Question

The recent default rate on all student loans is 6.2 percent. In a recent random sample of 314 loans at private universities, there were 8 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 = .010?

(a-1)	Choose the appropriate hypothesis.

a.	H0: ππ ≥ .062; H1: ππ < .062. Accept H0 if z < –2.326
b.	H0: ππ ≥ .062; H1: ππ < .062. Reject H0 if z < –2.326

	a b

(a-2)

Calculate the z-score for the sample data using a left-tailed test at α = .01. (A negative value should be indicated by a minus sign. Round your answer to 3 decimal places.)

  

zcalc

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

(b)	Calculate the p-value. (Round your answer to 4 decimal places.)

  

p-value

(c)	The assumption of normality is justified.
	Yes No

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Hint #1

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Answer 1

Solution

Given,

n = 314

x = 8

= 0.010

Let be the sample proportion and  π be the population proportion.

= x/n = 8/314 = 0.0255

a 1) For left tailed test, the critical value is - and the rejection region is < -

Here, = 0.010 , Using z table

   = 2.326

so, - = -2.326

H₀: ππ ≥ .062; H₁: ππ < .062. Reject H₀ if z < –2.326

option b is correct.

a 2) z zcore for sample data (i.e.z test statistic)

zcalc =   

=

= -2.682

a 3)Since , zcalc = -2.682 is less than -2.326 , zcalc lies in the rejection region.

Yes, the null hypothesis is rejected.

b) p value

For left tailed test ,

p value = P(Z < z)

= P(Z < -2.682)  

= 0.0037

c) Here , n * = 314 * 0.0255 = 8

Also, n * (1 - ) = 314 * (1 - 0.0255) = 306

Both are greater than 5

the assumption of normality is justified. YES