Answer to Question #200213 in Statistics and Probability for John

Question

USE ONE SAMPLE Z TEST 

A researcher reports that the average salary of College Deans is more than P63,000. A sample , test the claim that the College Deans earn more than P63,000 a month. The standard deviation of the population is P5,250.

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:

$\mu=P63,000\\n=35\\\bar x=P65,700\\\sigma=P5,250$

STEP 1: State the hypothesis and identify them

$H_0:\mu \leq P63,000\\H_1:\mu>P63,000,$ claim(One-tailed test)

STEP2: The level of significance is $\alpha=0.05$

STEP3: Determine the critical value using the table

$z_{critical}=+1.645$

STEP4: Compute the one sample z test value using the formula $z=\dfrac{\bar x -\mu}{\sigma/\sqrt n}$

$z_{computed}=\dfrac{\bar x -\mu}{\sigma/\sqrt n}=\dfrac{65,700-63,000}{5250/\sqrt {35}}=(2,700)[\dfrac{\sqrt{35}}{5250}]\\\ \\z_{computed}=3.043$

STEP5: Decision rule. Compare the computed and critical value of z

$\because |3.043|>|1.645|\implies z_{computed}>z_{critical}$

So, we $reject\ H_0\ and \ accept\ H_1$

STEP6: Conclusion:

There is evidence to support the claim that the monthly salary of the college dean is more than P63,000