/*! 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 28 Minor surgery on horses under fi... [FREE SOLUTION] | 91Ó°ÊÓ

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Minor surgery on horses under field conditions requires a reliable short-term anesthetic producing good muscle relaxation, minimal cardiovascular and respiratory changes, and a quick, smooth recovery with minimal aftereffects so that horses can be left unattended. The article "A Field Trial of Ketamine Anesthesia in the Horse" (Equine Vet. J., 1984: 176-179) reports that for a sample of \(n=73\) horses to which ketamine was administered under certain conditions, the sample average lateral recumbency (lying-down) time was \(18.86 \mathrm{~min}\) and the standard deviation was \(8.6 \mathrm{~min}\). Does this data suggest that true average lateral recumbency time under these conditions is less than \(20 \mathrm{~min}\) ? Test the appropriate hypotheses at level of significance .10.

Short Answer

Expert verified
Do not reject the null hypothesis; there is insufficient evidence that the average time is less than 20 minutes.

Step by step solution

01

Define the Hypotheses

Begin by defining the null and alternative hypotheses. The null hypothesis \( H_0 \) presumes that the true average lateral recumbency time \( \mu \) is equal to 20 minutes. The alternative hypothesis \( H_a \) is that the true average time is less than 20 minutes. Thus, we have:\[H_0: \mu = 20 \H_a: \mu < 20\]
02

Choose the Significance Level

The problem specifies that the level of significance is 0.10. This means we are using \( \alpha = 0.10 \) to determine whether to reject the null hypothesis.
03

Compute the Test Statistic

Use the sample data to calculate the test statistic. The formula for the test statistic in this case is:\[z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}\]Where \( \bar{x} = 18.86 \), \( \mu = 20 \), \( s = 8.6 \), and \( n = 73 \). Substituting these values in, we get:\[z = \frac{18.86 - 20}{\frac{8.6}{\sqrt{73}}} \approx \frac{-1.14}{1.004} \approx -1.136\]
04

Determine the Critical Value

Since this is a one-tailed test with \( \alpha = 0.10 \), we look up the critical z-value for a left-tailed test. The critical z-value is approximately -1.28.
05

Compare Test Statistic with Critical Value

Compare the computed test statistic to the critical value. If the test statistic is less than the critical value, we reject the null hypothesis. In this case, the test statistic \(-1.136\) is not less than the critical value \(-1.28\).
06

Conclusion

Since the test statistic is not less than the critical value, we do not reject the null hypothesis at a 0.10 significance level. This indicates that there is not enough statistical evidence to conclude that the true average lateral recumbency time is less than 20 minutes.

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Key Concepts

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

Null Hypothesis
The null hypothesis, often represented as \( H_0 \), is a starting assumption in hypothesis testing. In essence, it is the hypothesis that states there is no effect or no difference. It's a neutral statement, often assumed to be true until evidence suggests otherwise.
In the context of our exercise, the null hypothesis presumes that the true average lateral recumbency time, \( \mu \), for horses is 20 minutes. This is expressed as:
  • \( H_0: \mu = 20 \)
Understanding the null hypothesis is crucial because it sets the stage for statistical testing. It's the hypothesis we test directly against the alternative hypothesis. If the collected data provides sufficient evidence against the null hypothesis, we may reject it. If not, we do not reject it and infer that there isn't enough evidence to suggest a difference or effect exists.
Alternative Hypothesis
The alternative hypothesis, denoted as \( H_a \) or sometimes \( H_1 \), is the statement we aim to support through evidence from our data. It indicates the presence of an effect or a difference compared to the null hypothesis.
For our ketamine study on horses, the alternative hypothesis suggests that the true average lateral recumbency time is less than 20 minutes. This can be expressed as:
  • \( H_a: \mu < 20 \)
In hypothesis testing, if the data shows enough evidence to support this claim, the null hypothesis may be rejected in favor of the alternative hypothesis.
The alternative hypothesis is essential because it guides the direction of the statistical test. In this one-sided test, we are specifically looking for evidence to show that the average time is less than 20 minutes.
Significance Level
The significance level, symbolized by \( \alpha \), is a threshold that determines how confident we need to be to reject the null hypothesis. It represents the probability of making a Type I error, which is rejecting the null hypothesis when it is actually true.
In this exercise, the significance level is set at \( 0.10 \), or 10%. This means there's a 10% risk of concluding that the average lateral recumbency time is less than 20 minutes when in reality, it isn't.
  • A lower \( \alpha \) value indicates more stringent evidence is needed to reject the null hypothesis, reducing the chance of a Type I error.
  • A higher \( \alpha \) allows for a higher probability of making this error, but might also detect a true effect more easily if it exists.
Setting a clear \( \alpha \) level at the beginning helps us in deciding later whether the observed data is statistically significant.
Test Statistic
The test statistic is a crucial value calculated from the sample data. It is used to decide whether to reject the null hypothesis. For this particular problem, we employ a z-test since the sample size is reasonably large and the population standard deviation is not known.
The formula for the z-test statistic is:
  • \[ z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \]
where:
  • \( \bar{x} \) = sample mean
  • \( \mu \) = mean under the null hypothesis
  • \( s \) = sample standard deviation
  • \( n \) = sample size
Using the values provided (\( \bar{x} = 18.86 \), \( \mu = 20 \), \( s = 8.6 \), \( n = 73 \)), our computed z is approximately -1.136.
The test statistic is then compared against a critical value, determined by the chosen significance level and nature of the test (one-tailed or two-tailed), to reach a conclusion on the hypothesis test.

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Most popular questions from this chapter

Exercise 33 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 \(.05\).

For a fixed alternative value \(\mu^{\prime}\), show that \(\beta\left(\mu^{\prime}\right) \rightarrow 0\) as \(n \rightarrow \infty\) for either a one-tailed or a two-tailed \(z\) test in the case of a normal population distribution with known \(\sigma\).

One method for straightening wire before coiling it to make a spring is called "roller straightening." The article "The Effect of Roller and Spinner Wire Straightening on Coiling Performance and Wire Properties" (Springs, 1987: 27-28) reports on the tensile properties of wire. Suppose a sample of 16 wires is selected and each is tested to determine tensile strength \(\left(\mathrm{N} / \mathrm{mm}^{2}\right)\). The resulting sample mean and standard deviation are 2160 and 30 , respectively. a. The mean tensile strength for springs made using spinner straightening is \(2150 \mathrm{~N} / \mathrm{mm}^{2}\). What hypotheses should be tested to determine whether the mean tensile strength for the roller method exceeds 2150 ? b. Assuming that the tensile strength distribution is approximately normal, what test statistic would you use to test the hypotheses in part (a)? c. What is the value of the test statistic for this data? d. What is the \(P\)-value for the value of the test statistic computed in part (c)?

For the following pairs of assertions, indicate which do not comply with our rules for setting up hypotheses and why (the subscripts 1 and 2 differentiate between quantities for two different populations or samples): a. \(H_{0}: \mu=100, H_{\mathrm{a}}: \mu>100\) b. \(H_{0}: \sigma=20, H_{\mathrm{a}}: \sigma \leq 20\) c. \(H_{0}: p \neq .25, H_{\mathrm{a}}: p=.25\) d. \(H_{0}: \mu_{1}-\mu_{2}=25, H_{\mathrm{a}}: \mu_{1}-\mu_{2}>100\) e. \(H_{0}: S_{1}^{2}=S_{2}^{2}, H_{a}: S_{1}^{2} \neq S_{2}^{2}\) f. \(H_{0}: \mu=120, H_{\mathrm{a}}: \mu=150\) g. \(H_{0}: \sigma_{1} / \sigma_{2}=1, H_{\mathrm{a}}: \sigma_{1} / \sigma_{2} \neq 1\) h. \(H_{0}: p_{1}-p_{2}=-.1, H_{\mathrm{a}}: p_{1}-p_{2}<-.1\)

The article "Caffeine Knowledge, Attitudes, and Consumption in Adult Women" \((J\). Nutrit. Ed., 1992: 179-184) reports the following summary data on daily caffeine consumption for a sample of adult women: \(n=47, \bar{x}=215 \mathrm{mg}, s=235\) \(\mathrm{mg}\), and range \(=5-1176\). a. Does it appear plausible that the population distribution of daily caffeine consumption is normal? Is it necessary to assume a normal population distribution to test hypotheses about the value of the population mean consumption? Explain your reasoning. b. Suppose it had previously been believed that mean consumption was at most \(200 \mathrm{mg}\). Does the given data contradict this prior belief? Test the appropriate hypotheses at significance level \(.10\) and include a \(P\)-value in your analysis.

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