Kavita has been assigned the task of studying the average customer receipt for a branch of a major restaurant chain. The average receipt for the chain is $[tex]$72.00$[/tex] with a standard deviation of $[tex]$11.00$[/tex]. The branch she is studying has an average bill of $[tex]$67.00$[/tex] for the last 40 receipts. She needs to know if this falls below the chain's average. She will use a 1% level for significance because she does not want to inadvertently report the restaurant's income as below average.
Which choice depicts the result for Kavita's hypothesis test?
A. She should reject [tex]$H _0: \mu=72$[/tex] and accept [tex]$H _{ a }: \mu< 72$[/tex].
B. She should reject [tex]$H _0: \mu=72$[/tex] and accept [tex]$H _{ a }: \mu \neq 72$[/tex].
C. She should accept [tex]$H _0: \mu=72$[/tex] and reject [tex]$H _{ a }: \mu \neq 72$[/tex].
D. She should reject [tex]$H _{ a }: \mu< 72$[/tex] but cannot accept [tex]$H _0: \mu=72$[/tex].
Answer (2)
Kavita should reject the null hypothesis H 0 : μ = 72 and accept the alternative hypothesis H a : μ < 72 . This indicates that the average receipt for the branch she is studying is significantly below the chain's average at the 1% significance level. The chosen option is A.
;
State null hypothesis H 0 : μ = 72 and alternative hypothesis H a : μ < 72 .
Calculate the z-score: z = 40 11 67 − 72 ≈ − 2.8748 .
Compare the calculated z-score (-2.8748) to the critical z-value (-2.58).
Since -2.8748 < -2.58, reject H 0 and accept H a : μ < 72 . The answer is: She should reject H 0 : μ = 72 and accept H a : μ < 72 .
Explanation
Understand the problem and provided data We are given the task of determining whether the average customer receipt for a branch of a restaurant chain falls below the chain's average. We have the following information:
The population average receipt for the chain is μ = $72.00 .
The population standard deviation is σ = $11.00 .
The sample size is n = 40 receipts.
The sample average is x ˉ = $67.00 .
The significance level is α = 1% = 0.01 .
Our goal is to perform a hypothesis test to determine if the branch's average receipt is significantly lower than the chain's average.
Outline the hypothesis test We will perform a hypothesis test with the following steps:
State the null hypothesis H 0 : μ = 72 and the alternative hypothesis H a : μ < 72 . This is a left-tailed test because we are checking if the branch's average is below the chain's average.
Calculate the test statistic (z-score) using the formula: z = n σ x ˉ − μ .
With a significance level of 1%, find the critical z-value for a left-tailed test. Since the table provides upper-tail values, we use the fact that the z-distribution is symmetric. The critical z-value for a 1% significance level in the lower tail is -2.58.
Compare the calculated z-score to the critical z-value. If the calculated z-score is less than the critical z-value, reject the null hypothesis in favor of the alternative hypothesis.
If the calculated z-score is greater than the critical z-value, we fail to reject the null hypothesis.
Calculate the z-score Now, let's calculate the z-score using the given data:
z = n σ x ˉ − μ = 40 11 67 − 72
z = 40 11 − 5 ≈ 1.7369 − 5 ≈ − 2.8748 The calculated z-score is approximately -2.8748.
Compare the z-score with the critical value and make a decision We compare the calculated z-score (-2.8748) to the critical z-value (-2.58). Since -2.8748 < -2.58, we reject the null hypothesis H 0 : μ = 72 in favor of the alternative hypothesis H a : μ < 72 . This means that the branch's average receipt is significantly lower than the chain's average at the 1% significance level.
State the conclusion Based on our analysis, Kavita should reject the null hypothesis H 0 : μ = 72 and accept the alternative hypothesis H a : μ < 72 . This indicates that the average receipt for the branch she is studying is significantly below the chain's average.
Examples
In business, hypothesis testing is often used to determine if a new marketing strategy has led to a significant increase in sales, or if a manufacturing process change has resulted in a decrease in defects. For example, a company might want to test if a new advertising campaign has increased average customer spending. They would collect data on customer spending before and after the campaign and perform a hypothesis test to see if the increase is statistically significant. This helps them make informed decisions about whether to continue or modify the campaign.
