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

Null hypothesis:
`p_(1) - p_(2)=0`

Alternative hypothesis:
`p_(1) - p_(2) \\eq 0`

`z=\frac{0.389-0.418}{\sqrt{0.403(1-0.403)((1)/(800)+(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.

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)=(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)=(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) \\eq 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)((1)/(n_(1))+(1)/(n_(2)))}}` (1)

Where
`\hat p=(X_(1)+X_(2))/(n_(1)+n_(2))=(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)((1)/(800)+(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.