Answer :
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.
