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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

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. \(\begin{array}{lrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } \\ \text { Groups } & 3 & 450.0 & 150.0 & 0.75 \\ \text { Error } & 16 & 3200.0 & 200.0 & \\ \text { Total } & 19 & 3650.0 & & \end{array}\)

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?

Exercises 8.41 to 8.45 refer to the data with analysis shown in the following computer output: \(\begin{array}{lrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } \\ \text { A } & 5 & 10.200 & 2.864 \\ \text { B } & 5 & 16.800 & 2.168 \\ \text { C } & 5 & 10.800 & 2.387\end{array}\) \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 2 & 133.20 & 66.60 & 10.74 & 0.002 \\ \text { Error } & 12 & 74.40 & 6.20 & & \\ \text { Total } & 14 & 207.60 & & & \end{array}\) Test for a difference in population means between groups \(A\) and \(C\). Show all details of the test.

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.

Drug Resistance and Dosing Exercise 8.39 on page 561 explores the topic of drug dosing and drug resistance by randomizing mice to four different drug treatment levels: untreated (no drug), light ( \(4 \mathrm{mg} / \mathrm{kg}\) for 1 day), moderate \((8 \mathrm{mg} / \mathrm{kg}\) for 1 day), or aggressive ( \(8 \mathrm{mg} / \mathrm{kg}\) for 5 or 7 days). Exercise 8.39 found that, contrary to conventional wisdom, higher doses can actually promote drug resistance, rather than prevent it. Here, we further tease apart two different aspects of drug dosing: duration (how many days the drug is given for) and amount per day. Recall that four different response variables were measured; two measuring drug resistance (density of resistant parasites and number of days infectious with resistant parasites) and two measuring health (body mass and red blood cell density). In Exercise 8.39 we don't find any significant differences in the health responses (Weight and \(R B C)\) so we concentrate on the drug resistance measures (ResistanceDensity and DaysInfectious) in this exercise. The data are available in DrugResistance and we are not including the untreated group. (a) Investigate duration by comparing the moderate treatment with the aggressive treatment (both of which gave the same amount of drug per day, but for differing number of days). Which of the two resistance response variables (ResistanceDensity and DaysInfectious) have means significantly different between these two treatment groups? For significant differences, indicate which group has the higher mean. (b) Investigate amount per day by comparing the light treatment with the moderate treatment (both of which lasted only 1 day, but at differing amounts). Which of the two resistance response variables have means significantly different between these two treatment groups? For significant differences, indicate which group has the higher mean. (c) Does duration or amount seem to be more influential (at least within the context of this study)? Why?

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