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.

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