Question

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
critical z - values	1.65	1.96	2.58

which choice depicts the result for kavita’s hypothesis test?

• she should reject ( h_0: mu = 72 ) and accept ( h_a: mu < 72 ).
• she should reject ( h_0: mu = 72 ) and accept ( h_a: mu

eq 72 ).

• she should accept ( h_0: mu = 72 ) and reject ( h_a: mu

eq 72 ).

• she should reject ( h_a: mu < 72 ) but cannot accept ( h_0: mu = 72 ).

Explanation:

Step 1: Define Hypotheses

Null hypothesis \( H_0: \mu = 72 \) (chain's average), alternative hypothesis \( H_a: \mu < 72 \) (branch's average is below chain's).

Step 2: Calculate Test Statistic

Use z - test formula: \( z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}} \)
Given \( \bar{x} = 67 \), \( \mu = 72 \), \( \sigma = 11 \), \( n = 40 \)
\( z=\frac{67 - 72}{\frac{11}{\sqrt{40}}}=\frac{- 5}{\frac{11}{6.3246}}\approx\frac{-5}{1.74}\approx - 2.87 \)

Step 3: Determine Critical Value

For 1% significance level (left - tailed test), critical z - value is - 2.58 (since critical values in table are for upper tail, left - tailed critical value is negative of upper tail value).

Step 4: Compare Test Statistic and Critical Value

Test statistic \( z=-2.87 \), critical value \( z_{critical}=-2.58 \)
Since \( - 2.87<-2.58 \), we reject \( H_0 \) and accept \( H_a \).

Answer:

She should reject \( H_0:\mu = 72 \) and accept \( H_a:\mu < 72 \)