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Cheaper Airfares Looking for a great airfare? Consumer Reports \(^{8}\) has several hints about how to minimize your costs, which include being flexible about travel dates and times, checking multiple websites, and knowing when to book your flight. One suggestion involved checking fares at "secondary" airports - airports that might be slightly farther from your home, but where fares are lower. For example, the average of all domestic ticket prices at Los Angeles International Airport (LAX) was quoted as \(\$ 349\) compared to an average price of \(\$ 287\) at nearby Ontario International Airport (ONT). Suppose that these estimates were based on random samples of 1000 domestic tickets at each airport and that the standard deviation of the prices at both airports was \(\$ 200 .\) a. Is there sufficient evidence to indicate that the mean ticket prices differ for these two airports at the \(\alpha=\) .05 level of significance? Use the large-sample ztest. What is the \(p\) -value of this test? b. Construct a \(95 \%\) confidence interval for \(\left(\mu_{1}-\mu_{2}\right) .\) Does this interval confirm your conclusions in part a?

Step by step solution

01

State the null and alternative hypothesis

\(H_0 : \mu_1 = \mu_2\) The alternative hypothesis is that the mean ticket prices differ significantly between LAX and ONT: \(H_a : \mu_1 \neq \mu_2\)

02

Perform the large-sample z-test

We use the following z-test formula to calculate the test statistic: \(z = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{(\sigma_1^2)}{n_1} + \frac{(\sigma_2^2)}{n_2}}}\) Given that both airports have a standard deviation of $200 and sample sizes of 1000 domestic tickets: \(z = \frac{(349 - 287) - (0)}{\sqrt{\frac{(200)^2}{1000} + \frac{(200)^2}{1000}}} = 5.18\)

03

Calculate the p-value

Using the z-distribution table, we find the p-value associated with our calculated test statistic: \(p = 2 \times (1 - P(Z \leq 5.18)) \approx 0\) Since the p-value is less than the \(\alpha = 0.05\) level of significance, we reject the null hypothesis, indicating that there is sufficient evidence to believe that the mean ticket prices differ significantly for these two airports. b. Confidence interval:

04

Calculate the margin of error

Using the margin of error formula: \(E = z \cdot \sqrt{\frac{(\sigma_1^2)}{n_1} + \frac{(\sigma_2^2)}{n_2}}\) Using our previously-calculated test statistic, \(z = 5.18\): \(E = 5.18 \cdot \sqrt{\frac{(200)^2}{1000} + \frac{(200)^2}{1000}} = 61.55\)

05

Calculate the confidence interval

Our 95% confidence interval for \((\mu_1 - \mu_2)\) is: \((\bar{x}_1 - \bar{x}_2) \pm E = (349 - 287) \pm 61.55 = (62 \pm 61.55)\) The confidence interval is approximately \((0.45, 123.55)\). Since this interval does not contain 0, this confirms our conclusions from the hypothesis test that the mean ticket prices differ significantly for the two airports.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

z-test

The z-test is a statistical method used to determine if there is a significant difference between the means of two groups. In this case, we're comparing the average ticket prices at two different airports to see if they vary significantly.

• We start by setting up our hypotheses: the null hypothesis (\(H_0\)) suggests no difference in means (\(\mu_1 = \mu_2\)), while the alternative hypothesis (\(H_a\)) proposes a difference (\(\mu_1 eq \mu_2\)).
• We use the z-test when the standard deviation is known, and the sample size is large (typically \(n > 30\)). This makes the calculation robust.

To perform the z-test, you calculate the z-score, which tells you how many standard deviations away your sample mean is from the null hypothesis mean. For our example, the formula used is:\[z = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{(\sigma_1^2)}{n_1} + \frac{(\sigma_2^2)}{n_2}}}\]Here, \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means, and \(\sigma\) is the known standard deviation, while \(n\) denotes the number of samples.
In this scenario, the calculated z-score is 5.18. This sharp z-value signifies a significant departure from the assumed mean difference of zero, thereby suggesting different prices indeed.

confidence interval

A confidence interval provides a range of values that is likely to contain the true difference between two population means. It gives us a sense of the precision and reliability of our estimate.
In this example, we're looking at a 95% confidence interval for the difference in ticket prices between LAX and ONT. Here’s how it’s done:

• Calculate the margin of error (E), using a formula that includes our previous z-value and standard deviation: \(E = z \times \sqrt{\frac{(\sigma_1^2)}{n_1} + \frac{(\sigma_2^2)}{n_2}}\).
• Next, compute the interval using the sample mean difference and the margin of error: \((\bar{x}_1 - \bar{x}_2) \pm E\).

The computed confidence interval from our example is approximately (0.45, 123.55). Because this confidence interval does not include zero, it strengthens our evidence that there is indeed a significant difference in the mean ticket prices between the two airports.

significance level

The significance level, denoted by \(\alpha\), represents the threshold for deciding when to reject the null hypothesis. It is usually set at 0.05, meaning there is a 5% risk of concluding that a difference exists when there is none.
Choosing \(\alpha = 0.05\) implies that we are willing to accept a 5% chance of mistakenly rejecting a true null hypothesis (committing a Type I error). This level is a standard threshold in many scientific studies.
In the context of our airfare example, our z-test produced a p-value of approximately 0, which is far less than our \(\alpha\) of 0.05. Thus, we confidently reject the null hypothesis and infer that the mean ticket prices at the two airports are significantly different.

p-value

The p-value is a key concept in hypothesis testing that helps us determine the strength of our evidence against the null hypothesis. It tells us the probability of observing our data, or something more extreme, if the null hypothesis were true.
For our airfare case, the p-value is calculated from the z-score, which was found to be 5.18. For such a high z-score, the p-value is essentially zero (or very close to it).
This extremely low p-value indicates that the observed difference in mean prices is highly unlikely to be due to random chance. As a result, we have strong evidence to reject the null hypothesis, confirming that the ticket prices at LAX and ONT are indeed different.

• If the p-value is less than or equal to the significance level (\(\alpha = 0.05\)), we reject the null hypothesis.
• If it is greater than \(\alpha\), we do not reject it.

Since our p-value is significantly lower than 0.05, we have a solid basis to say that the price difference is not coincidental.