Sampling distro: Sample mean X(bar)=19,1, sample SD S=6, sample N= 344; Pop. n= 5 800

[Aminalek]

Hi Guys,


I hope you can help me out to figure what I am doing wrong... I have the answer to the exercise and I don't get how I find something completely different.


The informations I have :


Sample mean X(bar)=19,1 / Sample SD S=6 / Sample Size N= 344
Population n= 5 800


Question: If population mean is 18,5 what is P(X(bar)>19,1)


If population mean is 18,5 than sample mean is the same. (Property)


Thus: P(X(bar)>19,1)=P(Z>(19,1-18,5)/6)=P(Z>0,1)
P(Z>0,1)=1-P(Z<0,1)
In my table Z=0,1 is P=0,54


Thus P(Z>0,1)= 0,46


But the answer I have is 0.0322...


Can you help me ? What is wrong in my calculations ?


Thanks a lot !!

 

[Aminalek]

I found the issue : I have to divide not by 6 but with 6/sqrt(344) yay

 

[ksdhart2]

I found the issue : I have to divide not by 6 but with 6/sqrt(344) yay

Yes, that's correct, but there's also a sign error. In this problem, it won't manifest itself, but in the future it could mess you up big time. The general formula for converting from a given score to a z-score is:

\(\displaystyle z = \dfrac{x - \mu}{\sigma}\)

Where x is the given score (the population mean in this case), \(\displaystyle \mu\) is the sample mean, and \(\displaystyle \sigma\) is the standard error. The way you've done it is backwards:

\(\displaystyle z = \dfrac{\mu - x}{\sigma}\)

This will result not in z, but in -z. For this specific problem, due to what's being asked, you get the same answer whether the z-score is positive or negative, but in other problems it will matter a lot.

Intuitively, it also makes sense that your formula is producing the wrong answer. Your way suggests that the z-score for x = 18.5 is ~1.85. This means that 18.5 is about 1.85 standard errors greater than 19.1. Oops! That absolutely cannot be right, because 18.5 is not greater than 19.1.