d2jsp
Log InRegister
d2jsp Forums > Off-Topic > General Chat > Homework Help > Hate Maths So Much.. > My Last Quantitative Methods Question;)
Add Reply New Topic New Poll
Member
Posts: 2,446
Joined: Jul 16 2008
Gold: 2,615.00
Jun 17 2014 04:49am
You have been asked to investigate what is the mean starting annual salary of MAcc graduates from Flinders University. University course brochures claim it is $54,000 and that it has a standard deviation of σ = $6,000, stable over the long term. The Accountants’ union claims it is much less. They say they have figures showing it varies between $46,000 and $57,000.

(a) Describe the claim you would test and why. Explain how you would go about it. Give all details of your reasoning, without doing any calculations.

(b) You have now been instructed by your supervisor to test whether the mean annual salary of graduates is different from what the Flinders brochures claim. Suppose you have a sample of 45 graduate salaries that indicates the sample mean starting salary is $48,000. What is your answer and why? Show all work and explain your reasoning at every step. What do you think about the Accountants’ union claim?

(c) Would your approach to question (b) have changed if you had managed to obtain a sample of 36 graduates and calculated the sample mean at = $52,161 and the sample standard deviation at s = $6,961? What would you have done differently and why? What would you have concluded then?
Member
Posts: 71
Joined: Jan 9 2013
Gold: 0.01
Jun 17 2014 06:06am
a ) test the null hypothesis mean (use the greek letter myu here) =54,000 against the alternative hypothesis mean =/=54,000. Set an alpha = .05, which results in z = 1.96. Calculate the z score of the problem by doing (mean-54,000)/SE
b ) (48,000-54,000)/[6,000/sqrt(45)]. plug that into a calculator, if the result falls between -1.96 and positive 1.96, not enough evidence to reject the null hypothesis (stated in the previous problem)
c ) (52,161-54,000)/(6,961/sqrt(36), our standard error would be s/sqrt(n), because out s is known. Once you plug that shit in and see how it compares to 1.96

This post was edited by silentio on Jun 17 2014 06:07am
Member
Posts: 2,446
Joined: Jul 16 2008
Gold: 2,615.00
Jun 17 2014 06:41am
Quote (silentio @ Jun 17 2014 10:06pm)
a ) test the null hypothesis mean (use the greek letter myu here) =54,000 against the alternative hypothesis mean =/=54,000. Set an alpha = .05, which results in z = 1.96. Calculate the z score of the problem by doing (mean-54,000)/SE
b ) (48,000-54,000)/[6,000/sqrt(45)]. plug that into a calculator, if the result falls between -1.96 and positive 1.96, not enough evidence to reject the null hypothesis (stated in the previous problem)
c ) (52,161-54,000)/(6,961/sqrt(36), our standard error would be s/sqrt(n), because out s is known. Once you plug that shit in and see how it compares to 1.96


follow you for b

i got a very very low probability of sample mean being 48000 with that population and sd... and hence i would doubt the population mean of 54000

but for c are you sure you use the standard deviation from the sample and not the population?

and would you not use the 46000 and figures 57000 given at all?

Member
Posts: 71
Joined: Jan 9 2013
Gold: 0.01
Jun 17 2014 05:29pm
You would use the population standard deviation if you knew it was right. The brochure claims it is, but the reason for the test is questioning whether the brochure is right or not. Generally, the sample standard deviation is used to estimate the population standard deviation, and although in this case you already have the population standard deviation, it is not necessarily the correct one. The sample one is definitely true for the sample (based off of the problem's wording) and so might be a better estimate for the population. The problem above you weren't given an alternative, so you sorta had to use what you had.

Those figures are just the estimate of the accountants. The last part of b. actually sorta asks about them and how it compares, where you could mention that the number 48000 does fall between their estimate, so they were kind of right, although if we assume their estimate to have a symmetrical distribution, their mean would be closer to 51,500 rather than your sample mean of 48,000. So based on your sample, they seemed to have a good guess. Now, if you wanna determine whether that's within the mean confidence interval, perform the same calculation as before, just replace 54,000 by 51,500. Personally I wouldn't go as far as to do that since the problem seems to be asking more for just your general feel and not a test for this, but if your teacher is an asshole when it comes to grading, it might be worth doing. Then you could state more confidently whether or not their estimate was actually good.
Member
Posts: 2,446
Joined: Jul 16 2008
Gold: 2,615.00
Jun 17 2014 07:26pm
Quote (silentio @ Jun 18 2014 09:29am)
You would use the population standard deviation if you knew it was right. The brochure claims it is, but the reason for the test is questioning whether the brochure is right or not. Generally, the sample standard deviation is used to estimate the population standard deviation, and although in this case you already have the population standard deviation, it is not necessarily the correct one. The sample one is definitely true for the sample (based off of the problem's wording) and so might be a better estimate for the population. The problem above you weren't given an alternative, so you sorta had to use what you had.

Those figures are just the estimate of the accountants. The last part of b. actually sorta asks about them and how it compares, where you could mention that the number 48000 does fall between their estimate, so they were kind of right, although if we assume their estimate to have a symmetrical distribution, their mean would be closer to 51,500 rather than your sample mean of 48,000. So based on your sample, they seemed to have a good guess. Now, if you wanna determine whether that's within the mean confidence interval, perform the same calculation as before, just replace 54,000 by 51,500. Personally I wouldn't go as far as to do that since the problem seems to be asking more for just your general feel and not a test for this, but if your teacher is an asshole when it comes to grading, it might be worth doing. Then you could state more confidently whether or not their estimate was actually good.


thanks heaps homey!

my teacher is a dick haha
he never gives partial marks ... always 0 for full
Member
Posts: 1
Joined: Jun 18 2014
Gold: 0.00
Jun 18 2014 08:28pm
Hi, why would you do a two-tail test? Why not just do H0=54000. H1<54000? Also, what is SE?
Thanks
Member
Posts: 71
Joined: Jan 9 2013
Gold: 0.01
Jun 19 2014 07:39pm
Quote (roy99 @ Jun 18 2014 08:28pm)
Hi, why would you do a two-tail test? Why not just do H0=54000. H1<54000? Also, what is SE?
Thanks


I used the two tailed test because although the question said that accountants claimed it to be less, their range was 46k to 57k, and 57k >54k. So to be safe, I used two-tailed. Additionally b. asks specifically about it being different, not smaller, so that implies two tailed.
SE = standard error.
Go Back To Homework Help Topic List
Add Reply New Topic New Poll