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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 \(A\) and \(B\). Show all details of the test.

Short Answer

Expert verified
The decision on whether there is a significant difference in the means of groups A and B depends on the computed t-value from step 2 and the critical t-value from step 3. If \( |t_{computed}| > |t_{critical}| \), then the conclusion would be that there is a significant difference in means between the two groups.

Step by step solution

01

Compute t-statistic

First, compute the t-statistic for the difference in means between group A and group B. This can be computed using the formula \( t = \frac{{M_A - M_B}}{{\sqrt{{MSE(1/n_A + 1/n_B)}}}} \) where MSE (Mean Squares Within) is the within-group variance, n is the group size, M is the group mean. So, \( M_A = 86.833 \), \( M_B = 76.167 \), \( MSE = 48.3 \), \( n_A = n_B = 6 \).
02

Calculate t-statistic

Now, substitute the known values into the formula: \( t = \frac{{86.833 - 76.167}}{{\sqrt{{48.3(1/6 + 1/6)}}}} \). Solve this equation to get the computed t-value.
03

Determine the critical value

It's necessary to first find the degrees of freedom for the t-distribution. In this case, the degrees of freedom (df) would be: \( df = n_A + n_B - 2 = 6+6-2 = 10 \). Using a significance level of 0.05 (a standard value if none is given), the critical value of t for a two-tailed test from the t-distribution table is approximately ±2.228.
04

Compare t-statistic to critical value

Finally, compare the computed t-value from step 2 with the critical t-value from step 3. If the absolute value of the computed t-value is greater than the absolute value of the critical t-value, then there is a significant difference in means between the two groups.

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Key Concepts

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

Mean Comparison
When analyzing the difference between two groups, like group A and group B, it is common to use a mean comparison technique. This is particularly useful when you want to determine if the groups have different average values on a specific measurement.
This exercise involves comparing the means of two different sample groups: Group A with a mean of 86.833 and Group B with a mean of 76.167. To assess whether this difference is statistically significant, a t-test is performed. The t-test helps in understanding whether the observed differences between the sample means could have happened by random chance or whether they're statistically significant.
  • The mean is a measure of central tendency that represents the average of a set of values.
  • Comparing means can indicate whether one group experiences a higher or lower average outcome than another.
By performing the t-test, you essentially check if the mean difference observed is statistically reliable. This helps in making inferences about the population based on sample data.
Degrees of Freedom
Degrees of freedom (df) are a crucial part of statistical tests, particularly in hypothesis testing. In the context of a t-test, degrees of freedom refer to the number of independent values or quantities which can vary in an analysis without breaking any constraints.
For this specific t-test comparing the two groups, the degrees of freedom are calculated as the sum of sample sizes from each group minus two: \( df = n_A + n_B - 2 = 6 + 6 - 2 = 10 \). This formula helps estimate the number of observations that were used to calculate the means and variances that go into the t-test calculation.
  • Degrees of freedom impact the critical values used to determine statistical significance.
  • Having the correct degrees of freedom helps ensure accurate results in hypothesis testing.
In simpler terms, think of degrees of freedom as a way to account for the number of variables available to ensure the precision of your test outcomes.
Critical Value
The critical value in any t-test is vitally important because it helps establish the threshold for statistical significance. Essentially, it serves as a cutoff point to decide whether the null hypothesis can be rejected.
In this exercise, the significance level is chosen at 0.05, which is standard for many tests if not otherwise specified. This level represents a 5% chance that any differences observed between groups are due to random sampling error alone. For a two-tailed test and given the degrees of freedom are 10, the critical value is approximately \( \pm 2.228 \).
  • The critical value is determined using a t-distribution table.
  • It's compared against the computed t-statistic to assess significance.
If the computed t-value exceeds the critical value in absolute terms, it suggests there is a significant difference between the sample means, leading to the rejection of the null hypothesis. This conclusion helps in uncovering valuable insights that can guide practical decisions based on data.

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Most popular questions from this chapter

Exercises 8.41 to 8.45 refer to the data with analysis shown in the following computer output: \(\begin{array}{lrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } \\ \text { A } & 5 & 10.200 & 2.864 \\ \text { B } & 5 & 16.800 & 2.168 \\ \text { C } & 5 & 10.800 & 2.387\end{array}\) \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 2 & 133.20 & 66.60 & 10.74 & 0.002 \\ \text { Error } & 12 & 74.40 & 6.20 & & \\ \text { Total } & 14 & 207.60 & & & \end{array}\) What is the pooled standard deviation? What degrees of freedom are used in doing inferences for these means and differences in means?

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

In addition to monitoring weight gain, food consumed, and activity level, the study measured stress levels in the mice by measuring corticosterone levels in the blood (higher levels indicate more stress). Conditions for ANOVA are met and computer output for corticosterone levels for each of the three light conditions is shown. \(\begin{array}{lrrr}\text { Level } & \mathrm{N} & \text { Mean } & \text { StDev } \\ \text { DM } & 10 & 73.40 & 67.49 \\ \text { LD } & 8 & 70.02 & 54.15 \\ \text { LL } & 9 & 50.83 & 42.22\end{array}\) (a) What is the conclusion of the analysis of variance test? (b) One group of mice in the sample appears to have very different corticosterone levels than the other two. Which group is different? What aspect of the data explains why the ANOVA test does not find this difference significant? How is this aspect reflected in both the summary statistics and the ANOVA table?

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?

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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