kewlbeer’d

Case
studies: Election times and Speedyx

David
Vaughn

Qnt/561

11/26/2019

Dr.
Heidi Carty

Case
studies: Election times and Speedyx

            Introduction into the two sampled testing began this week
and using the formulas learned from the text book (black). I will use the information to
complete a one sample test that will determine if the exiting times will change
by one minute for the competing networks. This week’s assignment will also ask
me to conduct the sample means formula and test the hypotheses of the new bill
payment times given to me. With the excel data and the information I will test
the hypotheses formed to interpret the answer.

Conduct a one sample hypothesis test
for the election times.

  • Step1.establish
    hypothesis:
    is it necessary to change the time of the
    winning ballots? To set up the problem it will take the use of the t test with
    a unknown standard deviation to find rather it is necessary to do so. Beginning
    with this; Using the alpha as a .10 margin of error I will now test the
    hypothesis. Using formula;

             df = n – 1, df stands for degrees of freedom
which is a difference between the z score and the t test. That is how the
formula will be used to test the time difference and the probable change. Now
to plug in the appropriate data to construct the intervals. I also must note
that with a single mean rather it be time 2401 hours or anything else
the sample can be any size to represent the . I am choosing time or
(2401 hrs.), that is also

  • Step
    2. Determine appropriate statistical test.
    Now
  • Step
    3. Specify error rate,
    Alpha .10.  is the error rate

  • Step
    4 establish the decision rule.
    Since I am using the
    time as the sample or  =  the degree of freedom will be 1 – 2401 = 2400
  • Step
    5.
    Gather sample
    data:
    the information given says that I am to use the sample
    of 765 voters. This is split between the votes for presidential candidate Al
    Gore 358 and eventual winner former president George W. Bush with 407 votes.
    The exiting ballots are supposed at 8pm which should end the voting.

  • Step
    6.
    ) Analyze the
    data:
    Using the formula  df = n – 1 and with the information in place I
    used it in this form  The t test table on appendix A.6 says that t =
    6.314 based upon the alpha .10/2 subtract .05 from both ends to account for the
    two tailed test. 6.314 – .05 = 6.264. if I were to change and
    insert former president Bush’s total the answer would equal 405.5, both total
    being similar to the original figure. Taking the amount of time and subtracting
    one minute from 2400, and then using the total of  or 48.97. With that in place of the first
    total square rooted from 2400. I can say  = 355.50.

  • Step
    7.) Reach a statistical conclusion:
    If I looked over the
    totals and placed the means of the votes according to the time now tabulated
    they would look like this = 355.5 and  = 405.5, under the hypotheses test. Having
    seen no statistical difference in the new ballot based upon time I feel that
    the results are  = null accepted
    and  = rejected.

Step 8.)
Make a business decision:
Because the voting ends when the
candidates reach half the votes. I can see that the polls may not have closed
early, but through adding and leaving the announcing times the same. I could
see that there were no plausible differences that would be worthy of a change
Steps are according to (black).

Case 2 Speedyx:

      Using a significance level of .10 then
performing a one sample hypothesis test. I can determine rather the new ideas
will be profitable. By using the sample means formula and then conducting the
test in confidence levels of α, in this case .10, will look this way.  or new mean of 22 days because of the supposed
change and population mean of 220. Standard deviation of 6 and the square root
of the number 220.  says it is now z = 19.52 or 19.52 days.

      I wanted to check against the significance
levels or confidence levs which took another formula. In algebra form it will
look this way. Wit the numbers in place µ =  giving the answer 21.207. which is
a significant difference considering that the mean was said to be 24 days.

So, with the new mean µ = 21.207 I want to say that
it is the null or , accepted and = µ > 24 is rejected.

Conclusion

            The work being done is important to market analysis and I enjoy doing it. Being able to identify the proper formulas for a given situation will make the study of statistics a strong tool. When the numbers are what matters it is important to be able to do the math.

Published by David Vaughn

I am finished with my Graduate degree now I am hoping to enter into Doctoral college this fall now after starting and restarting transferring then starting again, but now unsure of my funds. I am in College now for my Phd and doing ok. This was my first website and now I have a radio station connected to it. The radio station is fanbaseradio and is apart of another website I now own. I am hoping to continue to grow my network and develop a marketing firm. kewlbeer'd marketing affiliate of fanbase.

Leave a comment

Leave a Reply

Discover more from Kewlbeer'd.com

Subscribe now to keep reading and get access to the full archive.

Continue reading