kavita has been assigned the task of studying the average customer receipt for a branch of a major…

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.\n\n| upper - tail values | | | |\n| ---- | ---- | ---- | ---- |\n| a | 5% | 2.5% | 1% |\n| critical z - values | 1.65 | 1.96 | 2.58 |\n\nwhich choice depicts the result for kavitas hypothesis test?\no she should reject h₀: μ = 72 and accept hₐ: μ < 72.\no she should reject h₀: μ = 72 and accept hₐ: μ ≠ 72.\no she should accept h₀: μ = 72 and reject hₐ: μ ≠ 72.\no she should reject hₐ: μ < 72 but cannot accept h₀: μ = 72.

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.\n\n| upper - tail values | | | |\n| ---- | ---- | ---- | ---- |\n| a | 5% | 2.5% | 1% |\n| critical z - values | 1.65 | 1.96 | 2.58 |\n\nwhich choice depicts the result for kavitas hypothesis test?\no she should reject h₀: μ = 72 and accept hₐ: μ < 72.\no she should reject h₀: μ = 72 and accept hₐ: μ ≠ 72.\no she should accept h₀: μ = 72 and reject hₐ: μ ≠ 72.\no she should reject hₐ: μ < 72 but cannot accept h₀: μ = 72.

Answer

Explanation:

Step1: State the hypotheses

$H_0:\mu = 72$ (null - hypothesis, the branch's average is the same as the chain's average), $H_a:\mu<72$ (alternative - hypothesis, the branch's average is less than the chain's average)

Step2: Calculate the test - statistic

The formula for the z - test statistic 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). $z=\frac{67 - 72}{\frac{11}{\sqrt{40}}}=\frac{- 5}{\frac{11}{\sqrt{40}}}\approx\frac{-5}{1.74}\approx - 2.87$

Step3: Compare with the critical 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 $z_{\alpha}=- 2.33$ (since it's a left - tailed test). Our calculated $z=-2.87<-2.33$. Since the calculated z - value is less than the critical z - value, we reject the null hypothesis $H_0:\mu = 72$ and accept the alternative hypothesis $H_a:\mu<72$.

Answer:

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