/*! 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 56 The times of first sprinkler act... [FREE SOLUTION] | 91Ó°ÊÓ

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The times of first sprinkler activation for a series of tests with fire prevention sprinkler systems using an aqueous film-forming foam were (in sec) \(\begin{array}{lllllllllll}27 & 41 & 22 & 27 & 23 & 35 & 30 & 33 & 24 & 27 & 28 & 22 & 24\end{array}\) (see "Use of AFFF in Sprinkler Systems," Fire Tech., 1976: 5). The system has been designed so that true average activation time is at most \(25 \mathrm{~s}\) under such conditions. Does the data strongly contradict the validity of this design specification? Test the relevant hypotheses at significance level \(.05\) using the \(P\)-value approach.

Short Answer

Expert verified
Since the P-value is greater than 0.05, we do not reject the null hypothesis. The data does not strongly contradict the design specification.

Step by step solution

01

State the Hypotheses

We are testing if the true average activation time is more than 25 seconds, so we set up our hypotheses. The null hypothesis is \( H_0: \mu \leq 25 \) and the alternative hypothesis is \( H_a: \mu > 25 \).
02

Calculate Sample Statistics

Compute the sample mean \( \bar{x} \) and sample standard deviation \( s \) from the data. The data is: 27, 41, 22, 27, 23, 35, 30, 33, 24, 27, 28, 22, 24. The sample mean is given by \( \bar{x} = \frac{\sum x_i}{n} \) and the sample standard deviation \( s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \).
03

Perform the T-test

With the sample mean \( \bar{x} \), the hypothesized mean \( \mu_0 = 25 \), sample standard deviation \( s \), and sample size \( n = 13 \), perform a one-sample t-test. The test statistic is given by \[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \].
04

Determine the P-value

Using the calculated t-statistic and degrees of freedom \( n-1 = 12 \), determine the one-tailed P-value from the t-distribution table.
05

Conclusion

Compare the P-value to the significance level \( \alpha = 0.05 \). If the P-value is less than \( \alpha \), we reject the null hypothesis; otherwise, we do not reject it.

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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 test that's used to compare the mean of a single group against a known value or compare the means of two different groups. It's especially useful when the sample size is small and the population standard deviation is unknown.
In the context of hypothesis testing for the sprinkler system, a T-test helps us determine if the average time for the sprinkler activation truly exceeds the specified design time of 25 seconds. We perform a one-sample T-test here, which means we're comparing the sample mean to a known value, in this case, 25 seconds.
  • The formula for the T-test statistic is: \[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \] where- \( \bar{x} \) is the sample mean,- \( \mu_0 \) is the hypothesized population mean,- \( s \) is the sample standard deviation,- \( n \) is the sample size.

The resulting T-statistic tells us how much the sample mean deviates from the hypothesized mean in terms of the standard error. A larger absolute value of \( t \) indicates a bigger difference.
P-value
The P-value helps us ascertain the strength of the evidence against the null hypothesis. It tells us the probability of obtaining a test statistic as extreme as, or more extreme than, the observed statistic, assuming the null hypothesis is true.
In the sprinkler system example, once we compute the T-statistic, we reference the t-distribution table with the calculated degrees of freedom (which is the sample size minus one, so here it's 12) to find the P-value. This is a one-tailed test because we're interested in whether the true mean is greater than 25.
  • If the resulting P-value is less than the predetermined significance level (\( \alpha = 0.05 \)), it implies there's enough evidence to reject the null hypothesis in favor of the alternative.
This means that there's a statistically significant indication that the average activation time is more than 25 seconds.
Sample Statistics
Sample statistics are numerical values that summarize the important features of a sample taken from a population. In this problem, we're concerned with the sample mean and the sample standard deviation.
  • The sample mean (\( \bar{x} \)) is calculated by summing all the observed values and dividing by the number of observations (). This gives us the average activation time in our sample.
  • The sample standard deviation (\( s \)) measures the amount of variation or dispersion in the sample. It's calculated by taking the square root of the variance (average of the squared deviations from the mean).
As an example:\[\bar{x} = \frac{\sum x_i}{n}\] and \[s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}\]where each \( x_i \) is an observed value in the sample.
Sample statistics provide a simplified view of the data and are crucial for computing the T-test and interpreting the results.
Significance Level
The significance level, often denoted by \( \alpha \), is the threshold for determining the statistical significance of a test. It represents the probability of rejecting the null hypothesis when it is actually true, also known as the Type I error.
In the exercise, we set the significance level at 0.05. This means that we are willing to accept a 5% chance of incorrectly rejecting the null hypothesis.
  • If the P-value is less than or equal to the significance level, we reject the null hypothesis.
  • If the P-value is greater, we do not have enough evidence to dismiss the null hypothesis.
Choosing an appropriate significance level is crucial as it affects the likelihood of making a mistake in hypothesis testing. Generally, a lower \( \alpha \) reduces the risk of a Type I error, but it may increase the chance of a Type II error – not rejecting a false null hypothesis.

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Most popular questions from this chapter

A new design for the braking system on a certain type of car has been proposed. For the current system, the true average braking distance at 40 mph under specified conditions is known to be \(120 \mathrm{ft}\). It is proposed that the new design be implemented only if sample data strongly indicates a reduction in true average braking distance for the new design. a. Define the parameter of interest and state the relevant hypotheses. b. Suppose braking distance for the new system is normally distributed with \(\sigma=10\). Let \(\bar{X}\) denote the sample average braking distance for a random sample of 36 observations. Which of the following rejection regions is appropriate: \(R_{1}=\\{\bar{x}: \bar{x} \geq 124.80\\}, R_{2}=\) \(\\{\bar{x}: \bar{x} \leq 115.20\\}, R_{3}=\\{\bar{x}:\) either \(\bar{x} \geq 125.13\) or \(\bar{x} \leq 114.87\\} ?\) c. What is the significance level for the appropriate region of part (b)? How would you change the region to obtain a test with \(\alpha=.001\) ? d. What is the probability that the new design is not implemented when its true average braking distance is actually \(115 \mathrm{ft}\) and the appropriate region from part (b) is used? e. Let \(Z=(\bar{X}-120) /(\sigma / \sqrt{n})\). What is the significance level for the rejection region \(\\{z\) : \(z \leq-2.33\\}\) ? For the region \(\\{z: z \leq-2.88\\}\) ?

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