Answer:
The difference in the sample proportions is not statistically significant at 0.05 significance level.
Step-by-step explanation:
Significance level is missing, it is α=0.05
Let p(public) be the proportion of alumni of the public university who attended at least one class reunion
p(private) be the proportion of alumni of the private university who attended at least one class reunion
Hypotheses are:
: p(public) = p(private)
: p(public) ≠ p(private)
The formula for the test statistic is given as:
z=
where
- p1 is the sample proportion of public university students who attended at least one class reunion (
)
- p2 is the sample proportion of private university students who attended at least one class reunion (
)
- p is the pool proportion of p1 and p2 (
)
- n1 is the sample size of the alumni from public university (1311)
- n2 is the sample size of the students from private university (1038)
Then z=
=-0.207
Since p-value of the test statistic is 0.836>0.05 we fail to reject the null hypothesis.
A librarian has 150 books to be placed on shelves with an equal number of books on each shelf. If k is the number of shelves to be used, then 150/k is the number of books on each shelf.
Example : lets say there are 10 shelves.....so k = 10
150/10 = 15 books on each shelf.
Could you past the graphs. I would like to help:)
Im Guessing this is a true and fasle
So the answer would be False