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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].

Sagot :

GinnyAnsweridn
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].
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