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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. \(\begin{array}{lrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } \\ \text { Groups } & 2 & 540.0 & 270.0 & 8.60 \\ \text { Error } & 27 & 847.8 & 31.4 & \\ \text { Total } & 29 & 1387.8 & & \end{array}\)

Short Answer

Expert verified
The number of groups is 2. The null hypothesis is that all group means are equal and the alternative hypothesis is that at least one group mean differs. The p-value is not provided in the table, so can't be determined. The conclusion of the test based on a 5% significance level isn't available due to absence of the p-value.

Step by step solution

01

Identify the Number of Groups

Looking at the provided output table for the ANOVA test, there is a row labeled 'Groups' with a value '2' under the 'DF' (or Degree of Freedom) column. This indicates that there are 2 groups or treatments.
02

State the Null and Alternative Hypotheses

In an ANOVA test, the null hypothesis states that all group means are equal, while the alternative hypothesis posits that at least one group mean is different. So, \(H_0: \mu_1 = \mu_2\) and \(H_a: \mu_1 \neq \mu_2\) (where \( \mu_1\) and \( \mu_2\) are the means of the first and second group respectively).
03

Determine the p-value

The ANOVA test table does not provide the p-value directly; it is typically calculated using the F statistic and the degrees of freedom for both the groups and error (except for some specific software outputs where p-value could be included). Since the p-value isn't mentioned here, we can not provide it directly. In general, software like R and Excel can be used to compute the p-value.
04

Conclude the Test

Based on a 5% significance level, if the determined p-value is less than 0.05, we would reject the null hypothesis, indicating there's evidence that at least one group mean is different. However, since the p-value isn't provided, the conclusion on this is not available.

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Key Concepts

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

Degrees of Freedom
Understanding degrees of freedom (DF) is crucial when working with statistical tests like ANOVA (Analysis of Variance). Degrees of freedom refer to the number of values in a calculation that are free to vary. Put simply, they're the number of 'pieces' of information that go into the estimate of a parameter.

For an ANOVA test, degrees of freedom are partitioned into two parts: one for the groups and one for the error or within groups. In the case of our exercise, with 'Groups' DF being 2, this implies that there were 3 groups being compared (since DF is one less than the number of groups). The 'Error' DF, which in this case is 27, represents the total number of observations minus the number of groups.

Why does this matter? The degrees of freedom affect the shape of the F-distribution used to calculate the p-value, and thus influence the outcome of an ANOVA test.
Null and Alternative Hypotheses
In any hypothesis test, including ANOVA, two competing hypotheses are formulated: the null hypothesis (denoted as H0) and the alternative hypothesis (denoted as Ha).

The null hypothesis is a statement of no effect or no difference, and it's what we assume to be true before collecting any data. For an ANOVA test, the null hypothesis states that all group means are equal. In mathematical terms, it's expressed as H0: µ1 = µ2 = ... = µk, where µ represents the group mean and k is the number of groups.

The alternative hypothesis, on the other hand, posits that there's at least one significant difference among the group means. It's denoted as Ha and often expressed as Ha: Not all µ's are equal. Hypothesis testing is all about deciding whether data provides enough evidence to reject the null hypothesis in favor of the alternative hypothesis.
P-value
The p-value is a bedrock concept in statistical hypothesis testing. It represents the probability of observing the data, or something more extreme, assuming that the null hypothesis is true. A low p-value suggests that the observed data is unlikely under the null hypothesis, thus providing evidence against the null hypothesis.

In the context of our ANOVA exercise, the F-statistic of 8.60 would be used along with degrees of freedom (2 for groups, 27 for error) to calculate the p-value. Typically, this is done with statistical software. A p-value less than the chosen significance level would mean rejecting the null hypothesis, suggesting at least one group mean differs from the others.

It's important to note that the p-value is not the probability that the null hypothesis is true, nor is it the probability that the alternative hypothesis is true. It is merely an indicator of how compatible the collected data is with the null hypothesis.
Significance Level
The significance level, often denoted by the Greek letter α (alpha), is a threshold predetermined by the researcher to decide whether to reject the null hypothesis. It's a measure of how much evidence you require before acting as if the null hypothesis is false. The most common significance level used is 0.05.

In our ANOVA example, we use a significance level of 5%. If the p-value calculated from the F-statistic and degrees of freedom is less than 0.05, we conclude that there's a statistically significant difference between at least two group means, thus rejecting the null hypothesis.

However, setting a significance level is a subjective decision that balances the risk of making a Type I error (falsely rejecting the null hypothesis) against making a Type II error (not detecting a real effect). The lower the significance level, the less chance of a Type I error, but the greater the risk of failing to detect a real effect (increasing Type II error).

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Most popular questions from this chapter

A recent study \(^{2}\) examines the impact of a mother's voice on stress levels in young girls. The study included 68 girls ages 7 to 12 who reported good relationships with their mothers. Each girl gave a speech and then solved mental arithmetic problems in front of strangers. Cortisol levels in saliva were measured for all girls and were high, indicating that the girls felt a high level of stress from these activities. (Cortisol is a stress hormone and higher levels indicate greater stress.) After the stress-inducing activities, the girls were randomly divided into four equal-sized groups: one group talked to their mothers in person, one group talked to their mothers on the phone, one group sent and received text messages with their mothers, and one group had no contact with their mothers. Cortisol levels were measured before and after the interaction with mothers and the change in the cortisol level was recorded for each girl. (a) What are the two main variables in this study? Identify each as categorical or quantitative. (b) Is this an experiment or an observational study? (c) The researchers are testing to see if there is a difference in the change in cortisol level depending on the type of interaction with mom. What are the null and alternative hypotheses? Define any parameters used. (d) What are the total degrees of freedom? The \(d f\) for groups? The \(d f\) for error? (e) The results of the study show that hearing mother's voice was important in reducing stress levels. Girls who talk to their mother in person or on the phone show decreases in cortisol significantly greater, at the \(5 \%\) level, than girls who text with their mothers or have no contact with their mothers. There was not a difference between in person and on the phone and there was not a difference between texting and no contact. Was the p-value of the original ANOVA test above or below \(0.05 ?\)

More on Exercise and Stress Exercise 6.219 on page 465 introduces a study showing that exercise appears to offer some resiliency against stress, and Exercise 8.19 on page 556 follows up on this introduction. In the study, mice were randomly assigned to live in an enriched environment (EE), a standard environment (SE), or an impoverished environment (IE) for several weeks. Only the enriched environment provided opportunities for exercise. Half the mice then remained in their home cage \((\mathrm{HC})\) as control groups while half were subjected to stress (SD). The researchers were interested in how resilient the mice were in recovering from the stress. One measure of mouse anxiety is amount of time hiding in a dark compartment, with mice who are more anxious spending more time in darkness. The amount of time (in seconds) spent in darkness during one trial is recorded for all the mice and the means and the results of the ANOVA analysis are shown. There are eight mice in each of the six groups. \(\begin{array}{l}\text { Group: } & \text { IE:HC } \\ \text { Mean: } & 192 & 196 & 205 & 392 & 438 & \text { SE:HC } & \text { EE:HC } & \text { IE:SD } & \text { SE:SD EE:SD } \\ \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } & \\ \text { Light } & 5 & 481776 & 96355.2 & 39.0 & 0.000 & \\ \text { Error } & 42 & 177835 & 2469.9 & & \\\ \text { Total } & 47 & 659611 & & & \end{array}\) (a) Is there a difference between the groups in the amount of time spent in darkness? Between which two groups are we most likely to find a difference in mean time spent in darkness? Between which two groups are we least likely to find a difference? (b) By looking at the six means, where do you think the differences are likely to lie? (c) Test to see if there is a difference in mean time spent in darkness between the IE:HC group and the EE:SD group (that is, impoverished but not stressed vs enriched but stressed).

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. \(\begin{array}{lrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } \\ \text { Groups } & 3 & 360.0 & 120.0 & 1.60 \\ \text { Error } & 16 & 1200.0 & 75.0 & \\ \text { Total } & 19 & 1560.0 & & \end{array}\)

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

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