At IDNLearn.com, find answers to your most pressing questions from experts and enthusiasts alike. Our experts provide timely, comprehensive responses to ensure you have the information you need.
Sagot :
Let's go through the problem step by step.
### Part (a) Hypothesis Test:
1. State the null and alternative hypotheses:
- Null hypothesis ([tex]\(H_0\)[/tex]): The mean ticket price in 2018 is equal to the mean ticket price in 2017.
[tex]\[ H_0: \mu = \$107.14 \][/tex]
- Alternative hypothesis ([tex]\(H_a\)[/tex]): The mean ticket price in 2018 is not equal to the mean ticket price in 2017.
[tex]\[ H_a: \mu \neq \$107.14 \][/tex]
From the given options, the correct choice is:
- A. [tex]\( H_0: \mu = \$107.14 \)[/tex]
2. Check conditions for the hypothesis test:
We need to ensure the following conditions are met for the t-test:
- The sample should be random.
- The sample size should be large enough (n ≥ 30) or the population should be approximately normally distributed.
- Independence: Each ticket price is independent of the others.
Given that the sample size is 30 and assuming the tickets were chosen randomly, we can proceed with the t-test.
From the given options:
- A. Yes, all of the conditions have been met.
3. Calculate the test statistic:
The test statistic [tex]\( t \)[/tex] is calculated as:
[tex]\[ t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} \][/tex]
Where:
- [tex]\(\bar{x}\)[/tex] = sample mean = \[tex]$112.87 - \(\mu_0\) = population mean in 2017 = \$[/tex]107.14
- [tex]\(s\)[/tex] = sample standard deviation = \[tex]$42.25 - \(n\) = sample size = 30 Plugging in the values: \[ t = \frac{112.87 - 107.14}{\frac{42.25}{\sqrt{30}}} \approx 0.74 \] So, the test statistic \( t \) is 0.74 (rounded to two decimal places). 4. Find the p-value: Using the t-distribution with \( df = n - 1 = 30 - 1 = 29 \), we find the p-value for the calculated t-statistic. The p-value is 0.464 (rounded to three decimal places). 5. Interpret the results of the test: - Compare the p-value to the significance level \( \alpha = 0.05 \): Since \( \text{p-value} = 0.464 \) is greater than \( \alpha = 0.05 \), we do not reject the null hypothesis. - Conclusion: - There is insufficient evidence to conclude that theater ticket prices in the city changed from the 2017 price. ### Part (b) Confidence Interval: 1. Construct a 95% confidence interval for the sample mean: The confidence interval for the sample mean can be constructed using the formula: \[ \bar{x} \pm t^* \left( \frac{s}{\sqrt{n}} \right) \] Where \( t^* \) is the critical value from the t-distribution with \( df = 29 \), for a 95% confidence level. Plugging in the values, the 95% confidence interval is: \[ (97.09, 128.65) \] 2. Interpret the confidence interval: - The 95% confidence interval for the theater ticket prices in 2018 is \((97.09, 128.65)\). - Since the 2017 mean price of \$[/tex]107.14 falls within the confidence interval, this supports our conclusion from part (a) that there is insufficient evidence to suggest a significant change in theater ticket prices from 2017 to 2018.
### Part (a) Hypothesis Test:
1. State the null and alternative hypotheses:
- Null hypothesis ([tex]\(H_0\)[/tex]): The mean ticket price in 2018 is equal to the mean ticket price in 2017.
[tex]\[ H_0: \mu = \$107.14 \][/tex]
- Alternative hypothesis ([tex]\(H_a\)[/tex]): The mean ticket price in 2018 is not equal to the mean ticket price in 2017.
[tex]\[ H_a: \mu \neq \$107.14 \][/tex]
From the given options, the correct choice is:
- A. [tex]\( H_0: \mu = \$107.14 \)[/tex]
2. Check conditions for the hypothesis test:
We need to ensure the following conditions are met for the t-test:
- The sample should be random.
- The sample size should be large enough (n ≥ 30) or the population should be approximately normally distributed.
- Independence: Each ticket price is independent of the others.
Given that the sample size is 30 and assuming the tickets were chosen randomly, we can proceed with the t-test.
From the given options:
- A. Yes, all of the conditions have been met.
3. Calculate the test statistic:
The test statistic [tex]\( t \)[/tex] is calculated as:
[tex]\[ t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} \][/tex]
Where:
- [tex]\(\bar{x}\)[/tex] = sample mean = \[tex]$112.87 - \(\mu_0\) = population mean in 2017 = \$[/tex]107.14
- [tex]\(s\)[/tex] = sample standard deviation = \[tex]$42.25 - \(n\) = sample size = 30 Plugging in the values: \[ t = \frac{112.87 - 107.14}{\frac{42.25}{\sqrt{30}}} \approx 0.74 \] So, the test statistic \( t \) is 0.74 (rounded to two decimal places). 4. Find the p-value: Using the t-distribution with \( df = n - 1 = 30 - 1 = 29 \), we find the p-value for the calculated t-statistic. The p-value is 0.464 (rounded to three decimal places). 5. Interpret the results of the test: - Compare the p-value to the significance level \( \alpha = 0.05 \): Since \( \text{p-value} = 0.464 \) is greater than \( \alpha = 0.05 \), we do not reject the null hypothesis. - Conclusion: - There is insufficient evidence to conclude that theater ticket prices in the city changed from the 2017 price. ### Part (b) Confidence Interval: 1. Construct a 95% confidence interval for the sample mean: The confidence interval for the sample mean can be constructed using the formula: \[ \bar{x} \pm t^* \left( \frac{s}{\sqrt{n}} \right) \] Where \( t^* \) is the critical value from the t-distribution with \( df = 29 \), for a 95% confidence level. Plugging in the values, the 95% confidence interval is: \[ (97.09, 128.65) \] 2. Interpret the confidence interval: - The 95% confidence interval for the theater ticket prices in 2018 is \((97.09, 128.65)\). - Since the 2017 mean price of \$[/tex]107.14 falls within the confidence interval, this supports our conclusion from part (a) that there is insufficient evidence to suggest a significant change in theater ticket prices from 2017 to 2018.
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Accurate answers are just a click away at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.
