Question

At a large midwestern university, a simple random sample of 100 entering freshmen in 1993 found that 20 of the sampled freshmen finished in the bottom third of their high school class. Admission standards at the university were tightened in 1995. In 1997, a simple random sample of 100 entering freshmen found that only 10 finished in the bottom third of their high school class. Let p1 and p2 be the proportions of all entering freshmen in 1993 and 1997, respectively, who graduated in the bottom third of their high school class. Is there evidence that the proportion of freshmen who graduated in the bottom third of their high school classes in 1997 has been reduced, as a result of the tougher admission standards adopted in 1995, compared to the proportion in 1993? To determine this, test the hypotheses H0: p1 = p2 versus Ha: p1 > p2. What is the value of the z statistic for testing these hypotheses?A. Z = 1.20 B. Z = 1.98 C. Z = 1.96 D. Z = 1.92

Answer 1

Answer:

B. Z=1.98

Step-by-step explanation:

1) Data given and notation  

$X_{1}=20$ represent the number of sampled freshmen finished in the bottom third of their high school class in 1993

$X_{2}=10$ represent the number of sampled freshmen finished in the bottom third of their high school class in 1997

$n_{1}=100$ sample 1 selected

$n_{2}=100$ sample 2 selected

$p_{1}=\frac{20}{100}=0.20$ represent the proportion of sampled freshmen finished in the bottom third of their high school class in 1993

$p_{2}=\frac{10}{100}=0.0023$ represent the proportion ofsampled freshmen finished in the bottom third of their high school class in 1997

z would represent the statistic (variable of interest)  

$p_v$ represent the value for the test (variable of interest)

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the proportion graduated in the bottom third of their high school class is higher for 1993 than 1997 , the system of hypothesis would be:  

Null hypothesis:$p_{1} \leq p_{2}$  

Alternative hypothesis:$p_{1} > p_{2}$  

We need to apply a z test to compare proportions, and the statistic is given by:  

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$   (1)

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{20+10}{100+100}=0.15$

3) Calculate the statistic

Replacing in formula (1) the values obtained we got this:  

$z=\frac{0.20-0.10}{\sqrt{0.15(1-0.15)(\frac{1}{100}+\frac{1}{100})}}=1.98$  

4) Statistical decision

For this case we don't have a significance level provided $\alpha$, but we can calculate the p value for this test.  

Since is a one side test the p value would be:  

$p_v =P(Z>1.98)=0.024$  

If we compare the p value and using any significance level for example $\alpha=0.05$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion graduated in the bottom third of their high school classes in 1993 is not significant higher than the proportion in 1997.