The mean annual tuition and fees for a sample of 9 private colleges was with a s

Question

The mean annual tuition and fees for a sample of 9 private colleges was with a standard deviation of A dotplot shows that it is reasonable to assume that the 1- population is approximately normal. You wish to test whether the mean tuition and fees for private colleges is different from .

State a conclusion regarding H0. Use the level of significance.

Question options:

A)

B)

There is not enough information to draw a conclusion.

c) Reject H0.
The mean annual tuition and fees appears to be different from $34,300.

2- About 29% of all burglaries are through an open or unlocked door or window. A sample of 130 burglaries indicated that 87 were not via an open or unlocked door or window. At the 0.05 level of significance, can it be concluded that this differs from the stated proportion?

Question options:

A)

B)

C)

A)

B)

There is not enough information to draw a conclusion.

c) Reject H0.
The mean annual tuition and fees appears to be different from $34,300.

Explanation / Answer

As per company policies, I am answering question one ONLY.

Ho: p =.29 v/s h1: p=/=.29

X= 87
n= 130
p-hat = X/n= 0.669
po= 29%

ALPHA= 5.00%
z(a/2)
z(0.05/2)
1.960

Z = (phat-p)/sqrt(p*(1-p)/n)
=(0.669230769230769-0.29)/SQRT(0.29*(1-0.29)/130)
9.52899

P-value
2*(1-P(z<|z|)
2*(1-P(z<abs(9.528994))
normsdist(abs(9.528994))
0.0000

With (z=9.52, p<5%),the null hypothesis is rejected at 5% level of significance and I conclude that p=/=.29.

In other words, I can say that there is sufficient evidence to support the claim that the proportion of open Windows or doors Bulgaria is different from 29%.

a. None of the above.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.