Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Join our Q&A platform to get precise answers from experts in diverse fields and enhance your understanding. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
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].
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.
