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](https://tex.z-dn.net/?f=H_0%3A%20%5Cmu%3D100%5C%5C%5C%5CH_a%3A%5Cmu%3C%20100)
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](https://tex.z-dn.net/?f=s_M%3D%5Cdfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%3D%5Cdfrac%7B12%7D%7B%5Csqrt%7B14%7D%7D%3D3.2071)
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](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7BM-%5Cmu%7D%7Bs%2F%5Csqrt%7Bn%7D%7D%3D%5Cdfrac%7B92-100%7D%7B3.2071%7D%3D%5Cdfrac%7B-8%7D%7B3.2071%7D%3D-2.4944)
The degrees of freedom for this sample size are:
![df=n-1=14-1=13](https://tex.z-dn.net/?f=df%3Dn-1%3D14-1%3D13)
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](https://tex.z-dn.net/?f=P-value%3DP%28t%3C-2.4944%29%3D0.0134)
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.