Question #318792

One of the undersecretary of the Department of Labor and

Employment (DOLE) claims that the average salary of civil engineer

is Php18,000. A sample of 19 civil engineer’s salary has a mean of

Php17,350 and a standard deviation of Php1,230. Is there enough

evidence to reject the undersecretary’s claim at α = 0.01?


1
Expert's answer
2022-03-29T12:51:38-0400

Hypothesized Population Mean μ=18000\mu=18000

Sample Standard Deviation s=1230

Sample Size n=19

Sample Mean xˉ=17350\bar{x}=17350

Significance Level α=0.01\alpha=0.01

Null and Alternative Hypotheses


The following null and alternative hypotheses need to be tested:

H0:μ=18000H _ 0 ​ :μ=18000

H1:μ18000H _ 1 ​ :μ \not= 18000

This corresponds to two-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.


Based on the information provided, the significance level is α=0.01\alpha=0.01

df=n1=18df=n−1=18

degrees of fredom, and the critical value for two-tailed test is

tc=2.87844t_c=2.87844

The rejection region for this left-tailed test is R={t:t>2.87844}R=\{t:|t|>2.87844\}

The tt - statistic is computed as follows: t=xμs/n=17350180001230/19=2.30348t=\frac{x-\mu}{s/\sqrt{n}}=\frac{17350-18000}{1230/\sqrt{19}}=-2.30348

Since it is observed that

t=2.30348<2.87844=tc|t|=2.30348<2.87844=t_c

it is then concluded that the null hypothesis is not rejected.


Therefore, there is not enough evidence to claim that the population mean μ\mu is different than 18000, at the α=0.01\alpha=0.01

significance level.




Using the P-value approach: The p-value for two-tailed, the significance level

α=0.01,df=18,t=2.30348\alpha=0.01, df=18, t=-2.30348

is p=0.033425 and since

p=0.033425>0.01=αp=0.033425>0.01=\alpha


it is concluded that the null hypothesis is not rejected.


Therefore, there is not enough evidence to claim that the population mean μ\mu is different than 18000, at the α=0.01\alpha=0.01

significance level.


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