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Intepretation of 2 Proportion Z-Test and P-Value

At a 5% significance level, we fail to reject that the probability of having a a higher stress level is caused by a part time job. Assuming that the null hypothesis is correct, the probability that the results of our experiment will occur is 16.6%, which is about 1 standard deviation below the mean. However, we did not take into account all the other factors that could lead to an increase in stress level among teenagers. Therefore, we cannot conclude that a high level of stress is a direct result of having a part time job.

2 Proportion Z-Test

P1= 3/19
P2= 6/21

H0: P1 = P2
Ha: P1 ≠ P2

p = (p₁ * n₁ + p₂ * n₂) / (n₁ + n₂)
p = (3 + 6) / 40 = 0.225

Test for Independence

• Random Sample
• np=(40)(.225)=9
• nq=(40)(1-.225)=31

Because nq is greater than 10, the date is said to be independent and therefore a 2 proportion z-test can be applied.

SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }
SE = sqrt{0.225(1-0.225) [(1/19)+(1/21)] }
SE = 0.132

z = (p₁ - p₂) / SE
z = [(3/19) - (6/21)] / 0.132
z = -0.968

p value: .1660= 16.60%

Results from the Survey

The results show that there are more students without jobs than with jobs and that the students without jobs are more likely to rate their stress level has high.

Student Survey

Using Google Docs, we made a survey asking students whether or not they had a job and what their stress level is.
Student Survey

Overstressed?

Students say that they are more stressed than ever, but the national percentage of teen employment has decreased exponentially. Is there a relationship between student stress levels and part-time jobs?