SOLUTION: 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 f

Question 1188467: 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!
29/60=0.48333
Ho: p=0.46
Ha: p NE 0.46
alpha=0.05 p{reject Ho| Ho true}
test is a z
critical value is |z|> 1.96
z=(0.48333-0.46)/sqrt (0.46*0.54/60)
=0.36
failure to reject the null hypothesis, insufficient evidence to conclude there is a difference
p-value=0.72
-
the second would have a numerator of 0.6-0.46 or 0.14
a z-value would be 2.18, and we would reject Ho, a different conclusion.
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