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Exercise 36 in Chapter 1 gave \(n=26\) observations on escape time \((\mathrm{sec})\) for oil workers in a simulated exercise, from which the sample mean and sample standard deviation are \(370.69\) and \(24.36\), respectively. Suppose the investigators had believed a priori that true average escape time would be at most \(6 \mathrm{~min}\). Does the data contradict this prior belief? Assuming normality, test the appropriate hypotheses using a significance level of .05.

Short Answer

Expert verified
Yes, the data contradicts the prior belief at the 0.05 significance level.

Step by step solution

01

Define the Statistical Hypotheses

We start by defining the null hypothesis, \( H_0 \), and the alternative hypothesis, \( H_a \). The null hypothesis is \( H_0: \mu \leq 360 \) seconds, which means the true average escape time is at most 6 minutes. The alternative hypothesis is \( H_a: \mu > 360 \) seconds, indicating the belief is contradicted and the average escape time is greater than 6 minutes.
02

Choose the Significance Level and Test

The problem specifies a significance level \( \alpha = 0.05 \). Since we have a sample with \( n = 26 \), we use the t-test for the hypothesis test because the population standard deviation is unknown and the sample size is relatively small.
03

Calculate the Test Statistic

Use the formula for the t-test statistic: \[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \] where \( \bar{x} = 370.69 \) is the sample mean, \( \mu_0 = 360 \) is the claimed population mean under the null hypothesis, \( s = 24.36 \) is the sample standard deviation, and \( n = 26 \) is the sample size.Substituting these values: \[ t = \frac{370.69 - 360}{24.36 / \sqrt{26}} = \frac{10.69}{4.777} \approx 2.237 \]
04

Determine the Critical Value

For a one-tailed test with \( \alpha = 0.05 \) and \( n - 1 = 25 \) degrees of freedom, we refer to the t-distribution table to find the critical value. The critical value \( t_{0.05, 25} \approx 1.708 \).
05

Make a Decision

Compare the test statistic calculated in Step 3, \( t \approx 2.237 \), with the critical value from Step 4, \( t_{0.05, 25} = 1.708 \). Since \( t \approx 2.237 > 1.708 \), we reject the null hypothesis.
06

State the Conclusion

Since the test statistic exceeds the critical value, we reject the null hypothesis \( H_0 \) at the \( 0.05 \) significance level. This indicates that the data contradicts the prior belief that the true average escape time is at most \( 6 \) minutes.

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Key Concepts

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

t-test
A t-test is a statistical method used to determine if there is a significant difference between the means of two groups, particularly when the sample size is small, and the population standard deviation is unknown. In this particular exercise, the one-sample t-test is used since we're comparing the sample mean of escape times to a known value, that is, 360 seconds. We apply the t-test to check if the observed sample mean of 370.69 seconds significantly deviates from the hypothesized population mean.

To calculate the t-test statistic, use the formula:
  • \[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \]
Here, \( \bar{x} \) is the sample mean, \( \mu_0 \) is the population mean under the null hypothesis, \( s \) is the sample standard deviation, and \( n \) is the sample size. For our scenario, substituting the values gives us a t-value of approximately 2.237, indicating the significance of the difference between the hypothesized and observed mean.

When using a t-test, it is crucial to ensure that the data approximately follows a normal distribution, especially since the sample size is smaller, as it is here with 26 observations.
significance level
The significance level, denoted by \( \alpha \), is a threshold used to decide whether to reject the null hypothesis. It is typically set before conducting the test, and it indicates the probability of rejecting the null hypothesis when it is actually true. In hypothesis testing, the significance level provides a balance between sensitivity and the risk of drawing incorrect conclusions.

In this exercise, a significance level of 0.05 was chosen, meaning there is a 5% risk of incorrectly rejecting the null hypothesis. This is a common choice for many researchers as it offers a reasonable balance between too lenient and too strict criteria, ensuring that results are neither too easily dismissed nor too rigorously held.

The chosen significance level directly affects the outcome of the test, guiding whether the calculated t-test statistic warrants rejection of the null hypothesis. The smaller the significance level, the more stringent the test, thereby lowering the risk of Type I error but potentially increasing the chance of a Type II error.
null and alternative hypotheses
The null and alternative hypotheses are fundamental concepts in statistical hypothesis testing, defining what we aim to test in an experiment. They are the backbone of any hypothesis testing activity, acting as claims about the population.

For this particular exercise:
  • The null hypothesis \( H_0: \mu \leq 360 \) seconds posits that the true average escape time for these workers does not exceed 360 seconds (or 6 minutes). It represents the investors' prior belief.
  • The alternative hypothesis \( H_a: \mu > 360 \) seconds opposes this, suggesting that the actual average escape time is greater than 360 seconds.
These hypotheses help determine the statistical test's outcome. If evidence strongly supports the null hypothesis, we maintain it. Conversely, if the evidence contradicts it, evident through calculations and critical value comparison, we adopt the alternative hypothesis.
critical value
Critical values serve as crucial boundary markers in hypothesis testing. They're derived from the sampling distribution under the null hypothesis and are used to decide the test's outcome.

In our case, we conducted a one-tailed test, significant for testing if the average time is specifically greater than the set threshold (6 minutes). With an \( \alpha = 0.05 \) significance level and 25 degrees of freedom (since \( n-1 \)), we found the critical value using a t-distribution table: approximately 1.708.

The calculated t-statistic (approximately 2.237) is compared to this critical value. Since 2.237 exceeds 1.708, our test statistic falls into the critical region, dictating the rejection of the null hypothesis. Thus, the evidence suggests that the true average escape time indeed surpasses 360 seconds.

In general, if the test statistic is more extreme than the critical value, we reject the null hypothesis, indicating that the observed data contradict the assumption or belief highlighted by the null hypothesis.

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Most popular questions from this chapter

To determine whether the pipe welds in a nuclear power plant meet specifications, a random sample of welds is selected, and tests are conducted on each weld in the sample. Weld strength is measured as the force required to break the weld. Suppose the specifications state that mean strength of welds should exceed \(100 \mathrm{lb} / \mathrm{in}^{2}\); the inspection team decides to test \(H_{0}: \mu=100\) versus \(H_{a}: \mu>100\). Explain why it might be preferable to use this \(H_{\mathrm{a}}\) rather than \(\mu<100\).

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Show that for any \(\Delta>0\), when the population distribution is normal and \(\sigma\) is known, the two-tailed test satisfies \(\beta\left(\mu_{0}-\Delta\right)=\beta\left(\mu_{0}+\Delta\right)\), so that \(\beta\left(\mu^{\prime}\right)\) is symmetric about \(\mu_{0}\).

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