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Exercises 8.46 to 8.52 refer to the data with analysis shown in the following computer output: \(\begin{array}{lrrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } & \\ \text { A } & 6 & 86.833 & 5.231 & \\ \text { B } & 6 & 76.167 & 6.555 & \\ \text { C } & 6 & 80.000 & 9.230 & \\ \text { D } & 6 & 69.333 & 6.154 & \\ \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 3 & 962.8 & 320.9 & 6.64 & 0.003 \\ \text { Error } & 20 & 967.0 & 48.3 & & \\ \text { Total } & 23 & 1929.8 & & & \end{array}\) Find a \(99 \%\) confidence interval for the mean of population \(\mathrm{A}\). Is 90 a plausible value for the population mean of group \(\mathrm{A}\) ?

Short Answer

Expert verified
The two endpoints of the confidence interval can be calculated by evaluating the expressions from Step 3. Once the confidence interval is known, it's immediate to see if 90 is within this interval: if so, it's plausible; if not, it's not plausible.

Step by step solution

01

Calculate the Standard Error

The standard error of the mean (SEM) can be calculated using the formula: SEM= StDev / \sqrt{N}. In this case, we have the standard deviation (StDev) as 5.231 and the number of observations (N) as 6 for population A. By substituting these values into the formula, you find that SEM = 5.231 / \sqrt{6}.
02

Find the t-value

Next, we will find the t-value corresponding to a 99% confidence interval. Using a t-table or statistical software with \(df = N - 1 = 5\), and level of significance \(0.01/2 = 0.005\) (since it's 2-tailed), we find the t-value.
03

Construct the Confidence Interval

The formula to construct the confidence interval is: Mean ± (t-value * SEM). Substituting our previously calculated values, we can find the \(99\% \)confidence interval.
04

Determine Plausibility of Given Value

To determine if 90 is a plausible value for the population mean of group A, we simply need to check if it falls within our calculated confidence interval.

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Key Concepts

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

Standard Error
When we talk about standard error (often abbreviated as SE or SEM), we're referring to a measure that indicates the accuracy with which a sample mean represents the true population mean. The formula for standard error is given by SEM = StDev / \(\text{\sqrt{N}}\).

In practical terms, the smaller the standard error, the more representative the sample mean is of the population mean. This concept is pivotal when we analyze sample data and attempt to infer characteristics about the whole population. For instance, in the context of our example, a group of six observations (N = 6) from population A with a standard deviation (StDev) of 5.231 provides us with a standard error after calculation. By keeping our data precise, we can ensure that the confidence intervals meaningfully inform us about the population characteristics with the least amount of uncertainty we can achieve with the given sample size.
T-value
The t-value is a statistic that tells you how many standard errors your sample mean is away from the population mean, under the assumption that the null hypothesis of no effect is true. It's calculated during a t-test and is used to determine the p-value, which in turn helps gauge the statistical significance. The t-value is dependent on the confidence level and the degrees of freedom, which is calculated as the sample size minus one (df = N - 1).

In our example, the t-value is found by looking at the t-distribution table or using statistical software with specific degrees of freedom and the desired confidence level (99% in this case). This t-value can then be multiplied by the standard error to find the margin of error, which is used to calculate the confidence interval around the sample mean.
Statistical Significance
Statistical significance is a determination that the observed effect in the data is unlikely to have occurred due to random chance alone. It's measured in terms of a probability, known as a p-value, which assesses the strength of the evidence against the null hypothesis. A p-value lower than a pre-determined significance level (often 0.05 or 0.01) indicates that the result is statistically significant.

The p-value in our case is associated with the F-statistic from an ANOVA (analysis of variance), and here it is 0.003. This suggests that there is a statistically significant difference between the group means. We perform further analysis by constructing a confidence interval to understand if a specified value (like 90) is a plausible mean for population A, taking into account the variability and size of our sample.

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Most popular questions from this chapter

Researchers hypothesized that the increased weight gain seen in mice with light at night might be caused by when the mice are eating. (As we have seen in the previous exercises, it is not caused by changes in amount of food consumed or activity level.) Perhaps mice with light at night eat a greater percentage of their food during the day, when they normally should be sleeping. Conditions for ANOVA are met and computer output for the percentage of food consumed during the day for each of the three light conditions is shown.\(\begin{array}{lrrr}\text { Level } & \mathrm{N} & \text { Mean } & \text { StDev } \\ \text { DM } & 10 & 55.516 & 10.881 \\ \text { LD } & 8 & 36.028 & 8.403 \\ \text { LL } & 9 & 76.573 & 9.646\end{array}\) (a) For mice in this sample on a standard light/dark cycle, what is the average percent of food consumed during the day? What percent is consumed at night? What about mice that had dim light at night? (b) Is there evidence that light at night influences when food is consumed by mice? Justify your answer with a p-value. Can we conclude that there is a cause-and-effect relationship?

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? Four groups with \(n_{1}=10, n_{2}=10, n_{3}=10\), and \(n_{4}=10 .\) ANOVA table includes: $$ \begin{array}{|l|l|l|l|l|} \hline \text { Source } & \text { df } & \text { SS } & \text { MS } & \text { F-statistic } \\ \hline \text { Groups } & & 960 & & \\ \hline \text { Error } & & 5760 & & \\ \hline \text { Total } & & 6720 & & \\ \hline \end{array} $$

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 \(\mathrm{A}\).

Two sets of sample data, \(\mathrm{A}\) and \(\mathrm{B}\), are given. Without doing any calculations, indicate in which set of sample data, \(\mathrm{A}\) or \(\mathrm{B}\), there is likely to be stronger evidence of a difference in the two population means. Give a brief reason, comparing means and variability, for your answer. $$ \begin{array}{cc|cc} \hline {\text { Dataset A }} & {\text { Dataset B }} \\ \hline \text { Group 1 } & \text { Group 2 } & \text { Group 1 } & \text { Group 2 } \\ \hline 13 & 18 & 13 & 48 \\ 14 & 19 & 14 & 49 \\ 15 & 20 & 15 & 50 \\ 16 & 21 & 16 & 51 \\ 17 & 22 & 17 & 52 \\ \bar{x}_{1}=15 & \bar{x}_{2}=20 & \bar{x}_{1}=15 & \bar{x}_{2}=50 \end{array} $$

Color affects us in many ways. For example, Exercise C.92 on page 498 describes an experiment showing that the color red appears to enhance men's attraction to women. Previous studies have also shown that athletes competing against an opponent wearing red perform worse, and students exposed to red before a test perform worse. \(^{3}\) Another study \(^{4}\) states that "red is hypothesized to impair performance on achievement tasks, because red is associated with the danger of failure." In the study, US college students were asked to solve 15 moderately difficult, five-letter, single-solution anagrams during a 5-minute period. Information about the study was given to participants in either red, green, or black ink just before they were given the anagrams. Participants were randomly assigned to a color group and did not know the purpose of the experiment, and all those coming in contact with the participants were blind to color group. The red group contained 19 participants and they correctly solved an average of 4.4 anagrams. The 27 participants in the green group correctly solved an average of 5.7 anagrams and the 25 participants in the black group correctly solved an average of 5.9 anagrams. Work through the details below to test if performance is different based on prior exposure to different colors. (a) State the hypotheses. (b) Use the fact that sum of squares for color groups is 27.7 and the total sum of squares is 84.7 to complete an ANOVA table and find the F-statistic. (c) Use the F-distribution to find the p-value. (d) Clearly state the conclusion of the test.

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