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The public relations officer for a particular city claims the average monthly cost for childcare outside the home for a single child is $600. A potential resident is interested in whether the claim is correct. She obtains a random sample of 14 records and computes the average monthly cost of this type of childcare to be $589 with a standard deviation of $40. Perform the appropriate test of hypothesis for the potential resident using . Approximate the p-value for the test in (a).

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Answer:

Since the calculated value of z= -1.0198 does not lie in the critical region  z = ± 1.96 the null hypothesis is accepted and it is concluded that the  average monthly cost for childcare outside the home for a single child is $600.

P- value  ≈ 0.307823 which supports the null hypothesis .

Step-by-step explanation:

The null and alternate hypotheses are

H0: u = $600   the average monthly cost for childcare outside the home for a single child is $600.

against the claim

Ha:  u ≠ $600 the average monthly cost for childcare outside the home for a single child is  not equal to $600.

The significance level is set at 0.05

3) The critical region is  z =  ±1.96

4) The test statistic

Z=x- u /s/√n

z= 589-600/40/√14

z= -11/10.7681

z= -1.0198

6) Conclusion

Since the calculated value of z= -1.0198 does not lie in the critical region  z = ± 1.96 the null hypothesis is accepted and it is concluded that the  average monthly cost for childcare outside the home for a single child is $600.

P- value  ≈ 0.307823 which supports the null hypothesis .

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