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Chapter 10: Problem 6

Two plots at Rothamsted Experimental Station (see reference in Problem 5 ) were studied for production of wheat straw. For a random sample of years, the annual wheat straw production (in pounds) from one plot was as follows: \(\begin{array}{llllll}6.17 & 6.05 & 5.89 & 5.94 & 7.31 & 7.18\end{array}\) \(\begin{array}{lllll}7.06 & 5.79 & 6.24 & 5.91 & 6.14\end{array}\) Use a calculator to verify that, for the preceding data, \(s^{2} \approx 0.318\). Another random sample of years for a second plot gave the following annual wheat straw production (in pounds): \(\begin{array}{llllllll}6.85 & 7.71 & 8.23 & 6.01 & 7.22 & 5.58 & 5.47 & 5.86\end{array}\) Use a calculator to verify that, for these data, \(s^{2}=1.078\). Test the claim that there is a difference (either way) in the population variance of wheat straw production for these two plots. Use a \(5 \%\) level of significance.

Short Answer

Expert verified
There is no significant difference in population variances between the two plots at \(5\%\) significance level.

Step by step solution

01

Define the Hypotheses

We will conduct an F-test to compare variances. Define the null hypothesis as \(H_0: \sigma_1^2 = \sigma_2^2\) (the variances are equal) and the alternative hypothesis as \(H_a: \sigma_1^2 eq \sigma_2^2\) (the variances are different).
02

Calculate the Test Statistic

The formula for the F-test statistic is \( F = \frac{s_1^2}{s_2^2} \), where \(s_1^2\) and \(s_2^2\) are the sample variances of the two plots. Here, \(s_1^2 = 0.318\) and \(s_2^2 = 1.078\). Thus, \( F = \frac{0.318}{1.078} \approx 0.295\).
03

Determine the Critical Value

For a significance level of \(\alpha = 0.05\), we need to find the critical F-values from the F-distribution table. For sample sizes \(n_1 = 11\) and \(n_2 = 8\), the degrees of freedom are \( df_1 = 10\) and \( df_2 = 7\). Look up these degrees of freedom in the F-distribution table to find \( F_{critical} \approx 3.73 \) for two-tailed test.
04

Make the Decision

Compare the calculated F-statistic to the critical values. Since the calculated F-statistic \(0.295\) is less than \(1/3.73\) (the inverse of the critical value), we do not have enough evidence to reject the null hypothesis.
05

Conclusion

Since the F-statistic does not exceed the critical value range, we fail to reject the null hypothesis at \(\alpha = 0.05\). This indicates that there is not sufficient evidence to support the claim that there is a difference in population variances for the wheat straw production between the two plots.

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Key Concepts

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

F-test
In statistics, the F-test is a way to compare the variances of two populations to see if they are significantly different. It involves calculating an F-statistic based on the sample variances - the measure of data dispersion, from two groups.

For the F-test:
  • Define your null hypothesis (\(H_0\)) that the two variances are equal.
  • The alternative hypothesis (\(H_a\)) suggests that the variances are different.
  • The formula: \( F = \frac{s_1^2}{s_2^2} \) is where \(s_1^2\) and \(s_2^2\) are sample variances.
The test statistic, \( F \), is compared against critical values from the F-distribution at a chosen significance level, such as \(\alpha = 0.05\). If the calculated \( F \) is less than the critical value, you fail to reject the null hypothesis, meaning there's no significant difference between variances.
Hypothesis Testing
Hypothesis testing is a method used to decide whether enough evidence exists in a sample to infer a certain condition for a whole population. In our exercise, we're testing the hypothesis about population variances between wheat straw production in two plots.

The steps include:
  • Formulating the null hypothesis (\(H_0\)), which assumes no effect or difference.
  • The alternative hypothesis (\(H_a\)) claims the opposite of \(H_0\).
  • A significance level (\(\alpha\)), often set at 0.05, determines the probability of rejecting \(H_0\) if it's true.
  • Calculate a test statistic from the sample data.
  • Determine critical values from statistical tables for the test.
  • Compare the test statistic with critical values to make a decision to reject or fail to reject the null hypothesis.
This systematic approach helps ensure conclusions drawn from sample data are valid.
Sample Variance
Sample variance is a measure of how much the data in a sample vary from the mean of the sample, indicating how spread out the data points are. It is calculated using the formula:

\[ s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1} \]where:
  • \(x_i\) represents each data point.
  • \(\bar{x}\) is the sample mean.
  • \(n\) is the number of observations in the sample.
The sample variance gives an estimation of the population variance, especially when it's impractical to survey an entire population. In our case, the sample variance helps compare variability in wheat straw production across plots.
Population Variance
Population variance is a statistical measure that represents the average squared deviation of each data point in a population from the population mean. It provides insight into how data points are spread in the entire set rather than just a sample.

The formula for population variance (\(\sigma^2\)) is:\[ \sigma^2 = \frac{\sum (X_i - \mu)^2}{N} \]where:
  • \(X_i\) is an individual data point.
  • \(\mu\) is the population mean.
  • \(N\) represents the number of observations in the population.
Calculating the population variance gives a more comprehensive picture of data variability. In our comparison of wheat straw production, analyzing variances determines if one plot exhibits differing production patterns from another. Understanding population variance is crucial when making decisions about large groups based on smaller sampled subsets.

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Most popular questions from this chapter

The following problem is based on information taken from Accidents in North American Mountaineering (jointly published by The American Alpine Club and The Alpine Club of Canada). Let \(x\) represent the number of mountain climbers killed each year. The long-term variance of \(x\) is approximately \(\sigma^{2}=136.2 .\) Suppose that for the past 8 years, the variance has been \(s^{2}=115.1 .\) Use a \(1 \%\) level of significance to test the claim that the recent variance for number of mountain climber deaths is less than \(136.2 .\) Find a \(90 \%\) confidence interval for the population variance.

A sociologist studying New York City ethnic groups wants to determine if there is a difference in income for immigrants from four different countries during their first year in the city. She obtained the data in the following table from a random sample of immigrants from these countries (incomes in thousands of dollars). Use a \(0.05\) level of significance to test the claim that there is no difference in the earnings of immigrants from the four different countries. $$ \begin{array}{rrcr} \text { Country I } & \text { Country II } & \text { Country III } & \text { Country IV } \\ 12.7 & 8.3 & 20.3 & 17.2 \\ 9.2 & 17.2 & 16.6 & 8.8 \\ 10.9 & 19.1 & 22.7 & 14.7 \\ 8.9 & 10.3 & 25.2 & 21.3 \\ 16.4 & & 19.9 & 19.8 \end{array} $$

An economist wonders if corporate productivity in some countries is more volatile than that in other countries. One measure of a company's productivity is annual percentage yield based on total company assets. Data for this problem are based on information taken from Forbes Top Companies, edited by J. T. Davis. A random sample of leading companies in France gave the following percentage yields based on assets: \(\begin{array}{lllllllllll}4.4 & 5.2 & 3.7 & 3.1 & 2.5 & 3.5 & 2.8 & 4.4 & 5.7 & 3.4 & 4.1\end{array}\) \(\begin{array}{llllllllll}6.8 & 2.9 & 3.2 & 7.2 & 6.5 & 5.0 & 3.3 & 2.8 & 2.5 & 4.5\end{array}\) Use a calculator to verify that \(s^{2} \approx 2.044\) for this sample of French companies. Another random sample of leading companies in Germany gave the following percentage yields based on assets: \(\begin{array}{ccccccccc}3.0 & 3.6 & 3.7 & 4.5 & 5.1 & 5.5 & 5.0 & 5.4 & 3.2\end{array}\) \(\begin{array}{lllllllll}3.5 & 3.7 & 2.6 & 2.8 & 3.0 & 3.0 & 2.2 & 4.7 & 3.2\end{array}\) Use a calculator to verify that \(s^{2} \approx 1.038\) for this sample of German companies. Test the claim that there is a difference (either way) in the population variance of percentage yields for leading companies in France and Germany. Use a \(5 \%\) level of significance. How could your test conclusion relate to the economist's question regarding volatility (data spread) of corporate productivity of large companies in France compared with large companies in Germany?

You don't need to be rich to buy a few shares in a mutual fund. The question is, "How reliable are mutual funds as investments?" That depends on the type of fund you buy. The following data are based on information taken from Morningstar, a mutual fund guide available in most libraries. A random sample of percentage annual returns for mutual funds holding stocks in aggressive- growth small companies is shown below. \(\begin{array}{lllllllllll}-1.8 & 14.3 & 41.5 & 17.2 & -16.8 & 4.4 & 32.6 & -7.3 & 16.2 & 2.8 & 34.3\end{array}\) \(\begin{array}{llllllllll}-10.6 & 8.4 & -7.0 & -2.3 & -18.5 & 2.5 .0 & -9.8 & -7.8 & -24.6 & 22.8\end{array}\) Use a calculator to verify that \(s^{2}=348.43\) for the sample of aggressive- growth small company funds. Another random sample of percentage annual returns for mutual funds holding value (i.e., market underpriced) stocks in large companies is shown below. \(\begin{array}{llllllllllll}16.2 & 0.3 & 7.8 & -1.6 & -3.8 & 19.4 & -2.5 & 15.9 & 32.6 & 22.1 & 3.4\end{array}\) \(\begin{array}{lllllllll}-0.5 & -8.3 & 2.5 .8 & -4.1 & 14.6 & 6.5 & 18.0 & 21.0 & 0.2 & -1.6\end{array}\) Use a calculator to verify that \(s^{2} \approx 137.31\) for value stocks in large companies. Test the claim that the population variance for mutual funds holding aggressive-growth small stocks is larger than the population variance for mutual funds holding value stocks in large companies. Use a \(5 \%\) level of significance. How could your test conclusion relate to the question of reliability of returns for each type of mutual fund?

A researcher forms three blocks of students interested in taking a history course. The groups are based on grade point average (GPA). The first group consists of students with a GPA less than \(2.5\), the second group consists of students with a GPA between \(2.5\) and \(3.1\), and the last group consists of students with a GPA greater than 3.1. History courses are taught in three ways: traditional lecture, small-group collaborative method, and independent study. The researcher randomly assigns 10 students from each block to sections of history taught each of the three ways. Sections for each teaching style then have 10 students from each block. The researcher records the scores on a common course final examination administered to each student. Draw a flow chart showing the design of this experiment. Does the design fit the model for randomized block design?
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