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Chapter 10: Problem 65

In a study of copper deficiency in cattle, the copper values \((\mu \mathrm{g} / 100 \mathrm{~mL}\) blood) were detemined both for cattle grazing in an area known to have welldefined molybdenum anomalies (metal values in excess of the normal range of regional variation) and for cattle grazing in a nonanomalous area ("An Investigation into Copper Deficiency in Cattle in the Southem Pennines," J. Agric. Soc. Cambridge, 1972: 157-163), resulting in \(s_{1}=21.5(m=48)\) for the anomalous condition and \(s_{2}=19.45\) \((n=45)\) for the nonanomalous condition. Test for the equality versus inequality of population variances at significance level . 10 by using the \(P\)-value approach.

Short Answer

Expert verified
There is insufficient evidence to conclude that the variances are unequal at the 0.10 significance level.

Step by step solution

01

Define the Hypotheses

We start by defining the null and alternative hypotheses for testing the equality of variances. The null hypothesis \(H_0\) states that the population variances are equal, \(\sigma_1^2 = \sigma_2^2\). The alternative hypothesis \(H_a\) states that the population variances are not equal, \(\sigma_1^2 eq \sigma_2^2\).
02

Choose the Appropriate Test Statistic

To test the variances, we will use the F-test, which involves the test statistic \(F = \frac{s_1^2}{s_2^2}\), where \(s_1^2\) and \(s_2^2\) are the sample variances for the two groups.
03

Calculate the Test Statistic

Calculate the test statistic with the given sample standard deviations. Here, \(s_1 = 21.5\) for the anomalous area and \(s_2 = 19.45\) for the non-anomalous area. Therefore, the test statistic \(F = \left(\frac{21.5}{19.45}\right)^2 = 1.222\).
04

Determine the Degrees of Freedom

The degrees of freedom for the F-test are calculated based on the sample sizes. For group 1, it is \(df_1 = m - 1 = 48 - 1 = 47\), and for group 2, it is \(df_2 = n - 1 = 45 - 1 = 44\).
05

Find the P-value

Using an F-distribution table or software, find the P-value associated with the calculated F-statistic \(1.222\) and degrees of freedom \((47, 44)\). The P-value for this test is approximately 0.326.
06

Compare P-value with Significance Level

At the \(0.10\) significance level, compare the P-value \(0.326\) to \(0.10\). Since \(0.326 > 0.10\), we do not reject the null hypothesis.
07

Conclusion of Hypothesis Test

Since we do not reject the null hypothesis, we conclude that there is insufficient evidence at the 0.10 significance level to say that the population variances are unequal.

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Key Concepts

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

Understanding Hypothesis Testing
In statistics, hypothesis testing is a method used to make informed conclusions about the characteristics of a population. When conducting hypothesis testing, we set up two opposing hypotheses:
  • The null hypothesis (\(H_0\)), which is a statement of no effect or no difference. It's what we seek to test against.
  • The alternative hypothesis (\(H_a\)), which is a statement indicating the presence of an effect or a difference.
In the context of testing for population variances, our null hypothesis could be that the variances of two different populations are equal (\(\sigma_1^2 = \sigma_2^2\)). The alternative hypothesis would then state that these variances are not equal (\(\sigma_1^2 eq \sigma_2^2\)).

The objective of hypothesis testing is to determine if the observed data can indeed reject the null hypothesis in favor of the alternative. This involves calculating a test statistic and comparing it to a significance level, which helps us decide whether the evidence is strong enough to make such a conclusion.
Exploring Population Variances
Population variance is a measure that describes how data points in a population spread out from the mean. It's represented by the symbol \( \sigma^2 \) and plays a crucial role in statistical analyses.

Understanding how population variances compare can reveal differences in variability between groups. For example, when comparing cattle grazing in two different areas with one having a molybdenum anomaly, calculating and comparing their variances can inform us about the consistency or variability of copper levels in their blood.
  • A lower population variance indicates that the data points are closer to the mean, suggesting less variability.
  • A higher population variance means the data spread more widely around the mean, indicating greater variability.
When testing for equality of variances, we often use the F-test, which enables comparison of the ratio of two sample variances, providing insights into whether the two samples have similar variability.
Interpreting the P-value
The \(P\)-value is a critical concept in hypothesis testing, representing the probability of obtaining test results at least as extreme as the results observed, under the assumption that the null hypothesis is true.

A small \(P\)-value (typically 鈮� 0.05) indicates strong evidence against the null hypothesis, suggesting it be rejected in favor of the alternative.
  • If the \(P\)-value is less than or equal to our chosen significance level (\(\alpha\)), we reject the null hypothesis.
  • If the \(P\)-value is greater than the significance level, we do not reject the null hypothesis.
In our cattle study example, we got a \(P\)-value of 0.326. Since this value exceeds our significance level of 0.10, we conclude that there is insufficient evidence to reject the null hypothesis. Thus, we believe that the population variances between the two grazing areas are not significantly different.

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Most popular questions from this chapter

An experiment to determine the effects of temperature on the survival of insect eggs was described in the article "Development Rates and a TemperatureDependent Model of Pales Weevil" (Emiron. Entomol, 1987: 956-962). At \(11^{\circ} \mathrm{C}, 73\) of \(91 \mathrm{eggs}\) survived to the next stage of development. At \(30^{\circ} \mathrm{C}\), 102 of 110 eggs survived. Do the results of this experiment suggest that the survival rate (proportion surviving) differs for the two temperatures? Calculate the \(P\)-value and use it to test the appropriate hypotheses.

The following data refers to airborne bacteria count (number of colonies/ft \({ }^{3}\) ) both for \(m=8\) carpeted hospital rooms and for \(n=8\) uncarpeted rooms ("Microbial Air Sampling in a Carpeted Hospital," J. Environ. Health, 1968: 405). Does there appear to be a difference in true average bacteria count between carpeted and uncarpeted rooms? $$ \begin{array}{llllllllll} \text { Carpeted } & 11.8 & 8.2 & 7.1 & 13.0 & 10.8 & 10.1 & 14.6 & 14.0 \\ \text { Uncarpeted } & 12.1 & 8.3 & 3.8 & 7.2 & 12.0 & 11.1 & 10.1 & 13.7 \end{array} $$ Suppose you later learned that all carpeted rooms were in a veterans' hospital, whereas all uncarpeted rooms were in a children's hospital. Would you be able to assess the effect of carpeting? Comment.

A group of 115 University of Iowa students was randomly divided into a build- up condition group \((m=56)\) and a scale-down condition group \((n=59)\). The task for each subject was to build his or her own pizza from a menu of 12 ingredients. The build-up group was told that a basic cheese pizza costs \(\$ 5\) and that each extra ingredient would \(\operatorname{cost} 50\) cents. The scale-down group was told that a pizza with all 12 ingredients (ugh!!!) would cost \(\$ 11\) and that deleting an ingredient would save 50 cents. The article "A Tale of Two Pizzas: Building Up from a Basic Product Versus Scaling Down from a Fully Loaded Product" (Market. Lett., 2002: 335-344) reported that the mean number of ingredients selected by the scale-down group was significantly greater than the mean number for the build-up group: \(5.29\) versus \(2.71\). The calculated value of the appropriate \(t\) statistic was \(6.07\). Would you reject the null hypothesis of equality in favor of inequality at a significance level of \(.05\) ? \(.01 ? .001 ?\) Can you think of other products aside from pizza where one could build up or scale down? [Note: A separate experiment involved students from the University of Rome, but details were a bit different because there are typically not so many ingredient choices in Italy.]

A sample of 300 urban adult residents of a particular state revealed 63 who favored increasing the highway speed limit from 55 to \(65 \mathrm{mph}\), whereas a sample of 180 rural residents yielded 75 who favored the increase. Does this data indicate that the sentiment for increasing the speed limit is different for the two groups of residents? a. Test \(H_{0}: p_{1}=p_{2}\) versus \(H_{a}: p_{1} \neq p_{2}\) using \(\alpha=.05\), where \(p_{1}\) refers to the urban population. a b. If the true proportions favoring the increase are actually \(p_{1}=.20\) (urban) and \(p_{2}=.40\) (rural), what is the probability that \(H_{0}\) will be rejected using a level \(.05\) test with \(m=300, n=180 ?\)

A study seeks to compare hospitals based on the performance of their intensive care units. The dependent variable is the mortality ratio, the ratio of the number of deaths over the predicted number of deaths based on the condition of the patients. The comparison will be between hospitals with nurse staffing problems and hospitals without such problems. Assume, based on past experience, that the standard deviation of the mortality ratio will be around \(.2\) in both types of hospital. How many of each type of hospital should be included in the study in order to have both the type I and type II error probabilities be \(.05\), if the true difference of mean mortality ratio for the two types of hospital is .2? If we conclude that hospitals with nurse staffing problems have a higher mortality ratio, does this imply a causal relationship? Explain.
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