Salaries of newly hired workers at a major company are normally distributed with a mean of $59,000 and a standard deviation of $7,000. However, a recent survey from a random sample of 40 newly hired workers shows a mean salary of $57,500 during the past recession. You want to determine if salary levels have decreased during the past recession. The p-value is:

A) 0.0869
B) 0.001
C) 99%
D) 0.173
E) None of the above
None of the above.

Question

Salaries of newly hired workers at a major company are normally distributed with a mean of $59,000 and a standard deviation of $7,000. However, a recent survey from a random sample of 40 newly hired workers shows a mean salary of $57,500 during the past recession. You want to determine if salary levels have decreased during the past recession. The p-value is:

A)   0.0869 
B)   0.001 
C)   99% 
D)  0.173 
E) None of the above
  None of the above.

anonymous · Answer

I ran this through a Z-Test on the Ti-84 and got a P value of 0.0876671314
Am I getting something wrong here?

irishboy123 · Answer

https://i.gyazo.com/42de97e48e2cdb7a33ae1c2581596a9f.png http://www.socscistatistics.com/tests/ztest_sample_mean/Default2.aspx

anonymous · Answer

So the hypothesis test required is u < 59000

Using technology...
stderr=7000/Sqrt(40)
p = normdist(57500,59000,stderr,true) = 0.08766707189

But using the charts
Zdata = (57500-59000)/(7000/Sqrt(40)) = -1.355261854
then if we use the Standard Normal Distribution Charts.. 
and lookup a left tailed test for z=-1.36 
We get under row -1.3 ... column 0.06 = 0.0869

So the answer is A and the moral is if in doubt don't trust your calculators and spreadsheets, and if you have the time take a quick look at the charts.

irishboy123 · Answer

holy cow.

you've pasted that from excel?!

anonymous · Answer

nah I just wrote it out.. but yeah excel commands.. I suppose I could expand the normdist function, but thats' the gist of it.