Question

The recent default rate on all student loans is 5.2 percent. In a recent random sample...

The recent default rate on all student loans is 5.2 percent. In a recent random sample of 300 loans at private universities, there were 9 defaults.

Does this sample show sufficient evidence that the private university loan default rate is below the rate for all universities, using a left-tailed test at α = .01?

(a-1) Choose the appropriate hypothesis.
a. H0: ππ .052; H1: ππ < .052. Accept H0 if z < –2.326
b. H0: ππ .052; H1: ππ < .052. Reject H0 if z < –2.326
a
b
(a-2)

What is the z-score for the sample data? (A negative value should be indicated by a minus sign. Round your answer to 2 decimal places.)

  

  zcalc   
(a-3) Should the null hypothesis be rejected?
Yes
No
(b)

Calculate the p-value. (Round intermediate calculations to 2 decimal places. Round your answer to 4 decimal places.)

  

  p-value   
(c) The assumption of normality is justified.
Yes
No
0 0
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Answer #1

Given that,  In a recent random sample of 300 loans at private universities, there were 9 defaults.

sample proportion = 9/300 = 0.03

a-1) The null and alternative hypotheses are,

H0: π ≥ 0.052; H1: π < 0.052

critical value at α = .01 is, Z* = -2.326

Decision Rule: Reject H0 if z < –2.326

a-2) Test statistic is,

=> Zcalc = -1.72

a-3) Since, Z = -1.72 > -2.326

we do not reject the null hypothesis.

b) p-value = P(Z < -1.72) = 0.0427

=> p-value = 0.0427

c) np = 300 * 0.052 = 15.6 > 10 and

n(1-p) = 300 * (1 - 0.052) = 284.4 > 10

=> Yes, assumption og normality is justified

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