution table.]

= 2(0.5000 - 0.1950) = 0.61
P-value for a two tailed test = 2 × P(z < - 0.51)

The P-value 0.61 > 0.01, do not reject the null hypothesis.

So, there is not enough evidence to reject the manufacturer's report that 95% of electronic components have no defects.