A Random Sample Of 50 Passengers Between The Two Airports
A Random Sample Of 50 Passengers Between The Two Airports Flying By Xy
A random sample of 50 passengers between the two airports flying by XYZ Airlines yielded a mean fare of $165. The population standard deviation of the fare is known to be $50.
1. a) Develop a 94% confidence interval estimate on the fare between the two airports for XYZ.
2. b) Develop a 98% confidence interval estimate on the fare between the two airports for XYZ.
3. c) Based on the confidence interval from part (b), does it appear that the mean fare between the two airports by XYZ Airlines is different than $145? State the Hypothesis. Conduct the hypothesis test using the confidence Interval from part (b). What is your conclusion?
Paper For Above instruction
The analysis of airline fare data provides valuable insights into the fare structure and customer pricing expectations, especially when the population parameters such as the mean fare are known or hypothesized. This report addresses the construction of confidence intervals at 94% and 98% levels based on a sample and uses these intervals to assess whether the fare differs significantly from a specific value, in this case, $145, through hypothesis testing.
Introduction
Confidence intervals are statistical tools used to estimate the range within which a population parameter, such as the mean, is likely to lie with a certain probability or confidence level. In airline economics, understanding fare variations helps the airline industry optimize pricing strategies and assess market competitiveness. When the population standard deviation is known, as in this case, the confidence interval can be computed using the z-distribution, simplifying the process and providing precise estimates. The current scenario involves a sample of 50 passengers with a mean fare of $165 and a known standard deviation of $50, from which confidence intervals at 94% and 98% are constructed.
Constructing Confidence Intervals
The fundamental formula for a confidence interval when the population standard deviation is known is:
\[
\text{CI} = \bar{x} \pm z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}
\]
Where:
- \(\bar{x}\) = sample mean = $165
- \(\sigma\) = population standard deviation = $50
- \(n\) = sample size = 50
- \(z_{\alpha/2}\) = critical value from the standard normal distribution corresponding to the confidence level.
Calculating for 94% Confidence Interval
The z-value for a 94% confidence level is approximately 1.88, obtained from standard z-tables or statistical software. The standard error (SE) is:
\[
SE = \frac{50}{\sqrt{50}} = \frac{50}{7.071} \approx 7.07
\]
Thus, the margin of error (ME) is:
\[
ME = 1.88 \times 7.07 \approx 13.3
\]
The 94% confidence interval is:
\[
\$165 \pm \$13.3 \Rightarrow (\$151.7, \$178.3)
\]
Calculating for 98% Confidence Interval
The z-value for a 98% confidence level is approximately 2.33. The margin of error is:
\[
ME = 2.33 \times 7.07 \approx 16.5
\]
The 98% confidence interval is:
\[
\$165 \pm \$16.5 \Rightarrow (\$148.5, \$181.5)
\]
Hypothesis Testing Based on Confidence Interval
To determine whether the mean fare significantly differs from $145, the following hypotheses are formulated:
- Null hypothesis \(H_0\): \(\mu = 145\)
- Alternative hypothesis \(H_A\): \(\mu \neq 145\)
Using the confidence interval from part (b), which is at 98%, the interval is \((\$148.5, \$181.5)\). Since $145$ is not within this interval, we reject the null hypothesis at the 1% significance level. This indicates that there is statistically significant evidence that the true mean fare differs from $145%.
Alternatively, a formal z-test can be performed:
\[
z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} = \frac{165 - 145}{50 / \sqrt{50}} = \frac{20}{7.07} \approx 2.83
\]
The critical z-value at a 2-tailed test with \(\alpha = 0.05\) is approximately 1.96. Since 2.83 > 1.96, we reject \(H_0\), confirming the conclusion from the confidence interval method.
Conclusion
The constructed 94% and 98% confidence intervals suggest that the true mean fare between the two airports for XYZ Airlines likely falls between approximately $152 and $178, and $149 and $182, respectively. The hypothesis test based on the 98% confidence interval indicates that the mean fare differs significantly from $145, leading us to reject the null hypothesis. This analysis highlights that the average fare is probably higher than $145, reflecting possible fare adjustments, market strategies, or inflationary factors affecting airline pricing.
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