Hypothesis testing is a statistical method used to assess the validity of a proposed idea or hypothesis. It involves comparing the observed data against a null hypothesis (the status quo) to determine if there is sufficient evidence to support the alternative hypothesis (the new idea). The goal is to determine whether the observed data is likely to have occurred by chance under the null hypothesis, or if it provides strong enough evidence to reject the null hypothesis in favor of the alternative hypothesis (Mandanat, 2024).

How to choose between Z-test and T-test?

Use a Z-test when the population standard deviation is known, and the sample size is large.

Use a T-test when the population standard deviation is unknown, and the sample size is small.

Difference between Z-test and T-test:

Criteria	Z-test	T-test
Population standard deviation known	Yes	No
Sample Size	Large (n>30)	Small (n<30)
Distribution used	Normal distribution	t-distribution

Information

Industry standard queue time (0) = 150 seconds

Significance level () = 0.05

Sample size is large (n > 30), we can use the z-test.

Hypothesis
Null Hypothesis	H0	150	The average queue time is greater than or equal to the industry standard of 150 seconds.
Alternative Hypothesis	H1	< 150	The average queue time is less than the industry standard of 150 seconds.

Decision Making Rule

If the probability p is less that the level of significance =0.05 then we reject the null hypothesis

If p

Assumptions

The sample size is large enough (n > 30) to use the t-distribution.

Hypothesis Test 1:

Is the average queue time (QueueTime) lower than the industry standard of 2.5 minutes (150 seconds), using a significance level of =0.05?

Sum of QueueTime (=SUM(B2:B1680)) = 247622

Sample mean (x) (=AVERAGE(B2:B1680)) = 147.92

Sample standard deviation(s) (=STDEV.S(B2:B1680)) = 137.97

t = (x - ) / (s / n)

t = (147.92-150) / (137.97/1674

t = (-2.08) / (139.97 / 1674)

t = (-2.08) / (139.97 / 40.91)

t = (-2.08) / (3.42)

t = -0.61

Degrees of freedom (df) - one sample test

df = n-1

df = 1674 - 1

df = 1673

Critical value

Critical value in the one-tailed t-distribution table for df = 1673 and = 0.05, which is -1.645.

Confidence level = 1 -

Confidence level = 1 - 0.05

Confidence level = 0.95 (or 95%)

The 95% confidence level means there is a 5% chance of rejecting the null hypothesis when it is true (Type I error).

Evaluate if the company should allocate more resources to improve its average TiQ.

If the test statistic falls below the critical value of -1.645, the company would have evidence to reject the null hypothesis (H0: 0) in favor of the alternative.

Rejecting the null hypothesis in this case would indicate the current average TiQ is not satisfactory and could be improved.

Recommendation

The company's current average TiQ is not meeting expectations, and a one-tailed left-sided t-test results in a test statistic below the critical value of -1.645, then the company would have statistical evidence to support allocating more resources to improve the average TiQ.

The 95% confidence level suggests the company can be confident in this decision, as there is only a 5% chance of making a Type I error (incorrectly rejecting the null hypothesis).