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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 \(90 \%\) confidence interval for the difference in the means of populations \(\mathrm{B}\) and \(\mathrm{C}\).

Short Answer

Expert verified
The 90% confidence interval for the difference in means of populations B and C can be calculated as \( 16.8 - 10.8 \pm 1.860*SE_d \), substituting the calculated SE

Step by step solution

01

Compute Standard Error

The Standard Error (SE) for the difference in means can be obtained using the formula: \( SE_d = \sqrt{(s_{1}^{2}/n_{1}) + (s_{2}^{2}/n_{2})} \). Substituting the given values, \( s_{1} = 2.168 \), \( s_{2} = 2.387 \), \( n_{1} = n_{2} = 5 \) into the formula, we get \( SE_d = \sqrt{(2.168^2/5) + (2.387^2/5)} \)
02

Compute degrees of freedom

Since we are using t-distribution, we need degrees of freedom (DF), which is \(DF=n_{1}+n_{2}-2 = 5+5-2 = 8\). Using this DF, we look at the t-distribution table to find the t-value that corresponds to the 90% confidence level, which is \(1.860\).
03

Calculate Confidence Interval

Using the formula for confidence interval for difference in means \(\mu_{1} - \mu_{2} \pm t* SE_d \), and substituting the calculated and given values, \( \mu_{1} = 16.8 \), \( \mu_{2} = 10.800 \), \( t=1.860 \), we get the confidence interval as \( 16.8 - 10.8 \pm 1.860*SE_d \).

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Key Concepts

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

Understanding Standard Error (SE)
When conducting statistics, it's essential to have an accurate measure of the variability or precision of a sample mean compared to the population mean. This is where Standard Error (SE) comes into play. The SE provides insight into how much a sample mean is expected to differ from the true population mean purely by chance.

Mathematically, the SE for the difference between two independent means is calculated as \( SE_d = \sqrt{(s_{1}^{2}/n_{1}) + (s_{2}^{2}/n_{2})} \), where \(s_{1}^{2}\) and \(s_{2}^{2}\) are the variances (squared standard deviations) of the two samples, and \(n_{1}\) and \(n_{2}\) are the sample sizes. Lower SE values signify greater precision in estimating the population mean, as they indicate less variability between the sample means and the true population mean.

For our problem, by substituting the given values into the formula, we can find the SE required to further calculate the confidence interval for the difference in means between populations B and C. A smaller SE potentially implies that the difference between the sample means is a more reliable estimator of the difference between the population means.
Interpreting the t-distribution

Working with Small Sample Sizes

When it comes to small sample sizes, normal distribution assumptions do not always hold. That's where the t-distribution comes in handy. It's a type of probability distribution that is symmetric and bell-shaped, much like the normal distribution, but with thicker tails, which means it accounts for more variability.

The t-distribution is particularly useful when dealing with estimates derived from small sample sizes (<30) or when the population standard deviation is unknown. With smaller samples, there is more uncertainty, which the t-distribution helps to account for. By using the t-distribution, we can generate more accurate confidence intervals and hypothesis tests for our sample data.

In our exercise, with degrees of freedom (DF) of 8, we use the t-distribution to find the appropriate t-value for constructing the confidence interval for the difference in means. The t-value, determined from statistical tables or software, corresponds to the desired level of confidence—in this case, 90%.
Degrees of Freedom Explained
The concept of degrees of freedom (DF) is a fundamental aspect of statistical analysis, which corresponds to the number of independent pieces of information we have while estimating a statistic. In the context of comparing two means, DF is calculated as \(DF = n_{1} + n_{2} - 2\), where \(n_{1}\) and \(n_{2}\) are the sample sizes for the respective groups.

Why Are Degrees of Freedom Important?

DF play a crucial role in distributions like the t-distribution that adjusts its shape based on the number of observations in the data. The more degrees of freedom we have, the closer the t-distribution gets to the normal distribution. For smaller samples, fewer degrees of freedom lead to wider confidence intervals, reflecting increased uncertainty.

In our example, with 5 observations in each group, we have 8 degrees of freedom (10 total observations minus the 2 means we are estimating). This value is required when we refer to the t-distribution to obtain the accurate critical value needed for constructing the confidence interval.

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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?

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 \(\mathrm{A}\) ?

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?

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 \(0.05 ?\)

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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