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Chapter 8: Problem 38

According to one source, \(50 \%\) of plane crashes are due at least in part to pilot error (http://www.planecrashinfo.com). Suppose that in a random sample of 100 separate airplane accidents, 62 of them were due to pilot error (at least in part.) a. Test the null hypothesis that the proportion of airplane accidents due to pilot error is not \(0.50\). Use a significance level of \(0.05\). b. Choose the correct interpretation: i. The percentage of plane crashes due to pilot error is not significantly different from \(50 \%\). ii. The percentage of plane crashes due to pilot error is significantly different from \(50 \%\).

Short Answer

Expert verified
The null hypothesis is rejected. Hence, the proportion of airplane accidents due to pilot error is significantly different from \( 50 \% \).

Step by step solution

01

State the hypotheses

The null hypothesis, \( H_0 \), is that the proportion of airplane accidents due to pilot error is \( 0.50 \) or \( 50\% \). The alternative hypothesis, \( H_1 \), is that the proportion is not \( 0.50 \) or \( 50\% \). Symbolically, these are expressed as: \( H_0: p = 0.50 \) \( H_1: p \neq 0.50 \)
02

Compute the test statistic

The test statistic for a hypothesis test for a proportion is given by the formula: \( Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \) Where, \( \hat{p} \) is the observed proportion, \( \hat{p} = \frac{x}{n} = \frac{62}{100} = 0.62 \) \( p_0 \) is the assumed proportion in the null hypothesis, \( p_0 = 0.50 \) n is the sample size, \( n = 100 \) Substituting these values in, we get \( Z = \frac{0.62 - 0.50}{\sqrt{\frac{0.50(1-0.50)}{100}}} = 2.40 \)
03

Compute the p-value

The p-value is the probability of obtaining a test statistic as extreme as the observed one, assuming the null hypothesis is true. Here we're performing a two-tailed test since we're looking for a difference either way from \( 0.50 \). We can use a standard normal distribution table (or a calculator or computer software) to determine the probability of getting a Z value as extreme as 2.40. This value is \( 0.016 \)
04

Decision and conclusion

The decision rule for a significance level of \( 0.05 \) is to reject the null hypothesis if the p-value is less than \( 0.05 \). Here, the p-value of \( 0.016 \) is less than \( 0.05 \), so we reject the null hypothesis. The result is significant at the \( 0.05 \) level. The data provide enough evidence to conclude that the proportion of airplane accidents due to pilot error is significantly different from \( 50 \%\). The correct interpretation is: ii. The percentage of plane crashes due to pilot error is significantly different from \( 50 \%\).

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Key Concepts

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

Proportion Hypothesis Test
Hypothesis testing involves making a claim or assertion about a population parameter and then using sample data to test this claim. In the context of a proportion hypothesis test, we assess whether the proportion of a certain outcome in a population matches a specified value.
For this type of test, we start by establishing two hypotheses:
  • The null hypothesis (\( H_0 \)): which suggests there is no effect or difference, stating that the true population proportion (\( p \)) is equal to the hypothesized value (e.g., 0.50).
  • The alternative hypothesis (\( H_1 \)): which indicates that the true proportion differs from the hypothesized value.
To conduct this test, we use a sample from the population to compute an observed proportion. We then calculate a test statistic, like the Z-score, to see how far this observed proportion is from what was claimed in the null hypothesis. If there is strong evidence against\( H_0 \), we conclude that the actual population proportion is different from the fitted value.
P-value
The p-value offers a powerful tool in hypothesis testing. It tells us how likely we are to observe a test statistic as extreme, or even more extreme, than what we got, presuming the null hypothesis is true.
In simpler terms, the p-value quantifies the strength of evidence against the null hypothesis. A smaller p-value signals stronger evidence that the null hypothesis is probably not true.
  • If the p-value is less than the significance level (commonly \( 0.05 \)), we reject the null hypothesis. This means the results are statistically significant.
  • If the p-value is greater than or equal to the significance level, we do not reject the null hypothesis. The results are not deemed statistically significant.
In our example, a p-value of 0.016 indicated that there was enough evidence to reject the null hypothesis, concluding that the real proportion of plane crashes due to pilot error is indeed different from 50%.
Z-score
The Z-score in the context of hypothesis testing is a measure of how many standard deviations an observed sample proportion is away from the population proportion specified in the null hypothesis.
We compute the Z-score using this formula: \[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \] Where:
  • \( \hat{p} \) is the observed proportion from the sample.
  • \( p_0 \) is the proportion stated in the null hypothesis.
  • \( n \) is the sample size.
The calculated Z-score helps us determine how unusual the observed sample proportion is under the null hypothesis. In our problem, a Z-score of 2.40 showed that the sample proportion of plane crashes due to pilot error was 2.40 standard deviations away from the hypothesized 50%.
This Z-score, combined with a significant p-value, led us to reject the null hypothesis, indicating that there was a significant difference in the proportion.

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Most popular questions from this chapter

The percentage of female CEOs in 2013 in Fortune 500 companies was \(4 \%\), according to Time magazine (March 18,2013 ). Suppose a student did a survey of 250 randomly selected large companies (not Fortune 500 companies), and 15 of them had women CEOs. a. About how many of the 250 should we expect to have female CEOs if the proportion of female CEOs is \(4 \%\), as it is for the Fortune 500 companies? b. Carry out the first two steps of a hypothesis test that will test whether the population proportion of women CEOs is not \(0.04\).

A new food supplement is being tested to see whether it can reduce obesity in the population of concern. At present, the percentage of obese people is \(60 .\) The null hypothesis is that \(p\) (the proportion of the population using the food supplement that is still obese) is \(0.60 .\) Pick the correct alternative hypothesis. i. \(p \neq 0.60\) ii. \(p<0.60\) iii. \(p>0.60\)

An MCQ test has 75 questions with four options each. Suppose a passing grade is 50 or more correct answers. Test the claim that a student knows more than half of the answers and is not just guessing. Assume the student gets 50 answers correct out of 75 . Use a significance level of \(0.05\). Steps 1 and 2 of a hypothesis test procedure are given. Show steps 3 and \(4 .\) and be sure to write a clear conclusion. Step \(1: \mathrm{H}_{0}: p=0.25\) \(\mathrm{H}_{\mathrm{a}}: p>0.25\) Step 2: Choose the one-proportion \(z\) -test. Sample size is large enough, because \(n p_{0}\) is \(75(0.25)=18.75\), and \(n\left(1-p_{0}\right)=75(0.75)=56.25\), and both are more than \(10 .\) Assume the sample is random and \(\alpha=0.05\).

A researcher carried out a hypothesis test using a two-sided alternative hypothesis. Which of the following \(z\) -scores is associated with the smallest p-value? Explain. $$ \text { i. } z=0.50 $$ ii. \(z=1.00\) iii. \(z=2.00\) iv. \(z=3.00\)

Suppose you are testing someone to see whether he or she can tell goat cheese from cheddar cheese. You use many bite-sized cubes selected randomly, half from goat cheese and half from cheddar cheese. The taster is blindfolded. The null hypothesis is that the taster is just guessing and should get about half right. When you reject the null hypothesis when it is actually true, that is often called the first kind of error. The second kind of error is when the null is false and you fail to reject. Report the first kind of error and the second kind of error.
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