Question #202108

A researcher reports that the average salary of College Deans is more than P63,000. A sample of 35 College Deans has a mean salary of P65, 700. At , test the claim that the College Deans earn more than P63,000 a month. The standard deviation of the population is P5,250.


1
Expert's answer
2021-06-02T18:05:09-0400

actually exact question ;

A researcher reports that the average salary of College Deans is more than P63,000. A sample of 35 College Deans has a mean salary of P65, 700. At α=0.05\alpha =0.05 , test the claim that the College Deans earn more than P63,000 a month. The standard deviation of the population is P5,250.




solution:

given:

μ=P63000, n=35, xˉ=P65700, σ=P5250\mu =P63000, \space n=35,\space \bar x =P65700 ,\space \sigma =P5250\\

since we are given population standard deviation, we use z-test


step1:

state the hypothesis and identify the claim

H0:μP63000H1:μ>P63000,claim(onetailed right)H_0 : \mu \leq P63000\\ H_1: \mu > P63000, claim(one-tailed \space right)

stpe 2:

the level of significance is α=0.05\alpha =0.05

step 3:

determine the critical value using the table

zcritical=+1.645z_{critical}=+1.645

step 4:

compute the one sample z test value using the formula

zcomputed=xˉμσnzcomputed=6570063000525035zcomputed=2700525035zcomputed=2700525035zcomputed=3.043z_{computed}=\frac{\bar x- \mu}{\frac{\sigma }{\sqrt{n}}}\\ z_{computed}=\frac{65700- 63000}{\frac{5250}{\sqrt{35}}}\\ z_{computed}=\frac{2700}{\frac{5250}{\sqrt{35}}}\\ z_{computed}=\frac{2700}{5250} {\sqrt{35}}\\ z_{computed}=3.043

step 5:

decision rule ,compare the computed and critical value of z

3.043>1.645Reject H0 and accept H13.043>1.645 \\ Reject\space H_0 \space and \space accept \space H_1

step 6;

There is enough evidence to support the claim that the monthly salary of the college deans is more than P63000


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