Answer to Question #200214 in Statistics and Probability for James

Question

USE ONE SAMPLE Z TEST

The average baptismal cost includes 50 guest. A random sample of 32 baptismal during the past year in the National Capital Region had a mean of 53 guest and a standard deviation of 10. Is there sufficient evidence at the 0.05 level of significance that the average number of guest differs from the national average?

Given:

STEP 1: STATE THE HYPOTHESIS AND IDENTIFY THE CLAIM.

STEP 2: THE LEVEL OF SIGNIFICANCE 

STEP 3: THE Z CRITICAL VALUE

STEP 4: COMPUTE THE ONE SAMPLE Z TEST VALUE

 

STEP 5: DECISION RULE

STEP 6: CONCLUSION

Answer 1

Given:

$\bar{X}=53 \\ \sigma=10\\ n=32$

STEP 1: STATE THE HYPOTHESIS AND IDENTIFY THE CLAIM.

$H_0: \mu=50 \\ H_1: \mu ≠50$

STEP 2: THE LEVEL OF SIGNIFICANCE

α=0.05

STEP 3: THE Z CRITICAL VALUE

$Z=\frac{\bar{X}- \mu}{\sigma / \sqrt{n}}$

STEP 4: COMPUTE THE ONE SAMPLE Z TEST VALUE

$Z = \frac{53-50}{10/ \sqrt{32}}=\frac{3}{1.767}=1.697$

STEP 5: DECISION RULE

Two-tailed test.

Reject H₀ if z≤-1.95 or z≥1.95.

STEP 6: CONCLUSION

Since Z=1.697 < 1.95

Thus, fail to reject the H₀.

There is no sufficient evidence at the 0.05 level of significance that the average number of guest differs from the national average.