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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.

Short Answer

Expert verified
After performing these steps, the T statistic value will be computed, and the associated p-value for the T statistic will be calculated. If the p-value is less than 0.05, then this implies a significant difference between the means of groups B and D.

Step by step solution

01

Identify Required Data

Extract the necessary values from the table for groups B and D, including the sample size (N), mean, and standard deviation (StDev). For B, N = 6, mean = 76.167, and StDev = 6.555. For D, N = 6, mean = 69.333, and StDev = 6.154. Also note the error mean square (MS) from the ANOVA table, which is 48.3.
02

Calculate the Standard Error

We calculate the standard error for the mean difference using the formula: \[ SE = \sqrt{\frac{MS}{n_B} + \frac{MS}{n_D}} \] where n_B and n_D are the sizes of groups B and D respectively, and MS is the error mean square. Here, the standard error (SE) = \sqrt{\frac{48.3}{6} + \frac{48.3}{6}}.
03

Calculate the Test Statistic

The test statistic for a difference in means is computed as the difference in group means divided by the standard error of the difference. \[ T = \frac{mean_B - mean_D}{SE} \] Here, calculate T as \[ T = \frac{76.167 - 69.333}{SE} \] from Step 2.
04

Determine the Degrees of Freedom

The degrees of freedom for this test are df within groups, which are found in the ANOVA table under the 'Error' row. Here, the degrees of freedom (df) = 20.
05

Compare the Test Statistic to a Critical Value

Using the calculated T-value from Step 3 and df from Step 4, find the p-value associated with this T-value from the t-distribution table or calculator tool. Compare this p-value with the significance level (assumed to be 0.05 unless specified otherwise). A p-value less than 0.05 indicates a significant difference in means between groups B and D.

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Key Concepts

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

Statistical Hypothesis Testing
Statistical hypothesis testing is a structured method used to determine if there is enough evidence to reject a null hypothesis. The null hypothesis, often denoted as H0, usually states that there is no effect or no difference, and it's what researchers aim to test against. In the context of ANOVA (Analysis of Variance), we are typically testing if there are any statistically significant differences among group means.

To perform a hypothesis test, we calculate a test statistic that tells us how much the observed data deviates from what's expected under the null hypothesis. This test statistic is then compared to a critical value determined by the desired confidence level and the distribution the test statistic follows. If the test statistic is more extreme than the critical value, we have grounds to reject the null hypothesis, thus concluding that our sample provides sufficient evidence that the null hypothesis is not true.
Mean Comparison
Mean comparison involves examining the arithmetic averages of two or more groups to determine if they are different from one another. In the exercise provided, this involves comparing the means of groups B and D. This comparison is crucial in many areas, from business to healthcare, as it helps in making informed decisions.

However, a difference in sample means does not automatically imply a significant difference in the population means. That’s where the ANOVA comes in—it helps us evaluate whether the observed differences in sample means are large enough to infer a difference in the population means or if they could have occurred just by chance.
Standard Error
The standard error (SE) measures the accuracy with which a sample mean estimates the population mean. It represents the standard deviation of the mean's sampling distribution and is a critical component in calculating the margin of error for confidence intervals as well as for conducting hypothesis tests.

In our exercise, the SE is used to assess the deviation in the differences between the group means (group B and group D). It gives us an understanding of the variability between these group means, accounting for sample sizes. The lower the SE, the more certain we are about the accuracy of our mean difference estimate.
Degrees of Freedom
Degrees of freedom (df) represent the number of independent values that can vary in an analysis without breaking any constraints. It is an essential concept in statistical hypothesis testing because it helps define the shape of the distribution used to calculate the critical values of our test statistic.

In the provided ANOVA table, the 'Error' row lists the degrees of freedom associated with the variability within groups. These degrees of freedom are used in determining the critical value for our test statistic. Specifically, they allow us to find the right cutoffs in the T-distribution table, ensuring that we correctly interpret the significance of our test results. In our exercise, knowing the degrees of freedom as 20 allows us to accurately determine whether the mean difference between groups B and D is statistically significant.

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Most popular questions from this chapter

Perhaps the mice with light at night in Exercise 8.28 gain more weight because they are exercising less. The conditions for an ANOVA test are met and computer output is shown for testing the average activity level for each of the three light conditions. Is there a significant difference in mean activity level? State the null and alternative hypotheses, give the F-statistic and the p-value, and clearly state the conclusion of the test.\(\begin{array}{lrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } \\ \text { DM } & 10 & 2503 & 1999 \\ \text { LD } & 8 & 2433 & 2266 \\ \text { LL } & 9 & 2862 & 2418\end{array}\)

Exercises 8.41 to 8.45 refer to the data with analysis shown in the following computer output: \(\begin{array}{lrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } \\ \text { A } & 5 & 10.200 & 2.864 \\ \text { B } & 5 & 16.800 & 2.168 \\ \text { C } & 5 & 10.800 & 2.387\end{array}\) \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 2 & 133.20 & 66.60 & 10.74 & 0.002 \\ \text { Error } & 12 & 74.40 & 6.20 & & \\ \text { Total } & 14 & 207.60 & & & \end{array}\) 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. Body Mass Gain Computer output showing body mass gain (in grams) for the mice after four weeks in each of the three light conditions is shown, along with the relevant ANOVA output. Which light conditions give significantly different mean body mass gain? \(\begin{array}{lrrr}\text { Level } & \mathrm{N} & \text { Mean } & \text { StDev } \\ \text { DM } & 10 & 7.859 & 3.009 \\ \text { LD } & 9 & 5.987 & 1.786 \\ \text { LL } & 9 & 11.010 & 2.624\end{array}\) One-way ANOVA: BM4Gain versus Light \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\\ \text { Light } & 2 & 116.18 & 58.09 & 8.96 & 0.001 \\ \text { Error } & 25 & 162.10 & 6.48 & & \\ \text { Total } & 27 & 278.28 & & & \end{array}\)

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

The mice in the study had body mass measured throughout the study. Computer output showing body mass gain (in grams) after 4 weeks for each of the three light conditions is shown, and a dotplot of the data is given in Figure 8.6 . \(\begin{array}{lrrr}\text { Level } & \mathrm{N} & \text { Mean } & \text { StDev } \\ \text { DM } & 10 & 7.859 & 3.009 \\ \text { LD } & 8 & 5.926 & 1.899 \\ \text { LL } & 9 & 11.010 & 2.624\end{array}\) (a) In the sample, which group of mice gained the most, on average, over the four weeks? Which gained the least? (b) Do the data appear to meet the requirement of having standard deviations that are not dramatically different? (c) The sample sizes are small, so we check that the data are relatively normally distributed. We see in Figure 8.6 that we have no concerns about the DM and LD samples. However, there is an outlier for the LL sample, at 17.4 grams. We proceed as long as the \(z\) -score for this value is within ±3 . Find the \(z\) -score. Is it appropriate to proceed with ANOVA? (d) What are the cases in this analysis? What are the relevant variables? Are the variables categorical or quantitative?

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