/*! 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
Fail to reject the null hypothesis; insufficient evidence to claim the true average time is less than 20 minutes.

Step by step solution

01

Define the Hypotheses

First, we need to state the null and alternative hypotheses. - The null hypothesis \( H_0: \mu = 20 \text{ min} \)- The alternative hypothesis \( H_a: \mu < 20 \text{ min} \)Here, \( \mu \) represents the true average lateral recumbency time. We are testing if the average time is less than 20 minutes.
02

Determine the Test Statistic

We will use the one-sample t-test for the mean because the population standard deviation is unknown. The test statistic is calculated as:\[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \]where \( \bar{x} = 18.86 \text{ min} \) is the sample mean, \( \mu_0 = 20 \text{ min} \) is the population mean under the null hypothesis, \( s = 8.6 \text{ min} \) is the sample standard deviation, and \( n = 73 \) is the sample size.
03

Calculate the Test Statistic

Substitute the given values into the test statistic formula:\[ t = \frac{18.86 - 20}{8.6/\sqrt{73}} \]Calculating the denominator:\[ \frac{8.6}{\sqrt{73}} \approx 1.004 \]Now calculate the t-statistic:\[ t = \frac{18.86 - 20}{1.004} \approx -1.14 \]
04

Find the Critical Value

Since this is a left-tailed test at a significance level of 0.10, we use the t-distribution table to find the critical value for \( 72 \) degrees of freedom (\( n-1 \)). The critical t-value at 0.10 significance is approximately \( -1.294 \).
05

Compare the Test Statistic to the Critical Value

The calculated t-statistic is \( t = -1.14 \), and our critical t-value is \( -1.294 \). We compare to determine if our test statistic falls in the rejection region. Since \( -1.14 > -1.294 \), the test statistic is not in the rejection region.
06

Make a Decision

Since the test statistic \( -1.14 \) is greater than the critical value \( -1.294 \), we fail to reject the null hypothesis. There is not enough evidence at the 0.10 level of significance 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.

Hypothesis Testing
Hypothesis testing is a method in statistics used to make decisions based on data. When we perform a hypothesis test, we start by establishing two hypotheses: the null hypothesis and the alternative hypothesis. The null hypothesis \( H_0 \) represents the status quo or the assertion that any observed difference in a sample is due to chance.
In contrast, the alternative hypothesis \( H_a \) is what we want to prove. For example, in our horse anesthesia study, the null hypothesis is that the true average lateral recumbency time \( \mu \) equals 20 minutes. Meanwhile, the alternative hypothesis suggests that this time is less than 20 minutes.
Hypothesis testing involves comparing these hypotheses using sample data to draw conclusions. This process helps us determine whether our sample data provides enough evidence to reject the null hypothesis.
Critical Value
The critical value in hypothesis testing is a threshold that determines the boundary between regions where the null hypothesis can and cannot be rejected. It's derived from the probability distribution of the test statistic under the null hypothesis.
For the problem at hand, we use a t-distribution since the population standard deviation is not known. The critical value indicates the point beyond which lies an area of significance, typically tailing off into regions of highest improbability under the null hypothesis.
In our exercise, with 72 degrees of freedom and a 0.10 significance level, we find a critical value of approximately \(-1.294\). If the test statistic falls beyond this critical value in the negative direction, the null hypothesis can be rejected, pointing towards a significant finding.
Significance Level
The significance level \( \alpha \) is a probability threshold chosen by the researcher, representing the risk of rejecting the null hypothesis when it is actually true. It's a way to quantify the likelihood of making a type I error, or a false positive.
In the context of our horse anesthesia study, the significance level is set at 0.10, or 10%. This means that we are accepting a 10% chance of wrongly rejecting the null hypothesis when it's true.
The significance level dictates the critical value and helps guide the decision-making process in hypothesis testing. A higher significance level increases the critical region under the null hypothesis distribution, allowing for more room to detect a significant effect.
Test Statistic
A test statistic is a standardized value resulting from a sample data analysis, used to decide whether to reject the null hypothesis. It summarizes how far the sample mean deviates from the null hypothesis in units of standard error.
For our one-sample t-test, the formula 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, and \( n \) is the sample size.
In our example, we compute \( t \approx -1.14 \), indicating the number of standard errors the sample mean is below the hypothesized population mean. This calculated t-value is then compared to the critical value to decide if the observed effect is statistically significant.

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