Question 986138: The U.S. Department of Education reports that 46% of full-time college students are employed while attending college. A recent survey of 60 full-time students at Miami University found that 29 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 Miami University is different from the national norm of 0.46.
b) Assume that the study found that 36 of the 60 full-time students were employed and repeat (a). Are the conclusions the same?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Ho:p=0.46
Ha:p not = 0.46
alpha=0.05
test statistic is a 1 sample proportion.
Critical value is |z|>1.96
z=(phat-p)/sqrt { p*(1-p)/n}; = 0.02833/0.0643
=0.36
Fail to reject the null hypothesis, so insufficient evidence to say that there is a difference from that national norm.
===
With 60, and p hat =0.6, z=2.18, and we would reject the null hypothesis and say there is a difference at the 3% level of significance (p=0.0295). The conclusions are not the same.
|
|
|
