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 chains 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%

which choice depicts the result for kavitas hypothesis test?

she should reject h₀: μ = 72 and accept hₐ: μ < 72.
she should reject h₀: μ = 72 and accept hₐ: μ ≠ 72.
she should accept h₀: μ = 72 and reject hₐ: μ ≠ 72.
she should reject hₐ: μ < 72 but cannot accept h₀: μ = 72.

Explanation:

Step1: Calculate the z - score

The formula for the z - score in a one - sample z - test is $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$, where $\bar{x} = 67$ (sample mean), $\mu = 72$ (population mean), $\sigma = 11$ (population standard deviation), and $n = 40$ (sample size).
$z=\frac{67 - 72}{\frac{11}{\sqrt{40}}}=\frac{- 5}{\frac{11}{6.3246}}\approx\frac{-5}{1.74}= - 2.87$

Step2: Compare with critical z - value

The significance level is $\alpha=0.01$ for a one - tailed test. The critical z - value for a one - tailed test with $\alpha = 0.01$ is $z_{critical}=- 2.33$ (since it is a left - tailed test $H_a:\mu<72$). Since the calculated $z=-2.87<-2.33$, we reject the null hypothesis $H_0:\mu = 72$.

Answer:

She should reject $H_0:\mu = 72$ and accept $H_a:\mu<72$.