/*! 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 7 An economist wonders if corporat... [FREE SOLUTION] | 91Ó°ÊÓ

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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?

Short Answer

Expert verified
No significant difference in population variance between France and Germany; thus, corporate productivity volatility is similar.

Step by step solution

01

Define the Hypotheses

To test if there is a significant difference in the population variances of percentage yields for companies in France and Germany, we set up our hypotheses as follows: the null hypothesis is that the population variances are equal, i.e., \( H_0: \sigma^2_{France} = \sigma^2_{Germany} \), and the alternative hypothesis is that the population variances are not equal, i.e., \( H_a: \sigma^2_{France} eq \sigma^2_{Germany} \).
02

Identify the Test Statistic

To compare variances, we use an F-test. The test statistic is the ratio of the two sample variances. Let \( s^2_{1} \) be the sample variance for France and \( s^2_{2} \) be the sample variance for Germany. The F-statistic is calculated as: \[ F = \frac{s^2_{1}}{s^2_{2}} = \frac{2.044}{1.038}. \]
03

Calculate the F-statistic

Substitute the given variances into the formula to calculate the F-statistic: \[ F = \frac{2.044}{1.038} \approx 1.97. \]
04

Determine the Critical Value

At a 5% level of significance, this two-tailed test needs two critical F-values. We use degrees of freedom for numerator (France) as \( df_1 = n_1 - 1 = 20 - 1 = 19 \) and for the denominator (Germany) as \( df_2 = n_2 - 1 = 17 - 1 = 16 \). Using these, obtain the critical F-value from F-distribution tables or software for \( F_{19, 16} \).
05

Compare F-statistic to Critical Values

The critical F-values define the rejection regions for F-distribution at the given degrees of freedom. If the computed F-statistic falls outside these regions, we reject the null hypothesis. Assuming typical critical values, our F-statistic, \( 1.97 \), will not exceed these bounds, indicating we fail to reject \( H_0 \).
06

Draw the Conclusion

Since our computed F-statistic does not fall into the rejection region, we fail to reject the null hypothesis. There is insufficient evidence at the 5% significance level to claim a difference in variances between France and Germany's companies.

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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 statistical method used to make decisions about population parameters based on sample data. In our specific context, we aim to understand if there is a significant difference in the variability of productivity percentages (yields) between companies in France and Germany.

To begin, we formulate a null hypothesis (\(H_0\)): \(\sigma^2_{France} = \sigma^2_{Germany}\), which states that there is no difference in variances. Conversely, our alternative hypothesis (\(H_a\)) is \(\sigma^2_{France} eq \sigma^2_{Germany}\). This step helps set a clear criterion against which we will test our data's evidence.

In essence, hypothesis testing provides a framework to measure if observed data aligns with our initial assumptions or signals a departure from them.
Variance Analysis
Variance analysis involves the examination of why and how the spread or dispersion of data points around the mean differs. It is crucial when assessing the volatility or consistency of a particular dataset. Here, we look at the annual percentage yields of company assets.

A higher variance indicates that the data points are more spread out from the mean, implying higher volatility. On the other hand, a lower variance means most data points are closer to the mean, suggesting stability and less fluctuation.

For our company yield data, we've calculated the variances: for France, it is approximately 2.044, while for Germany, it is approximately 1.038. These values provide an initial glance into how yield variability might differ by country.
F-Test
The F-Test is a statistical test used to compare two variances to assess whether they come from populations with the same variance. It is particularly useful when we have data from two independent samples, as with the French and German companies.

To conduct an F-Test, we use the test statistic, which is the ratio of the two sample variances: \[ F = \frac{s^2_{France}}{s^2_{Germany}} = \frac{2.044}{1.038} \approx 1.97 \]This ratio tells us the proportionate difference in variance between the two groups.

If this F-value lies beyond the critical value threshold (taken from F-distribution tables based on degrees of freedom), it signifies a significant difference in variance, leading us to potentially reject the null hypothesis.
Level of Significance
The level of significance, often symbolized by \(\alpha\), is the probability threshold set by the researcher before conducting a hypothesis test. It determines the cut-off point for deciding whether to reject the null hypothesis.

In our exercise, we use a level of significance of 5% (or 0.05). This level implies that there is a 5% risk of concluding that there is a difference between the variances when, in fact, there is none. Thus, it controls the chance of making a Type I error (incorrectly rejecting a true null hypothesis).

By adhering to this fixed threshold, researchers ensure their decisions are made with a consistent standard of certainty, which, in this case, helps evaluate the variance in productivity across these countries reliably.

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Most popular questions from this chapter

A pathologist has been studying the frequency of bacterial colonies within the field of a microscope using samples of throat cultures from healthy adults. Long-term history indicates that there is an average of \(2.80\) bacteria colonies per field. Let \(r\) be a random variable that represents the number of bacteria colonies per field. Let \(O\) represent the number of observed bacteria colonies per field for throat cultures from healthy adults. A random sample of 100 healthy adults gave the following information. $$ \begin{array}{l|rrrrrr} \hline \boldsymbol{r} & 0 & 1 & 2 & 3 & 4 & 5 \text { or more } \\ \hline 0 & 12 & 15 & 29 & 18 & 19 & 7 \\ \hline \end{array} $$ (a) The pathologist wants to use a Poisson distribution to represent the probability of \(r\), the number of bacteria colonies per field. The Poisson distribution is $$ P(r)=\frac{e^{-\lambda} \lambda^{r}}{r !} $$ where \(\lambda=2.80\) is the average number of bacteria colonies per field. Compute \(P(r)\) for \(r=0,1,2,3,4\), and 5 or more. (b) Compute the expected number of colonies \(E=100 P(r)\) for \(r=0,1,2,3,4\) and 5 or more. (c) Compute the sample statistic \(\chi^{2}=\Sigma \frac{(O-E)^{2}}{E}\) and the degrees of freedom. (d) Test the statement that the Poisson distribution fits the sample data. Use a \(5 \%\) level of significance.

The following problem is based on information from an article by N. Keyfitz in the American Journal of Sociology (Vol. 53, pp. \(470-480\) ). Let \(x=\) age in years of a rural Quebec woman at the time of her first marriage. In the year 1941 , the population variance of \(x\) was approximately \(\sigma^{2}=5.1 .\) Suppose a recent study of age at first marriage for a random sample of 41 women in rural Quebec gave a sample variance \(s^{2}=3.3 .\) Use a \(5 \%\) level of significance to test the claim that the current variance is less than \(5.1 .\) Find a \(90 \%\) confidence interval for the population variance.

Does talking while walking slow you down? A study reported in the journal Physical Therapy (Vol. 72, No. 4 ) considered mean cadence (steps per minute) for subjects using no walking device, a standard walker, and a rolling walker. In addition, the cadence was measured when the subjects had to perform dual tasks. The second task was to respond vocally to a signal while walking. Cadence was measured for subjects who were just walking (using no device, a standard walker, or a rolling walker) and for subjects required to respond to a signal while walking. List the factors and the number of levels of each factor. How many cells are there in the data table?

In general, is the \(F\) distribution symmetrical? Can values of \(F\) be negative?

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.

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