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 $72.00 with a standard deviation of $11.00. The branch she is studying has an average bill of $67.00 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 restaurants income as below average.

Upper-Tail Values

a: 5% 2.5% 1%

Critical z-values: 1.65 1.96 2.58


Which choice depicts the result for Kavita’s hypothesis test?
A. She should reject H0 : µ = 72 and accept Ha : µ < 72.
B. She should reject H0 : µ = 72 and accept Ha : µ not equal to 72.
C. She should accept H0 : µ = 72 and reject Ha : µ not equal to 72.
D. She should reject Ha : µ < 72 but cannot accept H0 : µ = 72.

Respuesta :

Answer:

A. She should reject H0 : µ = 72 and accept Ha : µ < 72

Step-by-step explanation:

Just took it on edge

The choice that best depicts the result for Kavita’s hypothesis test is; A. She should reject H0 : µ = 72 and accept Ha : µ < 72.

What is the hypothesis conclusion?

We are told that the average receipt for the chain is $72.00 with a standard deviation of $11.00. Thus;

Population mean; µ = 72

Population standard deviation; σ = $67

Sample mean; x' = 67

sample size; n = 40

Since the study from the 40 receipts shows that the mean is less than the population mean, then we can define the hyypothesis as;

Null Hypothesis; H0: µ = 72

Alternative hypothesis; Ha : µ < 72

Since 1% significance level is used, thus; α = 0.01

p-value of z-score of 2.58 and significance level of 0.01 from online calculator is; p-value = 0.00494.

Since p-value is less than sinificance level, the we reject null hypothesis and conclude that alternative hypothesis is true.

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