One-Sample Hypothesis Testing Cases

One-Sample Hypothesis Testing Cases

QNT 561

One-Sample Hypothesis Testing Cases

The assignment contains two case studies that will be discussed in the sections provided below. The first case study involves a television network that wants to utilize a one-sample hypothesis test to understand if George W. Bush has above 50% of election votes in a sample, and also conclude if the network should air that he has won the state at 8:01PM. The second case study regards a courier company that is attempting to decide if including a stamped self-addressed envelope will reduce the mean amount of time it takes to obtain customers’ payments. Both cases include a one-sample hypothesis test, yet the first case is a right-tailed test whereas the second is a left-tailed test.

Case Study – Election Results

The television network will be able to announce that George W. Bush will win the state at 8:01PM if he has obtained more than 50% of votes. The hypothesis test will need to demonstrate that he has acquired more than 50% of votes from a sample of 765 voters. The sample size is large and n*p and n*q are greater than 5, therefore a z test for testing population proportion is appropriate for this test. The null hypothesis and alternative hypothesis are symbolized by H0 and Ha, correspondingly, and are statistically written as:

Expending the null hypothesis for George W. Bush has at least 50% of the votes and the alternative hypothesis has more than 50%. Hence, it is indicated by Ho: p = 0.5; and alternative H1: p > 0.5. Thus, we will use the z formula to ratify the evidence provided for the sample population. In the sample population the is the symbol for it so we will divide, in this scenario, George Bush number of votes by total of votes altogether of the sample votes to answer, which is 407 / 765. The population refers to the 50% is equal in value to p0.

z = (x / n – p0) / (p0 x (1 – p0) / n

z = (407 / 765 -0.5) / 0.5 x (1 – 0.5) / 765 = 1.7716

(0.0532 – 0.5) / (0.5 x (1-0.5) / 765 = 1.77

The calculation of the formula displays that the z statistic = 1.77. Because the z statistic of 1.77 is greater than the critical value of 1.28, the conclusion is prepared to reject the null hypothesis. This means that the test results conclude that George W. Bush has achieved more than 50% of the votes and can be broadcasted as the state winner at 8:01PM.

SpeedX Case Study

SpeedX is a huge courier company that propels invoices to customers requesting payment within 30 days. The bill lists an address, and customers are expected to use their own envelopes to return their payments. Thus, the customers must have their individual supply of envelopes to make payments by mail. Correspondingly, the mean and standard deviation of the amount of time taken to pay bills are 24 days and 6 days. The chief financial officer (CFO) considers including a stamped self-addressed envelope would decline the extent of time for their customers. Hence, by computing the supplemented cash flow from a 2-day diminution in the payment period would pay for the costs of the envelopes and stamps.

As a business analyst one of my duties is to track analytics and present the results to the senior management for critical decision-making. The CFO assumed the test would work, therefore, my decision to erratically select 220 customer and suggest to include a self-addressed envelope with their invoices. The CFO agrees to my proposal and allows me to run a pilot study. I’ve documented the number of days until the payment is received. Additionally, using my knowledge and skills by gaining understanding in class to conduct a one-sample hypothesis test and to determine whether I can persuade the CFO to achieve the plan will be cost-effective. To conclude whether the CFO will be successful by containing a self-addressed envelope would decrease the amount of time by testing the hypothesis. The hypothesis formula as follows:

H0: µ ≥ 22

H1: µ < 22

-z0.10 = -1.28

To make the calculation for the statistics necessary for z:

z = (means – µ0) / (σ / n)

z = (21.6318 – 22) / (6 / 220) = 0.9102

The z statistic of -0.91 is greater than the critical value of -1.28, consequently, the decision is to not reject the null hypothesis, specifying that including a stamped self-addressed envelope in invoices may not reduce the mean payment time to under 22 days.

References

Black, K. (2017). Business statistics: for contemporary decision making (9th Edition). Hoboken, NJ: Wiley.