Question

Evaluate how organizations can use one-sample hypothesis testing to determine if there are performance issues in the organization. Support your response with a specific example.

Answer 1

Answer:

Organizations can use one-sample hypothesis test to determine if there are performance issues in many ways.

It can be applied to the performance of a sector, a machine, a product, an advertising campaing, etc.

For example, we can take the example of a machine. It may be claimed that a specific machine performs significantly worse than the average.

This average would be the population mean: the average performance of the machines of the same type or process.

Then, a sample of the performance of the machine in study is taken and the hypothesis test can be performed to test the claim that this machine performs significantly worse.

Step-by-step explanation:

For example, we have an historic performance for this type of machine of 100 units a day. The machine A in study is sampled 14 days and have a performance of 92 units a day, with a sample standard deviation of 12 units/day. We have to test the claim that the machine A makes less units per day than the average.

Then, the null and alternative hypothesis are:

$H_0: \mu=100\\\\H_a:\mu< 100$

The significance level is 0.05.

The sample has a size n=14.

The sample mean is M=92.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=12.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{12}{\sqrt{14}}=3.2071$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{92-100}{3.2071}=\dfrac{-8}{3.2071}=-2.4944$

The degrees of freedom for this sample size are:

$df=n-1=14-1=13$

This test is a left-tailed test, with 13 degrees of freedom and t=-2.4944, so the P-value for this test is calculated as (using a t-table):

$P-value=P(t<-2.4944)=0.0134$

As the P-value (0.0134) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that machine A produces significantly less units per day than the average.