In: Statistics and Probability
9.55 The U.S. Department of Education reports that 40% of full-time college students are employed while attending college. (Data extracted from The Condition of Education 2012,ncesed.gov/pubs2012/2012045.pdf.) A recent survey of 60 full-time students at a university found that 25 were employed.
A) Use the five-step p-value approach to hypothesis testing and a 0.05 level of significance to determine whether the proportion of full-time students at the university is different from the national norm of 0.4
B) Assume that the study found that 32 of the 60 full-time students were employed and repeat (a). Are the conclusions the same?
Here we have to test that
Here
and
Conditions are satisfied.
A) Here x = number of full time students who are employed = 25
n = Total number of full time students at a university = 60
(Round to 4 decimal)
Test statistic:
z = 0.26 (Round to 2 decimal)
Test statistic = 0.26
P value for two tailed test is
p value = 2 * P(z > 0.26)
= 2 * (1 - P(z < 0.26))
= 2 * (1 - 0.6026)
= 2 * 0.3974
= 0.7948
P value = 0.7948
Here p value = 0.7948 > alpha = 0.05
So we do not reject null hypothesis H0.
Conclusion: There is insufficient evidence to conclude that the proportion of full time students at the university is different from the national norm of 0.4.
B) Here x = number of full time students who are employed = 32
n = Total number of full time students at a university = 60
Test statistic:
z = 2.06 (Round to 2 decimal)
P value for two tailed test is
p value = 2 * P(z > 2.06)
= 2 * (1 - P(z < 2.06))
= 2 * (1 - 0.9803)
= 2 * 0.0197
= 0.0394
P value = 0.0394
Here p value = 0.0394 < alpha = 0.05
So we reject null hypothesis H0.
Conclusion: There is sufficient evidence to conclude that the proportion of full time students at the university is different from the national norm of 0.4.
That means conclusions are not same.