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Chapter 8: Problem 12

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. \(\begin{array}{lrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } \\ \text { Groups } & 4 & 1200.0 & 300.0 & 5.71 \\ \text { Error } & 35 & 1837.5 & 52.5 & \\ \text { Total } & 39 & 3037.5 & & \end{array}\)

Short Answer

Expert verified
There are 4 groups. Null hypothesis: all group means are equal, alternative hypothesis: at least one group mean is different. The p-value cannot be determined from the information given. With an F-value of 5.71, there is a suggestion of possible variance among group means; however, without the p-value, a conclusive decision can't be drawn for the \(5 \% \) significance level.

Step by step solution

01

Identify Number of Groups

One can directly look at the 'DF' (Degrees of Freedom) value corresponding to 'Groups' in the table. This represents the number of groups in our analysis. In this case, it's '4' which means we have 4 groups.
02

Set up null and alternative hypothesis

In an Analysis of Variance test, the null hypothesis \(H_o \) claims that all group means are the same. The alternative hypothesis \(H_1 \) states that at least one group mean is different. Thus, we can write: \n- Null Hypothesis \((H_o) \): \(\mu_1 = \mu_2 = \mu_3 = \mu_4\)\n- Alternative Hypothesis \((H_1) \): at least one \(\mu_i\) is different, where \(i\) = 1, 2, 3, 4.
03

Calculate P-value

Unluckily, this table contains no information regarding the p-value. Typically, the p-value is calculated using the F-value and the relevant degrees of freedom. Without the proper statistical tables or software, we cannot calculate the p-value ourselves in here.
04

Determining Test Conclusion

Again, without the calculated p-value, it is impossible to conclusively say if we accept or reject the null hypothesis at a \(5 \% \) significance level. A general guideline is that if the F-value is larger that 1, it means that the between-group variance is higher than the within-group, suggesting could consider rejecting the null hypothesis. But it's an indication, the final conclusion is based on the comparison with the significance level. In this case, the F-value is 5.71.

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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 powerful statistical tool used to make decisions based on data analysis. It's like putting a statement to the test using evidence from data. In the context of Analysis of Variance (ANOVA), we form two hypotheses:
  • **Null Hypothesis ( $H_0$ )**: This suggests that there is no difference between the group means. It assumes any observed differences are due to random chance.
  • **Alternative Hypothesis ( $H_1$ )**: This indicates that at least one group mean is different. It proposes that not all group means are equal, implying that the differences are significant and not just by chance.
The process starts by collecting data, calculating a test statistic (like the F-value in ANOVA), and comparing it to a critical value determined by a significance level, usually 5% (or 0.05). The goal is to determine whether the observed data fits well with the null hypothesis or if the alternative hypothesis provides a better explanation.
Statistical Significance
Statistical significance is a key concept in hypothesis testing. It answers the question, 'Are the observed results meaningful or just due to random variation?' By convention, a 5% significance level is commonly used, denoted as 0.05. What that essentially means is, you're allowing a 5% risk of concluding that a difference exists when there is none. If a result is statistically significant, it means the observed data is unlikely to have occurred under the null hypothesis, or in simpler terms, the result is not purely due to random chance. This is often assessed using a p-value:
  • **P-value**: This helps measure the strength of evidence against the null hypothesis. If the p-value is less than or equal to the significance level (e.g., 0.05), we reject the null hypothesis in favor of the alternative.
In ANOVA, you often see significance expressed by comparing the F-value to critical values that take into account the degrees of freedom. A suitable F-value suggests a meaningful difference between means, hence, statistical significance.
Degrees of Freedom
Degrees of freedom (DF) is a fundamental concept in statistics, representing the number of independent values or observations in a data set that are free to vary when estimating statistical parameters. In ANOVA, degrees of freedom are used in calculating the F-statistic and consist of two components:
  • **Between-group degrees of freedom**: Calculated as the number of groups minus one ( $k-1$ , where $k$ is the number of groups). It represents the number of ways the group averages can vary.
  • **Within-group degrees of freedom**: Calculated as the total number of samples minus the number of groups ( $N-k$ , where $N$ is the total number of observations). It accounts for individual variability within each group.
These degrees of freedom are essential for determining critical values from statistical tables and help in the interpretation of the F-value in the context of the ANOVA test. They provide a frame of reference in assessing how likely the observed variability is to occur under the null hypothesis.

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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: \(\begin{array}{lrrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } & \\ \text { A } & 6 & 86.833 & 5.231 & \\ \text { B } & 6 & 76.167 & 6.555 & \\ \text { C } & 6 & 80.000 & 9.230 & \\ \text { D } & 6 & 69.333 & 6.154 & \\ \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 3 & 962.8 & 320.9 & 6.64 & 0.003 \\ \text { Error } & 20 & 967.0 & 48.3 & & \\ \text { Total } & 23 & 1929.8 & & & \end{array}\) Find a \(99 \%\) confidence interval for the mean of population \(\mathrm{A}\). Is 90 a plausible value for the population mean of group \(\mathrm{A}\) ?

Exercises 8.41 to 8.45 refer to the data with analysis shown in the following computer output: \(\begin{array}{lrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } \\ \text { A } & 5 & 10.200 & 2.864 \\ \text { B } & 5 & 16.800 & 2.168 \\ \text { C } & 5 & 10.800 & 2.387\end{array}\) \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 2 & 133.20 & 66.60 & 10.74 & 0.002 \\ \text { Error } & 12 & 74.40 & 6.20 & & \\ \text { Total } & 14 & 207.60 & & & \end{array}\) Is there sufficient evidence of a difference in the population means of the three groups? Justify your answer using specific value(s) from the output.

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 \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Light } & 2 & 7238.4 & 3619.2 & 39.01 & 0.000 \\ \text { Error } & 25 & 2319.3 & 92.8 & & \\ \text { Total } & 27 & 9557.7 & & & \end{array}\)

We give sample sizes for the groups in a dataset and an outline of an analysis of variance table with some information on the sums of squares. Fill in the missing parts of the table. What is the value of the F-test statistic? Three groups with \(n_{1}=5, n_{2}=5,\) and \(n_{3}=5\). ANOVA table includes:$$ \begin{array}{|l|l|c|l|l|} \hline \text { Source } & \text { df } & \text { SS } & \text { MS } & \text { F-statistic } \\ \hline \text { Groups } & & 120 & & \\ \hline \text { Error } & & 282 & & \\ \hline \text { Total } & & 402 & & \\ \hline \end{array} $$

Exercises 8.46 to 8.52 refer to the data with analysis shown in the following computer output: \(\begin{array}{lrrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } & \\ \text { A } & 6 & 86.833 & 5.231 & \\ \text { B } & 6 & 76.167 & 6.555 & \\ \text { C } & 6 & 80.000 & 9.230 & \\ \text { D } & 6 & 69.333 & 6.154 & \\ \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 3 & 962.8 & 320.9 & 6.64 & 0.003 \\ \text { Error } & 20 & 967.0 & 48.3 & & \\ \text { Total } & 23 & 1929.8 & & & \end{array}\) Test for a difference in population means between groups \(\mathrm{B}\) and \(\mathrm{D} .\) Show all details of the test.
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