91Ó°ÊÓ

Two competing airlines, Alpha and Beta, fly a route between Des Moines, Iowa, and Wichita, Kansas. Each airline claims to have a lower percentage of flights that arrive late. Let \(p_{1}\) be the proportion of Alpha's flights that arrive late and \(p_{2}\) the proportion of Beta's tlights that arrive late. a. You are asked to observe a random sample of arrivals for each airline to estimate \(p_{1}-p_{2}\) with a \(90 \%\) confidence level and a margin of error of estimate of \(.05 .\) How many arrivals for each airline would you have to observe? (Assume that you will observe the same number of arrivals, \(n\), for each airline. To be sure of taking a large enough sample, use \(p_{1}=p_{2}=.50\) in your calculations for \(n .\) ) b. Suppose that \(p_{1}\) is actually \(.30\) and \(p_{2}\) is actually \(.23 .\) What is the probability that a sample of 100 flights for each airline ( 200 in all) would yield \(\hat{p}_{1} \geq \hat{p}_{2}\) ?

Step by step solution

01

Find the Sample Size

Let the desired Margin of Error (ME) = 0.05, Confidence level = 90%, implying z-value(z) = 1.645 (from z-table) and \(p_1\)=\(p_2\)=0.50. The formula for the margin of error for a difference in proportions is ME = z * sqrt [ (\(p_1\)(1-\(p_1\)) / n) + (\(p_2\)(1-\(p_2\)) / n) ]. Inserting the values, we get 0.05 = 1.645 * sqrt [ (0.50*0.50 / n) + (0.50*0.50 / n) ], which simplifies to 0.05 = 1.645*sqrt(0.50/n). Squaring both sides and solving for n, we get n >=1067. This means you need to observe at least 1067 arrivals for each airline.

02

Calculate the Probability that p-hat1 >= p-hat2

For a sample size n=100 for each airline, the p1 and p2 are actually 0.30 and 0.23 respectively. The difference between the sample proportions is \(\hat{p}_1- \hat{p}_2\). Its standard deviation (std dev) is sqrt [ (p1*(1-p1) / n) + (p2*(1-p2) / n) ]. Substituting for p1, p2 and n, the std dev = sqrt [ (0.30*0.70/100) + (0.23*0.77/100) ] = 0.072. The z-score for this difference in proportions is given by (Differnce -0) / std dev. As \(\hat{p}_1 \geq \hat{p}_2\), the difference is greater or equal to zero. So min z-value is when the difference is zero, that is Z= 0/0.072 =0. From the Z-table, the area to the right of Z=0 (i.e., probability p-hat1 >= p-hat2) is 0.5.

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Key Concepts

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

Confidence Intervals

When we estimate statistics like the proportion of flights that arrive late, we often use what's called a confidence interval. A confidence interval gives us a range where we believe the true proportion lies. This range is determined based on the sample data we collect and how confident we want to be about our estimate.
For example, if we say we have a 90% confidence interval, it means if we took 100 different samples and calculated their intervals, approximately 90 of those intervals would contain the true proportion. To construct these intervals, we use a z-value, which is determined by our confidence level. For a 90% confidence level, the z-value is 1.645.

• This z-value comes from the standard normal distribution, often found in a z-table.
• The larger the z-value, the wider the confidence interval, which means we can be more confident about the range covering the true proportion.

The width of the interval depends on the sample size and variability in the data. A larger sample size will yield a narrower confidence interval, reflecting a more precise estimate.

Sample Size Determination

Determining the right sample size is crucial in any statistical analysis. It impacts the accuracy of our results and our confidence about them. In this context, we need to calculate how many sample flights to observe for each airline so our estimate of the proportion difference has the desired precision: a 90% confidence level with a margin of error of 0.05.
The margin of error is the amount of error we can tolerate in our estimate. It is determined by the formula ME = z * sqrt[ (p(1-p)/n) for both populations. Simplified, this formula shows that the margin of error decreases with a larger sample size.

• Larger samples make us more certain that our estimated proportion is close to the true proportion.
• Using an expected proportion of 0.5 assumes maximum variability, making sure the sample size n is enough in the worst-case scenario.
• Calculations show at least 1067 arrivals per airline should be observed for accuracy.

Knowing how to adequately calculate sample size helps ensure our study can make valid and reliable inferences from the data collected.

Proportions

Proportions are a familiar concept, referring to parts of a whole expressed as a fraction or percentage. In our context, proportions tell us about the parts of flights arriving late out of the total observed flights. For instance, if 30% of Alpha's flights are late, that's a proportion of 0.30.
Accurate understanding of proportions helps in comparing two different groups and making valid conclusions. We use the difference in proportions to compare two airlines here, represented mathematically as \(\hat{p}_1 - \hat{p}_2\).

• Difference in proportions allows us to see which airline has more late flights, based on samples.
• The calculation involves sample proportions (\(\hat{p}\)), which estimate the true proportions (\(p\)).
• The probability calculation shows whether observed differences in samples could have happened by chance.

Understanding sample and true proportions can help in various real-world scenarios, not just airline punctuality. It's a critical concept for comparing segments and understanding variances in different data sets.