Example

Two types of medication for hives are being tested to determine if there is a difference in the proportions of adult patient reactions. Twenty out of a random sample of 200 adults given medication A still had hives 30 minutes after taking the medication. Twelve out of another random sample of 200 adults given medication B still had hives 30 minutes after taking the medication. Test at a 1% level of significance.

Solution:

The problem asks for a difference in proportions, making it a test of two proportions.

Let A and B be the subscripts for medication A and medication B, respectively. Then p_A and p_B are the desired population proportions.

Random Variable: P′_A – P′_B = difference in the proportions of adult patients who did not react after 30 minutes to medication A and to medication B.

H₀: p_A = p_B    or  p_A – p_B = 0

H_a: p_A ≠ p_B    or p_A – p_B ≠ 0

The words “is a difference” tell you the test is two-tailed.

Distribution for the test: Since this is a test of two binomial population proportions, the distribution is normal:

[latex]\displaystyle{p_c}=\frac{x_A-x_B}{n_A-n_B}=\frac{20+12}{200+200}=0.08[/latex] 1 – p_c = 0.92

[latex]\displaystyle{P}\prime_{{A}}-{P}\prime_{{B}}~{N}{\Bigg[{0},\sqrt {{{({0.08})}{({0.92})}{\bigg(\frac{{1}}{{200}}+\frac{{1}}{{200}}\bigg)}}}}\Bigg][/latex]

P′_A – P′_B follows an approximate normal distribution.

Calculate the p-value using the normal distribution: p-value = 0.1404.

Estimated proportion for group A: [latex]\displaystyle{p}\prime_{{A}}=\frac{{x}_{{A}}}{{n}_{{A}}}=\frac{{20}}{{200}}={0.1}[/latex]

Estimated proportion for group B: [latex]\displaystyle{p}\prime_{{B}}=\frac{{x}_{{B}}}{{n}_{{B}}}=\frac{{12}}{{200}}={0.06}[/latex]

Graph:

P′_A – P′_B = 0.1 – 0.06 = 0.04.

Half the p-value is below –0.04, and half is above 0.04.

Compare α and the p-value: α = 0.01 and the p-value = 0.1404. α <p-value.

Make a decision: Since α < p-value, do not reject H0.

Conclusion: At a 1% level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the proportions of adult patients who did not react after 30 minutes to medication A and medication B.

Example

A research study was conducted about gender differences in “sexting.” The researcher believed that the proportion of girls involved in “sexting” is less than the proportion of boys involved. The data collected in the spring of 2010 among a random sample of middle and high school students in a large school district in the southern United States is summarized in the table. Is the proportion of girls sending sexts less than the proportion of boys “sexting?” Test at a 1% level of significance.

Solution:

This is a test of two population proportions. Let M and F be the subscripts for males and females. Then p_M and p_F are the desired population proportions.

Random variable: p′_F − p′_M = difference in the proportions of males and females who sent “sexts.”

H₀: p_F = p_M H₀: p_F – p_M = 0

H_a: p_F < p_M H_a: p_F – p_M < 0

The words “less than” tell you the test is left-tailed.

Distribution for the test: Since this is a test of two population proportions, the distribution is normal:

[latex]\displaystyle{p}_c=\frac{x_F+x_M}{n_F+n_M}=\frac{156+183}{2169+2231}=0.077[/latex]

Therefore,

[latex]\displaystyle{P}\prime_{{A}}-{P}\prime_{{B}}~{N}{\Bigg[{0},\sqrt {{{({0.077})}{({0.923})}{\bigg(\frac{{1}}{{2169}}+\frac{{1}}{{2231}}\bigg)}}}}\Bigg][/latex]

p′F – p′M follows an approximate normal distribution.

Calculate the p-value using the normal distribution:

p-value = 0.1045

Estimated proportion for females: 0.0719

Estimated proportion for males: 0.082

Graph:

Decision: Since α < p-value, Do not reject H0

Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that the proportion of girls sending “sexts” is less than the proportion of boys sending “sexts.”

Example

Researchers conducted a study of smartphone use among adults. A cell phone company claimed that iPhone smartphones are more popular with whites (non-Hispanic) than with African Americans. The results of the survey indicate that of the 232 African American cell phone owners randomly sampled, 5% have an iPhone. Of the 1,343 white cell phone owners randomly sampled, 10% own an iPhone. Test at the 5% level of significance. Is the proportion of white iPhone owners greater than the proportion of African American iPhone owners?

Solution:

This is a test of two population proportions. Let W and A be the subscripts for the whites and African Americans. Then p_W and p_Aare the desired population proportions.

Random variable: p′_W – p′_A = difference in the proportions of Android and iPhone users.

H₀: p_W = p_A H₀: p_W – p_A = 0

H_a: p_W > p_A H_a: p_W – p_A > 0

The words “more popular” indicate that the test is right-tailed.

Distribution for the test: The distribution is approximately normal:

[latex]\displaystyle{p}_c=\frac{x_W+x_A}{n_W+n_A}=\frac{134+12}{1343+232}=0.0927[/latex]

Therefore,

[latex]\displaystyle{P}\prime_{{A}}-{P}\prime_{{B}}~{N}{\Bigg[{0},\sqrt {{{({0.0927})}{({0.9073})}{\bigg(\frac{{1}}{{1343}}+\frac{{1}}{{232}}\bigg)}}}}\Bigg][/latex]

follows an approximate normal distribution.

Calculate the p-value using the normal distribution:

p-value = 0.0077

Estimated proportion for group A: 0.10

Estimated proportion for group B: 0.05

Graph:

Decision: Since α > p-value, reject the H0.

Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that a larger proportion of white cell phone owners use iPhones than African Americans.