“Would you marry a person from a lower social class than your own?” Researchers asked this question of a random sample of $385$black, never married students at two historically black colleges in the South. Of the $149$men in the sample, $91$said “Yes.” Among the $236$women, $117$said “Yes.”$14$Is there reason to think that different proportions of men and women in this student population would be willing to marry beneath their class?

Holly carried out the significance test shown below to answer this question. Unfortunately, she made some mistakes along the way. Identify as many mistakes as you can, and tell how to correct each one.

State: I want to perform a test of

$H_0:p_1=p_2$

$H_a:p_1\ne p_2$

at the $95\%$confidence level.

Plan: If conditions are met, I’ll do a one-sample $z$test for comparing two proportions.

• Random The data came from a random sample of 385 black, never-married students.
• Normal One student’s answer to the question should have no relationship to another student’s answer.
• Independent The counts of successes and failures in the two groups$—91,58,117$, and $119—$are all at least $10$

Do: From the data, ${\hat{p}}_1=\frac{91}{149}=0.61$and ${\hat{p}}_2=\frac{117}{236}=0.46$.

Test statistic

$z=\frac{(0.61-0.46)-0}{\sqrt{\frac{0.61(0.39)}{149}+\frac{0.46(0.54)}{236}}}=2.91$

$p$=value From Table A, role="math" localid="1650292307192" $P(z\ge 2.91)\equiv 1-0.3982\equiv 0.0018$.

Conclude: The $p$-value, $0.0018$, is less than $0.05$, so I’ll reject the null hypothesis. This proves that a higher proportion of men than women are willing to marry someone from a social class lower than their own.

Step by step solution

01

Given Information

Given

$x_1=91$

$n_1=149$

$x_2=117$

role="math" $n_2=236$

Determine the hypothesis

$H_0:p_1-p_2=0$

$H_a:p_1-p_2\ne 0$

02

Explanation

The sample proportion is the number of successes divided by the sample size:

${\hat{p}}_1=\frac{x_1}{n_1}=\frac{91}{149}\approx 0.611$

${\hat{p}}_2=\frac{x_2}{n_2}=\frac{117}{236}\approx 0.496$

${\hat{p}}_p=\frac{x_1+x_2}{n_1+n_2}=\frac{91+117}{385}=\frac{208}{385}\approx 0.540$

Determine the value of the test statistic:

localid="1650450515979" $$\begin{gathered} z=\frac{{\hat{p}}_1-{\hat{p}}_2}{\sqrt{{\hat{p}}_p\left(1-{\hat{p}}_p\right)}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \\ =\frac{0.611-0.496}{\sqrt{0.540(1-0.540)}\sqrt{\frac{1}{149}+\frac{1}{236}}} \\ \approx 2.21 \end{gathered}$$

The $p$-value is the probability of obtaining the value of the test statistic, or a value more extreme. Determine the $p$-value using table A:

localid="1650450537687" $$\begin{gathered} P=P(Z<-2.21\text{or}Z>2.21) \\ =2\times P(Z<-2.21) \\ =2\times 0.0136 \\ =0.0272 \end{gathered}$$

If the $p$-value is smaller than the significance level, reject the null hypothesis:

$P<0.05\Rightarrow \text{Reject}{H}_0$

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