10.3 Comparing Two Independent Population Proportions

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

p_A – p_B = 0

H_a: p_A ≠ p_B

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:

$p_c=\frac{x_A+x_B}{n_A+n_B}=\frac{20+12}{200+200}=0.081–p_c=0.92$

${P^{\prime }}_A–{P^{\prime }}_B~N\left[0,\sqrt{(0.08)(0.92)(\frac{1}{200}+\frac{1}{200})}\right]$

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: ${p^{\prime }}_A=\frac{x_A}{n_A}=\frac{20}{200}=0.1$

Estimated proportion for group B: ${p^{\prime }}_B=\frac{x_B}{n_B}=\frac{12}{200}=0.06$

Graph:

Figure 10.7

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.

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:

$p_c=\frac{x_F+x_M}{n_F+n_M}=\frac{156+183}{2169+2231}=\text{0}\text{.077}$
$1-p_c=0.923$
Therefore,
${p^{\prime }}_F–{p^{\prime }}_M\sim N\left(0,\sqrt{(0.077)(0.923)\left(\frac{1}{2169}+\frac{1}{2231}\right)}\right)$
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:

Figure 10.8

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.”

This is a test of two population proportions. Let W and A be the subscripts for the White people and African American people. Then p_W and p_A are the desired population proportions.

Random Variable: p′_W – p′_A = difference in the proportions of White and African American cell phone users who own iPhones.

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:

$p_c=\frac{x_W+x_A}{n_W+n_A}=\frac{134+12}{1343+232}=0.0927$

$1-p_c=0.9073$

Therefore,

${p^{\prime }}_W–{p^{\prime }}_A\backsim N\left(0,\sqrt{\left(0.0927\right)\left(0.9073\right)\left(\frac{1}{1343}+\frac{1}{232}\right)}\right)$

${p^{\prime }}_W–{p^{\prime }}_A$ 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:

Figure 10.9

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 compared to African Americans.