Answer :
To address Kavita's hypothesis test, we need to follow these steps:
1. Define the hypotheses:
- Null hypothesis ([tex]\(H_0\)[/tex]): The average receipt for the branch is the same as the chain's average: [tex]\( \mu = 72 \)[/tex].
- Alternative hypothesis ([tex]\(H_a\)[/tex]): The average receipt for the branch is less than the chain's average: [tex]\( \mu < 72 \)[/tex].
2. Identify the significance level:
- Given in the problem is a significance level ([tex]\( \alpha \)[/tex]) of [tex]\(0.05\)[/tex] or [tex]\(5\%\)[/tex].
3. Calculate the test statistic (z-score):
- Using the sample size ([tex]\( n = 40 \)[/tex]), the chain's mean ([tex]\( \mu = 72 \)[/tex]), the branch's mean ([tex]\( \bar{x} = 67 \)[/tex]), and the chain's standard deviation ([tex]\( \sigma = 11 \)[/tex]), we use the formula for the z-score:
[tex]\[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \][/tex]
- Plugging in the numbers:
[tex]\[ z = \frac{67 - 72}{\frac{11}{\sqrt{40}}} = \frac{-5}{\frac{11}{6.3246}} \approx -2.875 \][/tex]
Thus, the z-score is approximately [tex]\(-2.875\)[/tex].
4. Determine the critical z-value:
- For a one-tailed test at [tex]\( 0.05 \)[/tex] significance level, the critical z-value is [tex]\(-1.65\)[/tex] (negative because it is a lower-tail test).
5. Compare the test statistic to the critical value:
- The calculated z-score ([tex]\(-2.875\)[/tex]) is less than the critical z-value ([tex]\(-1.65\)[/tex]).
6. Make a decision:
- Since [tex]\(-2.875 < -1.65\)[/tex], we reject the null hypothesis ([tex]\(H_0\)[/tex]).
Thus, the test statistic falls into the rejection region, indicating that there is sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. The result is significant at the [tex]\(5\%\)[/tex] level, suggesting that the average receipt at the branch is indeed less than the chain's average.
Therefore, the correct choice is:
- She should reject [tex]\( H _0: \mu=72 \)[/tex] and accept [tex]\( H _{ a }: \mu<72 \)[/tex].
1. Define the hypotheses:
- Null hypothesis ([tex]\(H_0\)[/tex]): The average receipt for the branch is the same as the chain's average: [tex]\( \mu = 72 \)[/tex].
- Alternative hypothesis ([tex]\(H_a\)[/tex]): The average receipt for the branch is less than the chain's average: [tex]\( \mu < 72 \)[/tex].
2. Identify the significance level:
- Given in the problem is a significance level ([tex]\( \alpha \)[/tex]) of [tex]\(0.05\)[/tex] or [tex]\(5\%\)[/tex].
3. Calculate the test statistic (z-score):
- Using the sample size ([tex]\( n = 40 \)[/tex]), the chain's mean ([tex]\( \mu = 72 \)[/tex]), the branch's mean ([tex]\( \bar{x} = 67 \)[/tex]), and the chain's standard deviation ([tex]\( \sigma = 11 \)[/tex]), we use the formula for the z-score:
[tex]\[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \][/tex]
- Plugging in the numbers:
[tex]\[ z = \frac{67 - 72}{\frac{11}{\sqrt{40}}} = \frac{-5}{\frac{11}{6.3246}} \approx -2.875 \][/tex]
Thus, the z-score is approximately [tex]\(-2.875\)[/tex].
4. Determine the critical z-value:
- For a one-tailed test at [tex]\( 0.05 \)[/tex] significance level, the critical z-value is [tex]\(-1.65\)[/tex] (negative because it is a lower-tail test).
5. Compare the test statistic to the critical value:
- The calculated z-score ([tex]\(-2.875\)[/tex]) is less than the critical z-value ([tex]\(-1.65\)[/tex]).
6. Make a decision:
- Since [tex]\(-2.875 < -1.65\)[/tex], we reject the null hypothesis ([tex]\(H_0\)[/tex]).
Thus, the test statistic falls into the rejection region, indicating that there is sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. The result is significant at the [tex]\(5\%\)[/tex] level, suggesting that the average receipt at the branch is indeed less than the chain's average.
Therefore, the correct choice is:
- She should reject [tex]\( H _0: \mu=72 \)[/tex] and accept [tex]\( H _{ a }: \mu<72 \)[/tex].
