Question

In a city known for many tech start-ups, 311 of 800 randomly selected college graduates with outstanding student loans currently owe more than $50,000. In another city known for biotech firms, 334 of 800 randomly selected college graduates with outstanding student loans currently owe more than $50,000. Perform a two-proportion hypothesis test to determine whether there is a difference in the proportions of college graduates with outstanding student loans who currently owe more than $50,000 in these two cities. Use α=0.05. Assume that the samples are random and independent. Let the first city correspond to sample 1 and the second city correspond to sample 2. For this test: H0:p1=p2; Ha:p1≠p2, which is a two-tailed test. The test results are: z≈−1.17 , p-value is approximately 0.242

Answer 1

Answer:

Null hypothesis:$p_{1} - p_{2}=0$  

Alternative hypothesis:$p_{1} - p_{2} \neq 0$  

$z=\frac{0.389-0.418}{\sqrt{0.403(1-0.403)(\frac{1}{800}+\frac{1}{800})}}=-1.182$    

$p_v =2*P(Z<-1.182)=0.2372$  

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say that we don't have significant difference between the two proportions.  

Step-by-step explanation:

1) Data given and notation  

$X_{1}=311$ represent the number college graduates with outstanding student loans currently owe more than $50,000 (tech start-ups)

$X_{2}=334$ represent the number college graduates with outstanding student loans currently owe more than $50,000 ( biotech firms)

$n_{1}=800$ sample 1

$n_{2}=800$ sample 2

$p_{1}=\frac{311}{800}=0.389$ represent the proportion of college graduates with outstanding student loans currently owe more than $50,000 (tech start-ups)

$p_{2}=\frac{334}{800}=0.418$ represent the proportion of college graduates with outstanding student loans currently owe more than $50,000 ( biotech firms)

z would represent the statistic (variable of interest)  

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

$\alpha=0.05$ significance level given

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to check if is there is a difference in the two proportions, the system of hypothesis would be:  

Null hypothesis:$p_{1} - p_{2}=0$  

Alternative hypothesis:$p_{1} - p_{2} \neq 0$  

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 $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{311+334}{800+800}=0.403$  

3) Calculate the statistic  

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

$z=\frac{0.389-0.418}{\sqrt{0.403(1-0.403)(\frac{1}{800}+\frac{1}{800})}}=-1.182$    

4) Statistical decision

Since is a two sided test the p value would be:  

$p_v =2*P(Z<-1.182)=0.2372$  

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say that we don't have significant difference between the two proportions.