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Some computer output for an analysis of variance test to compare means is given. (a) How many groups are there? (b) State the null and alternative hypotheses. (c) What is the p-value? (d) Give the conclusion of the test, using a \(5 \%\) significance level. \(\begin{array}{lrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } \\ \text { Groups } & 3 & 450.0 & 150.0 & 0.75 \\ \text { Error } & 16 & 3200.0 & 200.0 & \\ \text { Total } & 19 & 3650.0 & & \end{array}\)

Short Answer

Expert verified
There are 4 groups. The null hypothesis suggests that all group means are the same while the alternative poses at least one group mean is different. The p-value cannot be directly derived from the given ANOVA table while without it, we cannot conclude the test at the 5% significance level.

Step by step solution

01

Number of groups

The number of groups is determined by the degrees of freedom (DF) for the 'Groups' row in the ANOVA table. The given number under the 'DF' heading for 'Groups' represents the number of groups. Here, the DF for Groups is 3, implying that there are 4 groups in total.
02

Hypothesis setting

In ANOVA analysis, the null hypothesis (H0) posits that all group means are equal, while the alternative hypothesis (HA) suggests that at least one group mean is different. In this case: \nNull Hypothesis, H0: µ1= µ2 = µ3 = µ4, \nAlternative Hypothesis, HA: At least one group mean (µi) is different.
03

Calculating the p-value

In typical ANOVA table output, the p-value is directly given. However, in the given table, an F statistic is provided instead of a p-value. To find the p-value from an F statistic, you have to refer to an F-distribution table or use statistical software. Without this supplementary material or software, we cannot directly find the p-value.
04

Test conclusion

Without a p-value or F-critical value, we can't make concrete conclusions about the test at a \(5\%\) significance level. Usually, if the p-value is less than the significance level (0.05), we reject the null hypothesis.

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Key Concepts

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

Null Hypothesis in ANOVA
Understanding the null hypothesis in the context of Analysis of Variance (ANOVA) is crucial for interpreting the results of an experiment comparing multiple groups. In ANOVA, the null hypothesis states that there are no significant differences between the group means, suggesting that any observed variation is due to random chance. For example, if you’re comparing the effectiveness of four different diets on weight loss, the null hypothesis would be that all diets result in the same average weight loss.

In the context of our exercise, the null hypothesis posits that the means of the four groups are equal (Null Hypothesis, H0: µ1= µ2 = µ3 = µ4). The alternative hypothesis, on the other hand, proposes that at least one group's mean is significantly different from the others (Alternative Hypothesis, HA: At least one group mean (µi) is different).

If the null hypothesis is true, it implies that any differences observed in our data are likely due to sampling error or random fluctuations, rather than a real effect. Testing this hypothesis allows researchers to make inferences about the populations from which their samples were drawn.
ANOVA p-value
The p-value in ANOVA analysis represents the probability of obtaining results at least as extreme as the ones observed during the test, assuming that the null hypothesis is true. It is a crucial piece of information for deciding whether to reject the null hypothesis. The smaller the p-value, the stronger the evidence against the null hypothesis.

In our exercise example, the p-value is not directly provided but instead, we’re given an F-statistic. Typically, you would compare the F-statistic to a critical value from an F-distribution table or use a statistical software to obtain the p-value. If the F-statistic corresponds to a p-value less than the chosen significance level (often set at 0.05), we would reject the null hypothesis. This signifies that there is a statistically significant difference between the group means.

Without the corresponding p-value or critical F-value, as in the step-by-step solution provided, we cannot conclusively determine the outcome of the ANOVA test for our exercise. Nonetheless, understanding the role of the p-value is imperative for interpreting ANOVA results.
Degrees of Freedom (DF)
Degrees of freedom (DF) are a vital component in statistical tests, including ANOVA. They represent the number of independent values that can vary within the data set. For an ANOVA test, degrees of freedom are calculated for both the groups and the error term.

The DF for groups is one less than the number of groups being compared. It accounts for the variability between the group means. In the example exercise, the DF for groups is given as 3, which indicates there are 4 groups (DF for Groups = Number of Groups - 1). The DF for the error, similarly, is the total number of observations minus the number of groups, representing the variability within the groups.

Determining the degrees of freedom is essential for finding the appropriate critical values in statistical tables, which in turn helps us evaluate the test statistic. This metric lays the groundwork for calculating the variability in the data, and subsequently, affects the robustness and accuracy of the ANOVA results.

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Most popular questions from this chapter

Exercises 8.46 to 8.52 refer to the data with analysis shown in the following computer output: \(\begin{array}{lrrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } & \\ \text { A } & 6 & 86.833 & 5.231 & \\ \text { B } & 6 & 76.167 & 6.555 & \\ \text { C } & 6 & 80.000 & 9.230 & \\ \text { D } & 6 & 69.333 & 6.154 & \\ \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 3 & 962.8 & 320.9 & 6.64 & 0.003 \\ \text { Error } & 20 & 967.0 & 48.3 & & \\ \text { Total } & 23 & 1929.8 & & & \end{array}\) Is there evidence for a difference in the population means of the four groups? Justify your answer using specific value(s) from the output.

Some computer output for an analysis of variance test to compare means is given. (a) How many groups are there? (b) State the null and alternative hypotheses. (c) What is the p-value? (d) Give the conclusion of the test, using a \(5 \%\) significance level. \(\begin{array}{lrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } \\ \text { Groups } & 4 & 1200.0 & 300.0 & 5.71 \\ \text { Error } & 35 & 1837.5 & 52.5 & \\ \text { Total } & 39 & 3037.5 & & \end{array}\)

Exercises 8.41 to 8.45 refer to the data with analysis shown in the following computer output: \(\begin{array}{lrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } \\ \text { A } & 5 & 10.200 & 2.864 \\ \text { B } & 5 & 16.800 & 2.168 \\ \text { C } & 5 & 10.800 & 2.387\end{array}\) \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 2 & 133.20 & 66.60 & 10.74 & 0.002 \\ \text { Error } & 12 & 74.40 & 6.20 & & \\ \text { Total } & 14 & 207.60 & & & \end{array}\) Find a \(95 \%\) confidence interval for the mean of population \(\mathrm{A}\).

Exercises 8.41 to 8.45 refer to the data with analysis shown in the following computer output: \(\begin{array}{lrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } \\ \text { A } & 5 & 10.200 & 2.864 \\ \text { B } & 5 & 16.800 & 2.168 \\ \text { C } & 5 & 10.800 & 2.387\end{array}\) \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 2 & 133.20 & 66.60 & 10.74 & 0.002 \\ \text { Error } & 12 & 74.40 & 6.20 & & \\ \text { Total } & 14 & 207.60 & & & \end{array}\) Test for a difference in population means between groups \(A\) and \(C\). Show all details of the test.

Data 4.1 introduces a study in mice showing that even low-level light at night can interfere with normal eating and sleeping cycles. In the full study, mice were randomly assigned to live in one of three light conditions: LD had a standard light/dark cycle, LL had bright light all the time, and DM had dim light when there normally would have been darkness. Exercises 8.28 to 8.34 in Section 8.1 show that the groups had significantly different weight gain and time of calorie consumption. In Exercises 8.55 and \(8.56,\) we revisit these significant differences. When Calories Are Consumed Researchers hypothesized that the increased weight gain seen in mice with light at night might be caused by when the mice are eating. Computer output for the percentage of food consumed during the day (when mice would normally be sleeping) for each of the three light conditions is shown, along with the relevant ANOVA output. Which light conditions give significantly different mean percentage of calories consumed during the day? \(\begin{array}{lrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } \\ \text { DM } & 10 & 55.516 & 10.881 \\ \text { LD } & 9 & 36.485 & 7.978 \\ \text { LL } & 9 & 76.573 & 9.646\end{array}\) One-way ANOVA: Day/night consumption versus Light \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Light } & 2 & 7238.4 & 3619.2 & 39.01 & 0.000 \\ \text { Error } & 25 & 2319.3 & 92.8 & & \\ \text { Total } & 27 & 9557.7 & & & \end{array}\)

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