Spring 2013 – MAT 1272 Statistics – Reitz

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  • Announcement Office hours on Thursday 5/9 will take place after class
  • #13351

    Jonas Reitz

    This Thursday, May 9th, office hours will take place after class from 3:45 – 4:45 (rather than the usual time, 11:30am – 12:30pm).

    NOTE: Office hours tomorrow (Tuesday May 7th) will take place at the usual time, 3:45 – 4:45.



    For Assignment12 – Sec 7.2: Problem 2

    I may need help with critical values for the positive and negative z scores. When i trying to find the z scores from what i learned, the webwork does not accept it and i used the z table for the hypthesis question since the “n” is larger than 30.
    Ho u = 31.4

    Ha u ≠ 31.4

    random sample, n= 120 sample mean x= 29.2, standardeviation, s= 3.7 significance, α= .05
    σx = s/√n = 3.7/(√120) = .3378

    It is necessary for an automobile producer to estimate the number of miles per gallon achieved by its cars. Suppose that the sample mean for a random sample of 120 cars is 29.2 miles and assume the standard deviation is 3.7 miles. Now suppose the car producer wants to test the hypothesis that (µx), the mean number of 31.4 miles per gallon, is against the alternative hypothesis that it is not . Conduct a test using α= .05 by giving the following:

    (a) positive critical score 1.645

    (b) negative critical score -1.645

    (c) test statistic -6.51

    The final conclustion is

    A. There is not sufficient evidence to reject the null hypothesis that .

    (this is the answer i chose ) B. We can reject the null hypothesis that and accept that .

    a 1.645 incorrect
    b -1.645 incorrect
    c -6.51 correct
    B correct

    At least one of the answers above is NOT correct



    for α= .05 significance

    i used used the z table and i had got 0.0495 for z at -1.65 and .0505 for z at -1.64

    and added the two z’s then divided by 2 and got -1.645 for the neagtive z score and the positive z score would be 1.645


    Jonas Reitz

    Hi Anil,
    I think I see what happened. Alpha = .05 represents the total area in the rejection region(s) — since this is a two-tailed test, the total alpha is split between the two tails. Because of this, you first must divide alpha by 2 (to find the area of a single tail), and then look that up in the table.
    Let me know if this doesn’t fix it for you,
    Mr. Reitz



    i divided the alpha fir 0.05 by 2 and got .025 which would have a z of -1.96 and then used the positive z value of 1.96 for the other critical value

    and yes it was correct Thank you.

    **i just fixed it the day after though…so i’m not sure if i will still receive a full grade for it…



    this is the result for the problem 2 of Web Work 12…….(see attached file)




    i tried to all the problem for web work 13..only part of problem 1 and 2 seems ok while i was taking too long solve problem 3 and 4 and still it was not accepted…i had then generated a pdf and i saw what i did wrong but i was too late to fix it…

    i have an excel sheet that i used also for part 4 since it had more data than part 3

    and do have to round to 4 decimal places for when i find the SSxx, SSyy SSxy with the part that is (sum of [ xx or yy or xy]/ N)

    and rounded to the nearest three decimal places for the “r” and it still would not be accepted.



    oh the pdf with the correct answers will only show when the webwork is has closed…lolz…so i will try better on the last webwork…. and i will bring the pdf to class on Thursday for webwork 13.



    i had tried solving for the frist question on webwork 14 though it was still a bit weird since i was not able to find an exact answer for the m and b values

    i had set up most of my data in excel and save it as a pdf.

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