anonymous
  • anonymous
A random sample of 10 one-bedroom apartments from your local newspaper has these monthly rents (dollars): 500 650 600 505 450 550 515 495 650 395 Do these data give good reason to believe that the mean rent of all advertised apartments is greater than $500 per month? Carry out a complete significance test.
Statistics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
can we assume a normal distribution? if so, then what criteria should we be looking for?
anonymous
  • anonymous
yes we can assume a normal distribution @amistre64
anonymous
  • anonymous
@Nnesha

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
@Zarkon
anonymous
  • anonymous
@dan915
anonymous
  • anonymous
okay start with the mean, what is the mean of this data
anonymous
  • anonymous
The mean is 531
anonymous
  • anonymous
now find the sample standard deviation
anonymous
  • anonymous
82.79157
anonymous
  • anonymous
thats quite a bit of deviation right
anonymous
  • anonymous
yes haha
anonymous
  • anonymous
|dw:1440116014027:dw|
anonymous
  • anonymous
right I got that too :)
anonymous
  • anonymous
|dw:1440116108715:dw|
anonymous
  • anonymous
you can find the exact probability
anonymous
  • anonymous
but i dont think we can use this one yet actually
anonymous
  • anonymous
okay so what do we do
anonymous
  • anonymous
I know the hypotheses
anonymous
  • anonymous
you have to use a correction factor we cannot assume this sample standard deviation is the same as the real standard deviation of the population
anonymous
  • anonymous
oh what have u studied so far actually depends on that
anonymous
  • anonymous
Have you done null hypothesis test or T-test,
anonymous
  • anonymous
well the lesson I am on is t distributions with the calculator
anonymous
  • anonymous
im sorry but you are better off watching one of these videos https://www.youtube.com/watch?v=Uv6nGIgZMVw
anonymous
  • anonymous
id have to review the rules for t test too, to help you
anonymous
  • anonymous
can you help me find the t value? this is what I did but it is not right. (531 - 500) / the standard deviation (82.79157) = 0.374
anonymous
  • anonymous
What you calculated there is the Z score, t values are different
anonymous
  • anonymous
shoot, you are right
anonymous
  • anonymous
did u watch the video, he is very clear
anonymous
  • anonymous
|dw:1440117355331:dw|
anonymous
  • anonymous
|dw:1440117452584:dw|
anonymous
  • anonymous
find your t and degrees of freedom and look at t chart

Looking for something else?

Not the answer you are looking for? Search for more explanations.