Question

A large university claims that the average cost of housing within 5 miles of the campus is $8900 per scholl year. A high shcool student is preparing her budget for her freshman year at the university. She is concerned that the university's esimate is too low. She obtained a random sample of 81 records and computes the average cost is $9050. Based on earlier data, the population stand deviation is $760. Use a=0.01 level of significance.

Step 1. State the null and alternative hypothese.

Step 2. Write doown the appropriate test statistic (formula) and the rejection region of your test (repot z critical(s)

Step 3. Compute the value of the test statistic (z observed)

Step 4. State your conclusin (in one sentence, state whether of not the test reject the null hypothesis and in another sentence apply the result to the problem).

Step 5. Compute the p-value for this test. Is this evidence strong or weak in spporting the alternative hypothesis.

Answer 1

1) H₀: = 8900

    H₁: 8900

2) The tst statistic z = ()/()

At = 0.01, the critical values are z_0.005 = +/- 2.58

Reject H₀, if z < -2.58 and z > 2.58

3) z = ()/()

        = (9050 - 8900)/(760/)

       = 1.78

4) As the test statistic value is not greater than the upper critical value (1.78 < 2.58), we should not reject the null hypothesis.

So there is sufficient evidence to support the University's claim that the average cost of of housing within 5 miles of the campus is $8900 per scholl year.

5) P-value = 2 * P(Z > 1.78)

                  = 2 * (1 - P(Z < 1.78))

                  = 2 * (1 - 0.9625)

                  = 0.075

The evidence is weak in supporting the alternative hypothesis.