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A random sample of companies in electric utilities (I), financial services (II), and food processing (III) gave the following information regarding annual profits per employee (units in thousands of dollars). $$ \begin{array}{ccc} \text { I } & \text { II } & \text { III } \\ 49.1 & 55.6 & 39.0 \\ 43.4 & 25.0 & 37.3 \\ 32.9 & 41.3 & 10.8 \\ 27.8 & 29.9 & 32.5 \\ 38.3 & 39.5 & 15.8 \\ 36.1 & & 42.6 \\ 20.2 & & \end{array} $$ Shall we reject or not reject the claim that there is no difference in population mean annual profits per employee in each of the three types of companies? Use a \(1 \%\) level of significance.

Short Answer

Expert verified
Reject \( H_0 \) if the computed F-statistic exceeds the critical F-value at \( 1\% \) significance.

Step by step solution

01

Define Hypotheses

To determine if there is a difference in the annual profits per employee among the three company types, we begin by establishing our hypotheses. The null hypothesis \( H_0 \) states that there are no differences in the population means: \( \mu_1 = \mu_2 = \mu_3 \). The alternative hypothesis \( H_a \) posits that at least one mean is different.
02

Choose the Significance Level

We will use a \(1\%\) level of significance, \( \alpha = 0.01 \). This indicates the risk of rejecting the null hypothesis when it is actually true. It also sets our critical value threshold for the test.
03

Organize the Data

Assemble the data provided into groups: Group I (electric utilities): 49.1, 43.4, 32.9, 27.8, 38.3, 36.1, 20.2; Group II (financial services): 55.6, 25.0, 41.3, 29.9, 39.5; Group III (food processing): 39.0, 37.3, 10.8, 32.5, 15.8, 42.6.
04

Conduct ANOVA Test

Use Analysis of Variance (ANOVA) to check for significant differences between group means. Calculate the overall mean, group means, sum of squares between (SSB), sum of squares within (SSW), and finally the F-statistic. Compare this against the critical F-value from the F-distribution table for \( \alpha = 0.01 \) with degrees of freedom df1 = 2 (groups - 1) and df2 = 15 (total observations - groups).
05

Calculate Test Statistic

First, compute the mean for each group and the total mean. Calculate SSB: sum of \( n_i(\bar{x}_i - \bar{x})^2 \) for each group. Then, calculate SSW: sum of each group's squared deviations from their mean. The F-statistic is \( F = \frac{(SSB/(k-1))}{(SSW/(N-k))} \).
06

Compare F-statistic to Critical Value

Determine the critical value for F at df1 = 2 and df2 = 15 from the F-distribution table at \( \alpha = 0.01 \). If the calculated F-statistic is greater than the critical F-value, reject the null hypothesis.
07

Decision

After calculating and comparing, if \( F_{calculated} > F_{critical} \), reject \( H_0 \), indicating there is a significant difference in mean profits per employee among the groups. If not, fail to reject \( H_0 \).

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Key Concepts

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

Hypothesis Testing
Hypothesis testing starts with forming two opposing hypotheses. The null hypothesis, denoted as \( H_0 \), asserts that there is no effect or no difference. In this exercise, \( H_0 \) claims that the mean annual profits per employee across all three company types (electric utilities, financial services, and food processing) are equal: \( \mu_1 = \mu_2 = \mu_3 \).

The alternative hypothesis, denoted as \( H_a \), suggests that there is a difference in at least one of the means. Essentially, it states that not all the group means are equal. This hypothesis is non-directional because it does not specify which mean is different, just that at least one is.

Understanding hypothesis testing helps us decide, based on statistical evidence, whether to reject \( H_0 \). We'll use data samples to calculate a test statistic, which will provide insight into whether the sample data could have come about under the assumption that the null hypothesis is true.
Significance Level
The significance level, \( \alpha \), represents the probability of rejecting the null hypothesis when it is true, also known as a Type I error. In this exercise, the significance level is set to 1%, or 0.01.

A lower significance level such as 1% indicates a more stringent criterion for rejecting \( H_0 \). This implies we demand stronger evidence from the data to claim a difference exists between the group means. By choosing a significance level, we balance the risk of finding a false positive.

The chosen \( \alpha \) determines the critical value in hypothesis testing. This is the threshold against which the test statistic is compared to decide whether to reject \( H_0 \). If the test statistic exceeds this critical value, \( H_0 \) is rejected in favor of \( H_a \).
F-distribution
The F-distribution is critical in ANOVA tests, which compare variances to determine if means are different. It is a family of distributions defined by two sets of degrees of freedom: numerator (df1) and denominator (df2).

In this exercise, the F-statistic is calculated from the variance between the group means (between-group variance) and within the groups (within-group variance). The formula for the F-statistic is:
  • \[ F = \frac{ \frac{SSB}{k-1} }{ \frac{SSW}{N-k} } \]
Where \( SSB \) is the sum of squares between groups, \( SSW \) is the sum of squares within groups, \( k \) is the number of groups, and \( N \) is the total number of observations.

After computing the F-statistic, it is compared to the critical value from the F-distribution table, identified by \( \alpha = 0.01 \), with df1 = 2 (number of groups minus 1) and df2 = 15 (total observations minus number of groups). This comparison helps determine if the observed variances are greater than could be expected just by chance.

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Most popular questions from this chapter

Jim Mead is a veterinarian who visits a Vermont farm to examine prize bulls. In order to examine a bull, Jim first gives the animal a tranquilizer shot. The effect of the shot is supposed to last an average of 65 minutes, and it usually does. However, Jim sometimes gets chased out of the pasture by a bull that recovers too soon, and other times he becomes worried about prize bulls that take too long to recover. By reading journals, Jim has found that the tranquilizer should have a mean duration time of 65 minutes, with a standard deviation of 15 minutes. A random sample of 10 of Jim's bulls had a mean tranquilized duration time of close to 65 minutes but a standard deviation of 24 minutes. At the \(1 \%\) level of significance, is Jim justified in the claim that the variance is larger than that stated in his journal? Find a \(95 \%\) confidence interval for the population standard deviation.

Investing: Mutual Funds How reliable are mutual funds that invest in bonds? Again, this depends on the bond fund you buy (see reference in Problem 9). A random sample of annual percentage returns for mutual funds holding shortterm U.S. government bonds is shown below. \(\begin{array}{lllllll}4.6 & 4.7 & 1.9 & 9.3 & -0.8 & 4.1 & 10.5\end{array}\) \(\begin{array}{llllll}4.2 & 3.5 & 3.9 & 9.8 & -1.2 & 7.3\end{array}\) Use a calculator to verify that \(s^{2}=13.59\) for the preceding data. A random sample of annual percentage returns for mutual funds holding intermediate-term corporate bonds is shown below. \(\begin{array}{rrrrrrrr}-0.8 & 3.6 & 20.2 & 7.8 & -0.4 & 18.8 & -3.4 & 10.5 \\\ 8.0 & -0.9 & 2.6 & -6.5 & 14.9 & 8.2 & 18.8 & 14.2\end{array}\) Use a calculator to verify that \(s^{2} \approx 72.06\) for returns from mutual funds holding intermediate-term corporate bonds. Use \(\alpha=0.05\) to test the claim that the population variance for annual percentage returns of mutual funds holding short-term government bonds is different from the population variance for mutual funds holding intermediate- term corporate bonds. How could your test conclusion relate to the question of reliability of returns for each type of mutual fund?

For chi-square distributions, as the number of degrees of freedom increases, does any skewness increase or decrease? Do chi-square distributions become more symmetric (and normal) as the number of degrees of freedom becomes larger and larger?

The following problem is based on information taken from Academe, Bulletin of the American Association of University Professors. Let \(x\) represent the average annual salary of college and university professors (in thousands of dollars) in the United States. For all colleges and universities in the United States, the population variance of \(x\) is approximately \(\sigma^{2}=47.1\). However, a random sample of 15 colleges and universities in Kansas showed that \(x\) has a sample variance \(s^{2}=83.2 .\) Use a \(5 \%\) level of significance to test the claim that the variance for colleges and universities in Kansas is greater than 47.1. Find a \(95 \%\) confidence interval for the population variance.

For the study regarding mean cadence (see Problem 1), two-way ANOVA was used. Recall that the two factors were walking device (none, standard walker, rolling walker) and dual task (being required to respond vocally to a signal or no dual task required). Results of two-way ANOVA showed that there was no evidence of interaction between the factors. However, according to the article, "The ANOVA conducted on the cadence data revealed a main effect of walking device." When the hypothesis regarding no difference in mean cadence according to which, if any, walking device was used, the sample \(F\) was \(30.94\), with \(d . f \cdot \mathrm{N}=2\) and \(d . f \cdot D=18\). Further, the \(P\) -value for the result was reported to be less than \(0.01\). From this information, what is the conclusion regarding any difference in mean cadence according to the factor "walking device used"?

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