/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Exercise 36 in Chapter 1 gave \(... [FREE SOLUTION] | 91Ó°ÊÓ

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Exercise 36 in Chapter 1 gave \(n=26\) observations on escape time (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 suggests the true average escape time is greater than 6 minutes.

Step by step solution

01

Define the hypothesis

We want to test if the true average escape time exceeds 6 minutes, which is 360 seconds. So, the null hypothesis (H_0) is that the mean escape time is less than or equal to 360 seconds, and the alternative hypothesis (H_a) is that the mean escape time is greater than 360 seconds. Mathematically, \( H_0: \mu \leq 360 \) and \( H_a: \mu > 360 \).
02

Set the significance level

The significance level, \( \alpha \), is provided as 0.05. This is the threshold for determining whether we have enough evidence to reject the null hypothesis.
03

Calculate the test statistic

To calculate the test statistic, we use the formula for the t-statistic: \[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \] where \( \bar{x} = 370.69 \) is the sample mean, \( \mu_0 = 360 \) is the hypothesized mean, \( s = 24.36 \) is the sample standard deviation, and \( n = 26 \) is the sample size. Substituting the values, \[ t = \frac{370.69 - 360}{24.36 / \sqrt{26}} \approx 2.155 \].
04

Determine the critical value

Since this is a one-tailed test with \( n - 1 = 25 \) degrees of freedom, we find the critical value \( t_{0.05, 25} \) from the t-distribution table. The critical value for a 0.05 significance level for 25 degrees of freedom is approximately \(1.708 \).
05

Make a decision

Compare the test statistic to the critical value. Since our test statistic \( t \approx 2.155 \) is greater than the critical value \(1.708\), we reject the null hypothesis.

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Key Concepts

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

t-test
The t-test is a statistical test used to determine if there is a significant difference between the means of two groups. It is particularly useful when dealing with small sample sizes and when the population standard deviation is unknown.
The t-test evaluates how the sample mean differs from the null hypothesis value by using the t-statistic.
  • For the t-test, we assume that the data comes from a normally distributed population.
  • The formula for the t-statistic is \( t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \), where \( \bar{x} \) is the sample mean, \( \mu_0 \) is the population mean, \( s \) is the sample standard deviation, and \( n \) is the sample size.
The calculated t-value tells us how many standard deviations the observed mean is from the hypothesized mean. In our example, the calculated t-value of approximately 2.155 indicates that the sample mean differs from the hypothesized mean by more than two standard deviations. If this calculated t-value is greater than the critical value, we reject the null hypothesis, indicating a statistically significant difference.
significance level
The significance level, often denoted as \( \alpha \), is a threshold that helps us decide whether to reject the null hypothesis in a statistical test. In simpler terms, it is the probability of making a Type I error, which means rejecting the null hypothesis when it is actually true.
  • A common significance level used is 0.05, meaning there is a 5% risk of incorrectly rejecting the null hypothesis.
  • The chosen significance level determines how strict the criteria are for judging the statistical evidence.
In our example, a significance level of 0.05 was used. This means that only if the probability of observing the sample data, given that the null hypothesis is true, is less than 5%, we will reject the null hypothesis. Using this level provides a balance between being too lenient and too strict in making statistical inferences.
null and alternative hypotheses
Formulating null and alternative hypotheses is a crucial step in hypothesis testing. The null hypothesis, denoted as \( H_0 \), represents a statement of no effect or no difference, and it assumes that any kind of difference or significance observed in a set of data is due to chance. Conversely, the alternative hypothesis, \( H_a \), suggests that there is an actual effect or a difference that is not due to random chance.
  • The null hypothesis usually includes "equal to" or "less than or equal to" statements.
  • The alternative hypothesis might include "greater than" or "not equal to," indicating the direction of the test.
In our oil workers' escape time example:
  • The null hypothesis \( H_0: \mu \leq 360 \) seconds means the average escape time is at most 360 seconds.
  • The alternative hypothesis \( H_a: \mu > 360 \) seconds proposes that the average escape time is more than 360 seconds.
By conducting the t-test, we aim to decide whether to reject the null hypothesis in favor of the alternative, based on the data.
critical value
The critical value is a point on the test statistic distribution that defines a threshold. It helps in deciding whether to reject the null hypothesis. This value corresponds to the chosen significance level \( \alpha \) and the degrees of freedom associated with the data.
  • For different tests and distributions, critical values are determined by the significance level and degrees of freedom.
  • The critical value divides the distribution into the region where we accept the null hypothesis and the region where we reject it.
In the context of our exercise, the critical value was approximately 1.708 for a one-tailed t-test with 25 degrees of freedom (calculated as \( n-1 \)). By comparing the test statistic to this critical value, we conclude that if the test statistic exceeds it, the null hypothesis is rejected. In our scenario, since the test statistic of 2.155 is greater than the critical value of 1.708, it indicates sufficient evidence against the null hypothesis, leading us to reject it.

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