Final answer:
The question revolves around hypothesis testing to determine if the average lunch cost exceeds $13.50.
Step-by-step explanation:
The question deals with the concept of hypothesis testing in statistics, a method by which an analyst tests an assumption regarding a population parameter. The scenario provided involves testing whether the mean amount spent on lunch (population mean) is at least $13.50, as the CEO of a restaurant chain conjectures. The marketing manager performs the test and obtains a p-value of 0.077. This value is used to decide if the null hypothesis can be rejected or not.
For instance, if the level of significance (α) chosen for the test is 0.05, since the p-value is higher than 0.05, we would fail to reject the null hypothesis, meaning that there's insufficient evidence to support that the average customer spends more than $13.50 on lunch. However, if a higher significance level such as 0.10 is chosen, the null hypothesis can be rejected, asserting that the average spending is indeed higher.
The information from the reference can be used to demonstrate the process of hypothesis testing.
For example, if our null hypothesis is about the mean salary of entry-level managers being $44,000, we would collect a sample, calculate the sample mean and compare it using a statistical test such as a t-test or z-test, depending upon the sample size and known variance. The result will inform whether to reject or fail to reject the null hypothesis.