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]$\$[/tex]72.00[tex]$ with a standard deviation of $[/tex]\[tex]$11.00$[/tex]. The branch she is studying has an average bill of [tex]$\$[/tex]67.00$ for the last 40 receipts. She needs to know if this falls below the chain's average. She will use a level of significance because she does not want to inadvertently report the restaurant's income as below average.

\begin{tabular}{|c|c|c|c|}
\hline \multicolumn{4}{|c|}{Upper-Tail Values} \\
\hline [tex]$a$[/tex] & [tex]$5\%$[/tex] & [tex]$2.5\%$[/tex] & [tex]$1\%$[/tex] \\
\hline \begin{tabular}{c}
Critical \\
[tex]$z$[/tex]-values
\end{tabular} & 1.65 & 1.96 & 2.58 \\
\hline \hline
\end{tabular}

Which choice depicts the result for Kavita's hypothesis test?

A. She should reject [tex]$H_0: \mu=72$[/tex] and accept [tex]$H_a: \mu\ \textless \ 72$[/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]$H_a: \mu\ \textless \ 72$[/tex].
D. She should reject [tex]$H_a: \mu\ \textless \ 72$[/tex] but cannot accept [tex]$H_0: \mu=72$[/tex].

To determine whether Kavita should accept or reject the null hypothesis $ H_0 $ that the average receipt for the branch is [tex]$72.00$[/tex], we need to perform a hypothesis test using the following steps:

1. State the Hypotheses:
- Null Hypothesis ($ H_0 $): $ \mu = 72 $
- Alternative Hypothesis ($ H_a $): $ \mu < 72 $

2. Set the Significance Level:
- Given the critical z-value for a 5% significance level (lower-tail test) is $ -1.65 $.

3. Calculate the Test Statistic:
- Given:
- Chain average ($ \mu $): 72.00
- Chain standard deviation ($ \sigma $): 11.00
- Branch average ($ \bar{x} $): 67.00
- Sample size ($ n $): 40
- Calculate the standard error ($ SE $):
[tex]\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{11.00}{\sqrt{40}} \approx 1.74 \][/tex]
- Calculate the z-statistic:
[tex]\[ z = \frac{\bar{x} - \mu}{SE} = \frac{67.00 - 72.00}{1.74} \approx -2.875 \][/tex]

4. Compare the Test Statistic to the Critical Value:
- The calculated z-statistic is $ -2.875 $.
- The critical value for a 5% significance level is $ -1.65 $.

5. Decision:
- Since $ -2.875 $ is less than $ -1.65 $, we reject the null hypothesis $ H_0 $.

Therefore, Kavita should reject $ H_0: \mu = 72 $ and accept $ H_a: \mu < 72 $. This means that the average customer receipt for the branch is significantly below the chain's average.

The correct choice is:
- She should reject [tex]$ H_0: \mu = 72 $[/tex] and accept [tex]$ H_a: \mu < 72 $[/tex].