Question

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

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

Calculate the pooled standard deviation S_p

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

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 \(\alpha=0.05\)).

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.

Key concepts

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.