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anonymous

  • one year ago

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.

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  1. amistre64
    • one year ago
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    can we assume a normal distribution? if so, then what criteria should we be looking for?

  2. anonymous
    • one year ago
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    yes we can assume a normal distribution @amistre64

  3. anonymous
    • one year ago
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    @Nnesha

  4. anonymous
    • one year ago
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    @Zarkon

  5. anonymous
    • one year ago
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    @dan915

  6. anonymous
    • one year ago
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    okay start with the mean, what is the mean of this data

  7. anonymous
    • one year ago
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    The mean is 531

  8. anonymous
    • one year ago
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    now find the sample standard deviation

  9. anonymous
    • one year ago
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    82.79157

  10. anonymous
    • one year ago
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    thats quite a bit of deviation right

  11. anonymous
    • one year ago
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    yes haha

  12. anonymous
    • one year ago
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    |dw:1440116014027:dw|

  13. anonymous
    • one year ago
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    right I got that too :)

  14. anonymous
    • one year ago
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    |dw:1440116108715:dw|

  15. anonymous
    • one year ago
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    you can find the exact probability

  16. anonymous
    • one year ago
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    but i dont think we can use this one yet actually

  17. anonymous
    • one year ago
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    okay so what do we do

  18. anonymous
    • one year ago
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    I know the hypotheses

  19. anonymous
    • one year ago
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    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

  20. anonymous
    • one year ago
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    oh what have u studied so far actually depends on that

  21. anonymous
    • one year ago
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    Have you done null hypothesis test or T-test,

  22. anonymous
    • one year ago
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    well the lesson I am on is t distributions with the calculator

  23. anonymous
    • one year ago
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    im sorry but you are better off watching one of these videos https://www.youtube.com/watch?v=Uv6nGIgZMVw

  24. anonymous
    • one year ago
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    id have to review the rules for t test too, to help you

  25. anonymous
    • one year ago
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    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

  26. anonymous
    • one year ago
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    What you calculated there is the Z score, t values are different

  27. anonymous
    • one year ago
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    shoot, you are right

  28. anonymous
    • one year ago
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    did u watch the video, he is very clear

  29. anonymous
    • one year ago
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    |dw:1440117355331:dw|

  30. anonymous
    • one year ago
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    |dw:1440117452584:dw|

  31. anonymous
    • one year ago
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    find your t and degrees of freedom and look at t chart

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