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 |
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(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? | ||||
|
(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. | ||||
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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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