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Magazine Chapter 8: Problem 22
In the past, \(20 \%\) of all airline passengers flew first class. In a sample of 15 passengers, 5 flew first class. At \(\alpha=0.10,\) can you conclude that the proportions have changed?
Short Answer
Expert verified
No, there is not enough evidence to conclude the proportions have changed at \(\alpha=0.10\).
Step by step solution
01
Define Hypotheses
First, we need to formulate our null and alternative hypotheses. The null hypothesis, denoted as \(H_0\), is that the proportion of first class passengers is still \(0.2\). The alternative hypothesis, \(H_a\), is that the proportion has changed. Thus, we have:\[ H_0: p = 0.2 \]\[ H_a: p eq 0.2 \]
02
Identify Parameters and Conditions
The sample size \(n\) is 15, and the sample proportion \( \hat{p} \) is \( \frac{5}{15} = \frac{1}{3} \approx 0.333 \). We check the conditions for using the normal approximation for the binomial distribution: \(np = 15 \times 0.2 = 3\) and \(n(1-p) = 15 \times 0.8 = 12\). Both are greater than 5, so the normal approximation is appropriate.
03
Calculate the Test Statistic
Using the sample proportion \( \hat{p} = 0.333 \), we calculate the test statistic \(z\) using the formula: \[ z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} \]Substitute the values: \[ z = \frac{0.333 - 0.2}{\sqrt{\frac{0.2 \times 0.8}{15}}} = \frac{0.133}{0.103} \approx 1.29 \]
04
Determine the Critical Value and Decision Rule
For a two-tailed test at \(\alpha = 0.10\), we split the alpha level into two tails. Hence, each tail is \(\alpha/2 = 0.05\). Using a standard normal distribution table, the critical z-scores are approximately \( \pm 1.645 \). If the test statistic is greater than 1.645 or less than -1.645, we reject the null hypothesis.
05
Conclusion
The calculated \(z\)-score is 1.29. Since 1.29 is between -1.645 and 1.645, we do not reject the null hypothesis. There is not enough evidence at the \(\alpha = 0.10\) significance level to conclude that the proportion of first-class passengers has changed.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Null Hypothesis
In hypothesis testing, the null hypothesis represents an initial assumption about a population parameter. Typically denoted as \(H_0\), it asserts that any observed effect or difference is due to random chance. It is a crucial step in testing as it provides a baseline for statistical comparison.
In our airline passenger example, the null hypothesis states that the proportion of passengers flying first-class has not changed from the original \(20\%\). We express this with the equation: \(H_0: p = 0.2\).
The purpose of the null hypothesis is to be challenged and possibly disproved by evidence found in the data sample. If the evidence (i.e., statistical data) supports the opposite of the null hypothesis, then we are prompted to reject it in favor of an alternative hypothesis.
In our airline passenger example, the null hypothesis states that the proportion of passengers flying first-class has not changed from the original \(20\%\). We express this with the equation: \(H_0: p = 0.2\).
The purpose of the null hypothesis is to be challenged and possibly disproved by evidence found in the data sample. If the evidence (i.e., statistical data) supports the opposite of the null hypothesis, then we are prompted to reject it in favor of an alternative hypothesis.
Alternative Hypothesis
The alternative hypothesis, denoted by \(H_a\), is the statement we accept when we reject the null hypothesis. It suggests that there is a statistically significant effect or difference - meaning that the initial assumption is incorrect.
For the airline study, our alternative hypothesis considers that the proportion of first-class passengers has changed, expressed as \(H_a: p eq 0.2\). This hypothesis includes the possibility that the percentage could have increased or decreased.
The alternative hypothesis is two-sided in this scenario because we are interested in detecting any change in the proportion, regardless of direction. Consequently, this requires splitting the level of significance \(\alpha\) equally between both tails of the normal distribution curve.
For the airline study, our alternative hypothesis considers that the proportion of first-class passengers has changed, expressed as \(H_a: p eq 0.2\). This hypothesis includes the possibility that the percentage could have increased or decreased.
The alternative hypothesis is two-sided in this scenario because we are interested in detecting any change in the proportion, regardless of direction. Consequently, this requires splitting the level of significance \(\alpha\) equally between both tails of the normal distribution curve.
Normal Approximation
Normal approximation allows us to simplify the analysis of binomial data by using the normal distribution鈥攅specially when dealing with large sample sizes, due to its simpler properties and wider use.
To use normal approximation, certain conditions must be met. Primarily, both \(np\) and \(n(1-p)\) need to be greater than 5, where \(p\) is the hypothesized population proportion and \(n\) is the sample size. In our scenario, the values \(np = 3\) and \(n(1-p) = 12\) satisfy this criterion.
When these conditions are satisfied, the binomial distribution of sample proportions can be approximated using a normal distribution. This approach simplifies calculating probabilities and critical values in hypothesis testing.
To use normal approximation, certain conditions must be met. Primarily, both \(np\) and \(n(1-p)\) need to be greater than 5, where \(p\) is the hypothesized population proportion and \(n\) is the sample size. In our scenario, the values \(np = 3\) and \(n(1-p) = 12\) satisfy this criterion.
When these conditions are satisfied, the binomial distribution of sample proportions can be approximated using a normal distribution. This approach simplifies calculating probabilities and critical values in hypothesis testing.
Binomial Distribution
The binomial distribution models binary outcomes, where there are only two possible results鈥攕uch as success or failure鈥攊n a fixed number of independent trials.
In the context of the exercise, each airline passenger can be viewed as having a binary outcome: they either fly first class (success) or not (failure). Here, the total number of trials equals the number of passengers sampled, which is 15.
The binomial distribution relies on two parameters: the number of trials \(n\) and the probability of success on each trial \(p\). When conditions allow, this distribution can be approximated with a normal distribution to facilitate straightforward analytical techniques, as we utilized in this hypothesis testing scenario.
In the context of the exercise, each airline passenger can be viewed as having a binary outcome: they either fly first class (success) or not (failure). Here, the total number of trials equals the number of passengers sampled, which is 15.
The binomial distribution relies on two parameters: the number of trials \(n\) and the probability of success on each trial \(p\). When conditions allow, this distribution can be approximated with a normal distribution to facilitate straightforward analytical techniques, as we utilized in this hypothesis testing scenario.
