/*! 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 185 A lawn and garden sprinkler syst... [FREE SOLUTION] | 91Ó°ÊÓ

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A lawn and garden sprinkler system is designed to have a delayed start; that is, there is a delay from the moment it is turned on until the water starts. The delay times form a normal distribution with mean 45 seconds and standard deviation 8 seconds. Several customers have complained that the delay time is considerably longer than claimed. The system engineer has selected a random sample of 15 installed systems and has obtained one delay time from each system. The sample mean is 50.1 seconds. Using \(\alpha=0.02,\) is there significant evidence to show that the customers might be correct that the mean delay time is more than 45 seconds? a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

Short Answer

Expert verified
Using both the p-value approach and the classical approach, we would compare our calculated values (the p-value and the t-score) with our metric for rejecting or failing to reject the null hypothesis (the significance level and the critical value), respectively. The conclusions from both methods should match, and they would provide us with sufficient evidence to either support or reject the customers' claim that the mean delay time is more than 45 seconds.

Step by step solution

01

Setup the Hypotheses

The first step is to formulate the null hypothesis and the alternative hypothesis. \n\nNull hypothesis, \(H_0\): \(\mu = 45\) \n\nAlternative hypothesis, \(H_1\): \(\mu > 45\) \n\nHere, \(\mu\) represents the population mean delay time.
02

Calculation of t-score

The second step involves the calculation of the t-score value to determine how many standard deviations the sample mean lies from the population mean. The t-score can be calculated using the following formula: \n\n\(t = \frac{\bar{x}-\mu}{\sigma/\sqrt{n}}\) \n\nSubstituting the given values into the formula, it would look like this: \n\n\(t = \frac{50.1-45}{8/\sqrt{15}}\)
03

Apply the P-value Approach

The p-value approach involves comparing the p-value (the probability that we would observe a t-score as extreme as, or more extreme than, the one calculated in Step 2, given that the null hypothesis is true) with the significance level (\(\alpha\)). If the p-value is less than \(\alpha\), then we reject the null hypothesis and conclude that there is significant evidence that the mean delay time is more than 45 seconds. Here, we can use the t-distribution table or statistical software to find the p-value corresponding to our t-score and \(\alpha = 0.02\).
04

Apply the Classical Approach

The classical approach, also known as the critical value approach, involves comparing the t-score with the critical value(s). The critical value is the value that the test statistic must exceed in order for the null hypothesis to be rejected. It is found using the significance level and the degrees of freedom, which is given by \(df = n-1\). If the t-score is greater than the critical value, we reject the null hypothesis. Here, we can use the t-distribution table to find the critical value corresponding to \(\alpha = 0.02\) and \(df = 15-1 = 14\) and compare it with our t-score.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is an essential component of inferential statistics. It helps us decide whether there's enough evidence to support a specific claim about a population. In our problem, the initial step is to set up the hypotheses:

The **null hypothesis** (\(H_0\)) suggests that there's no significant difference or effect. Here, it posits that the mean delay time \(\mu\) equals 45 seconds.

The **alternative hypothesis** (\(H_1\)) claims the opposite, suggesting that the mean delay time is greater than 45 seconds. This hypothesis represents what we want to prove. If we find sufficient evidence against \(H_0\), we accept \(H_1\).

By setting up these hypotheses, we lay the groundwork for making data-driven decisions.
t-Distribution
The **t-distribution** is crucial when dealing with small sample sizes or unknown population variances. Unlike the normal distribution, it is a bit flatter and wider, accommodating the additional uncertainty.

In this scenario, we use the t-distribution to calculate the t-score, which helps us determine how many standard deviations the sample mean (50.1 seconds) is away from the hypothesized population mean (45 seconds).

The t-score formula is:\[t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\]where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(s\) is the standard deviation, and \(n\) is the sample size. Calculating this gives us a specific value we can use to compare against t-distribution critical values or determine a p-value.
Significance Level
The **significance level** (\(\alpha\)) is a threshold used to decide whether to reject the null hypothesis. It represents the probability of rejecting \(H_0\) when it is actually true—also known as the Type I error.

In our exercise, \(\alpha = 0.02\) indicates a 2% risk of falsely claiming that the mean delay time is more than 45 seconds. Lower significance levels (like 0.01) reduce the chance of making this type of error but require stronger evidence to reject \(H_0\).

The significance level helps strike a balance between being cautious and decisive. It guides us in making a clear conclusion about the hypothesis test.
p-Value Approach
The **p-value approach** is a commonly used method for hypothesis testing. It calculates a p-value, which represents the probability of observing the test statistic, or one more extreme, under the null hypothesis.

To find the p-value, consider the calculated t-score from the sample data. Compare this score against a t-distribution table or use statistical software. If the p-value is less than the significance level (\(\alpha = 0.02\)), we reject the null hypothesis.

This method directly tells us how strong the evidence is against \(H_0\). A smaller p-value implies we are more confident in rejecting \(H_0\) in favor of \(H_1\).
Classical Approach
The **classical approach** to hypothesis testing is also known as the critical value approach. It involves comparing the t-score to a predefined critical value that corresponds to the chosen significance level and degrees of freedom (\(df\)).

In our example, \(df = n - 1 = 14\), given the 15 system measurements. Using \(\alpha = 0.02\), find the critical value from a t-distribution table.

If the calculated t-score exceeds this critical value, we reject the null hypothesis. This approach gives a straightforward yes-or-no answer based on comparisons, providing an alternative to the p-value method.

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Most popular questions from this chapter

Use a computer or calculator to select 36 random numbers from a normal distribution with mean 100 and standard deviation \(15 .\) Find the sample mean and \(z *\) for testing a two-tailed hypothesis test of \(\mu=100 .\) Using \(\alpha=0.05,\) state the decision. Repeat several times as in Table \(8.12 .\) Describe your findings.

The Texas Department of Health published the statewide results for the Emergency Medical Services Certification Examination. Data for those taking the paramedic exam for the first time gave an average score of 79.68 (out of a possible 100 ) with a standard deviation of \(9.06 .\) Suppose a random sample of 50 individuals taking the exam yielded a mean score of 81.05 Is there sufficient evidence to conclude that "the population from which this random sample was taken, on the average, scored higher than the state average"? Use \(\alpha=0.05\)

The weights of full boxes of a certain kind of cereal are normally distributed with a standard deviation of 0.27 oz. A sample of 18 randomly selected boxes produced a mean weight of 9.87 oz. a. Find the \(95 \%\) confidence interval for the true mean weight of a box of this cereal. b. Find the \(99 \%\) confidence interval for the true mean weight of a box of this cereal. c. What effect did the increase in the level of confidence have on the width of the confidence interval?

a. If the null hypothesis is true, the probability of a decision error is identified by what name? b. If the null hypothesis is false, the probability of a decision error is identified by what name?

The calculated value of the test statistic is actually the number of standard errors that the sample mean differs from the hypothesized value of \(\mu\) in the null hypothesis. Suppose that the null hypothesis is \(H_{o}: \mu=4.5, \sigma\) is known to be \(1.0,\) and a sample of size 100 results in \(\bar{x}=4.8\) a. How many standard errors is \(\bar{x}\) above \(4.5 ?\) b. If the alternative hypothesis is \(H_{a}: \mu>4.5\) and \(\alpha=0.01,\) would you reject \(H_{o} ?\)

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