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Refer to Exercise \(11.63 .\) The means of all observations, at the factor A levels \(\mathrm{A}_{1}\) and \(\mathrm{A}_{2}\) are \(\bar{x}_{1}=3.7\) and \(\bar{x}_{2}=1.4,\) respectively. Find a \(95 \%\) confidence interval for the difference in mean response for factor levels \(\mathrm{A}_{1}\) and \(\mathrm{A}_{2}\)

Short Answer

Expert verified
Question: Calculate the 95% confidence interval for the difference in mean response for factor levels A1 and A2 given the means of all observations for both factors, \(\bar{x}_{1}=3.7\), and \(\bar{x}_{2}=1.4\), and assuming known sample sizes, and standard deviations for the samples.

Step by step solution

01

Calculate the pooled standard deviation S_p

Start by calculating the pooled standard deviation using the formula: \(S_p=\sqrt{\frac{(n_{1}-1)S_{1}^2+(n_{2}-1)S_{2}^2}{n_{1}+n_{2}-2}}\).
02

Find the t-value

Determine the t-value with \(\alpha / 2\) level of significance. To find the t-value, use a t-table or a calculator with a t-distribution function. To find the critical value, look up the value that corresponds to the appropriate degrees of freedom (\((n_{1}+n_{2}-2)\)) and chosen significance level (typically \(\alpha=0.05\)).
03

Calculate the confidence interval

Now we can calculate the 95% confidence interval for the difference in mean response for factor levels A1 and A2. Use the following formula: \(CI = (\bar{x}_{1} - \bar{x}_{2}) \pm t_{\alpha / 2}S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}\). Plug in the values of \(\bar{x}_{1}\), \(\bar{x}_{2}\), \(t_{\alpha / 2}\), S_p, \(n_{1}\), and \(n_{2}\) to calculate the confidence interval for the difference in mean response. The resulting confidence interval will give us the range in which we can expect the true difference in mean response between the two factor levels, A1 and A2, with a 95% level of confidence.

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Key Concepts

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

Pooled Standard Deviation
Pooled standard deviation is a method used to estimate the overall standard deviation from two or more groups that have potentially different variances but are assumed to come from distributions with the same variance. It is particularly useful when dealing with two sample t-tests involving the means of the independent groups.

The formula for calculating the pooled standard deviation is as follows:\[\begin{equation}S_p = \sqrt{\frac{(n_{1}-1)S_{1}^2 +(n_{2}-1)S_{2}^2}{n_{1}+n_{2}-2}}\end{equation}\]Where:
  • \(n_{1}\) and \(n_{2}\) are the sample sizes of the two groups,
  • \(S_{1}^2\) and \(S_{2}^2\) are the sample variances of the two groups.
The denominator, \(n_{1}+n_{2}-2\), represents the total degrees of freedom in the data, which is the sum of the degrees of freedom for both groups. The pooled standard deviation is used because it provides a weighted average that accounts for the varying sample sizes and brings about a more accurate estimation by combining the variability information from both samples.
T-value
The t-value, often referred to as a t-score, is a type of standardized score that is derived from the t-distribution. It is used to determine the number of standard deviations a particular sample mean deviates from the hypothesized population mean in a t-test.

The t-value is calculated based on the sample mean, population mean (\(\mu\)), and standard deviation, as well as the sample size. In hypothesis testing, the calculated t-value is compared to a critical t-value from the t-distribution, which depends on the chosen level of significance \(\alpha\) and the degrees of freedom of the data. If the absolute value of the calculated t-value is greater than the critical t-value, the null hypothesis of no effect or no difference is typically rejected.

For calculating the confidence interval, we use the t-value to establish the range around the sample mean which, with a certain level of confidence, contains the population mean. The selected level of confidence corresponds to a certain percentile of the t-distribution, and the t-value for this percentile provides the critical value needed to set the bounds of the confidence interval.
Mean Response Difference
The mean response difference is a measure of the average difference between the responses in two groups or conditions. For example, it could be the difference between the average blood pressure reductions in two different medication groups. It is central to comparison studies because it quantifies the effect caused by changing the independent variable.

In the context of our exercise, the mean response difference is calculated by subtracting the mean response (\(\bar{x}_{2}\)) of factor level \(\mathrm{A}_{2}\) from the mean response (\(\bar{x}_{1}\)) of factor level \(\mathrm{A}_{1}\). The resulting value indicates whether there is a significant difference between the two factor levels and how large that difference might be.

The formula to find the mean response difference is:\[\begin{equation}\bar{x}_{1} - \bar{x}_{2}\end{equation}\]Generally, if the resulting difference is zero or very close to zero, it suggests that the two groups do not differ significantly in terms of their mean response. Otherwise, a non-zero difference might suggest that there is a significant effect occurring between the groups.
T-distribution
The t-distribution, also known as Student's t-distribution, plays a crucial role in small sample hypothesis testing and construction of confidence intervals when the population standard deviation is unknown. It is similar to the normal distribution in shape but has heavier tails because it accounts for the increased uncertainty that comes with smaller samples.

As the sample size increases, the t-distribution approaches the normal distribution. For each sample size, there is a different t-distribution, and the exact form is determined by the degrees of freedom (\(u\)=\(n-1 \)), with \(n \)representing the sample size. The degrees of freedom essentially correct for the sample size in estimating the population parameter.

In conducting a t-test or forming a confidence interval, analysts use the t-distribution to determine critical values known as t-scores. These values bound the acceptance region for the null hypothesis or the confidence interval for an estimate. Especially when the sample size is small, the t-distribution provides a more accurate estimation than the normal distribution would.

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Most popular questions from this chapter

A randomized block design was used to compare the means of three treatments within six blocks. Construct an ANOVA table showing the sources of variation and their respective degrees of freedom.

An experiment was conducted to compare the glare characteristics of four types of automobile rearview mirrors. Forty drivers were randomly selected to participate in the experiment. Each driver was exposed to the glare produced by a headlight located 30 feet behind the rear window of the experimental automobile. The driver then rated the glare produced by the rearview mirror on a scale of 1 (low) to 10 (high). Each of the four mirrors was tested by each driver; the mirrors were assigned to a driver in random order. An analysis of variance of the data produced this ANOVA table: $$ \begin{array}{lcc} \text { Source } & d f & \text { SS } & \text { MS } \\ \hline \text { Mirrors } & 46.98 & \\ \text { Drivers } & & 8.42 \\ \text { Error } & & & \\ \hline \text { Total } & 638.61 & \end{array} $$ a. Fill in the blanks in the ANOVA table. b. Do the data present sufficient evidence to indicate differences in the mean glare ratings of the four rearview mirrors? Calculate the approximate \(p\) -value and use it to make your decision. c. Do the data present sufficient evidence to indicate that the level of glare perceived by the drivers varied from driver to driver? Use the \(p\) -value approach. d. Based on the results of part b, what are the practical implications of this experiment for the manufacturers of the rearview mirrors?

Suppose you were to conduct a two-factor factorial experiment, factor A at four levels and factor \(\mathrm{B}\) at five levels, with three replications per treatment. a. How many treatments are involved in the experiment? b. How many observations are involved? c. List the sources of variation and their respective degrees of freedom.

Refer to Exercise \(11.46 .\) The means of two of the factor-level combinations- say, \(\mathrm{A}_{1} \mathrm{~B}_{1}\) and \(\mathrm{A}_{2} \mathrm{~B}_{1}-\) are \(\bar{x}_{1}=8.3\) and \(\bar{x}_{2}=6.3,\) respectively. Find a \(95 \%\) confidence interval for the difference between the two corresponding population means.

Swampy Sites An ecological study was conducted to compare the rates of growth of vegetation at four swampy undeveloped sites and to determine the cause of any differences that might be observed. Part of the study involved measuring the leaf lengths of a particular plant species on a preselected date in May. Six plants were randomly selected at each of the four sites to be used in the comparison. The data in the table are the mean leaf length per plant (in centimeters) for a random sample of ten leaves per plant. The MINITAB analysis of variance computer printout for these data is also provided. $$ \begin{array}{lllllll} \text { Location } & {\text { Mean Leaf Length (cm) }} \\ \hline 1 && 5.7 & 6.3 & 6.1 & 6.0 & 5.8 & 6.2 \\ 2 && 6.2 & 5.3 & 5.7 & 6.0 & 5.2 & 5.5 \\ 3 && 5.4 & 5.0 & 6.0 & 5.6 & 4.9 & 5.2 \\ 4 && 3.7 & 3.2 & 3.9 & 4.0 & 3.5 & 3.6 \end{array} $$ a. You will recall that the test and estimation procedures for an analysis of variance require that the observations be selected from normally distributed (at least, roughly so) populations. Why might you feel reasonably confident that your data satisfy this assumption? b. Do the data provide sufficient evidence to indicate a difference in mean leaf length among the four locations? What is the \(p\) -value for the test? c. Suppose, prior to seeing the data, you decided to compare the mean leaf lengths of locations 1 and \(4 .\) Test the null hypothesis \(\mu_{1}=\mu_{4}\) against the alternative \(\mu_{1} \neq \mu_{4}\) d. Refer to part c. Construct a \(99 \%\) confidence interval for \(\left(\mu_{1}-\mu_{4}\right)\) e. Rather than use an analysis of variance \(F\) -test, it would seem simpler to examine one's data, select the two locations that have the smallest and largest sample mean lengths, and then compare these two means using a Student's \(t\) -test. If there is evidence to indicate a difference in these means, there is clearly evidence of a difference among the four. (If you were to use this logic, there would be no need for the analysis of variance \(F\) -test.) Explain why this procedure is invalid.

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