91Ó°ÊÓ

A driving school wants to find out which of its two instructors is more effective at preparing students to pass the state’s driver’s license exam. An incoming class of $100$students is randomly assigned to two groups, each of size $50$. One group is taught by Instructor A; the other is taught by Instructor B. At the end of the course, $30$of Instructor A’s students and $22$of Instructor B’s students pass the state exam. Do these results give convincing evidence that Instructor A is more effective?

Min Jae carried out the significance test shown below to answer this question. Unfortunately, he 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

$H_0:p_1-p_2=0$

$H_a:p_1-p_2>0$

where $p_1=$the proportion of Instructor A's students that passed the state exam and $p_2=$the proportion of Instructor B's students that passed the state exam. Since no significance level was stated, I'll use $\mathrm{σ}=0.05$

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

Random The data came from two random samples of $50$students.

- Normal The counts of successes and failures in the two groups $-30,20,22$, and $28-$are all at least $10$.

- Independent There are at least 1000 students who take this driving school's class.

Do: From the data, ${\hat{p}}_1=\frac{20}{50}=0.40$and ${\hat{p}}_2=\frac{30}{50}=0.60$. So the pooled proportion of successes is

${\hat{p}}_{\mathrm{C}}=\frac{22+30}{50+50}=0.52$

- Test statistic

localid="1650450621864" $$\begin{gathered} z=\frac{(0.40-0.60)-0}{\sqrt{\frac{0.52(0.48)}{100}+\frac{0.52(0.48)}{100}}} \\ =-2.83 \end{gathered}$$

- $p$-value From Table A, localid="1650450641188" $$\begin{gathered} P(zâ‰\yen -2.83)=1-0.0023 \\ =0.9977 \end{gathered}$$.

Conclude: The $p$-value, $0.9977$, is greater than $\mathrm{Î\pm }=0.05$, so we fail to reject the null hypothesis. There is no convincing evidence that Instructor A's pass rate is higher than Instructor B's.

Step by step solution

01

Given Information

Given

$x_1=30$

$n_1=50$

$x_2=22$

$n_2=50$

Determine the hypothesis

$H_0:p_1-p_2=0$

$H_a:p_1-p_2>0$

02

Explanation

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

${\hat{p}}_1=\frac{x_1}{n_1}=\frac{30}{50}=0.6$

${\hat{p}}_2=\frac{x_2}{n_2}=\frac{22}{50}=0.44$

${\hat{p}}_p=\frac{x_1+x_2}{n_1+n_2}=\frac{30+22}{100}=\frac{52}{100}=0.52$

Determine the value of the test statistic:

localid="1650450696625" $$\begin{gathered} z=\frac{{\hat{p}}_1-{\hat{p}}_2}{\sqrt{{\hat{p}}_p\left(1-{\hat{p}}_p\right)}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \\ =\frac{0.6-0.44}{\sqrt{0.52(1-0.52)}\sqrt{\frac{1}{50}+\frac{1}{50}}} \\ ≈1.60 \end{gathered}$$

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="1650450713795" $$\begin{gathered} P=P(Z>1.60) \\ =P(Z<-1.60) \\ =0.0548 \end{gathered}$$

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

$P>0.05⇒\text{Fail to reject}{H}_0$

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