Question

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 \(\mathrm{A}\).

Short answer

The 95% confidence interval for the mean of population A is approximately [7.12, 13.28].

Step-by-step solution

Identify necessary data

From the data tables, you extract the necessary values. For population A: N (number of elements) = 5, Mean (average value of the population) = 10.2, and StDev (standard deviation) = 2.864. Moreover, you obtain the Error MS (Mean square of errors) from the ANOVA table which equals 6.2.

Calculate standard error

In most cases, the standard deviation of the population is not known so we use the standard error instead. The standard error can be calculated with the formula: \(\sqrt{MS_{error}/N}\), where \(MS_{error}\) is the mean square error from the ANOVA table and N the number of observations in the group. This yields: \(\sqrt{6.2/5} = 1.111\).

Find the value Tα/2,ν from T-distribution table

We want a confidence level of \(95\%\) which means an alpha level of 0.05. Since it's a two-tailed test, you look up the value for alpha divided by 2 (0.025) in the T-distribution table. The degrees of freedom (ν), in this case, is N-1 = 5-1 = 4. The T value corresponding to 0.025 with 4 degrees of freedom is approximately 2.776.

Compute the Confidence Interval

The confidence interval (CI) is calculated by the formula: \(\[Mean \pm T_{α/2,ν} \times SE]\), where Mean is the mean of group A, \(T_{α/2,ν}\) is the T-value from step 3, and SE is the standard error calculated in step 2. Substituting these values, we get \[10.2 \pm 2.776 \times 1.111\]. This implies our CI ranges from \(10.2 - 2.776 \times 1.111\) to \(10.2 + 2.776 \times 1.111\)

Evaluate the results

Performing the calculation, the 95% confidence interval for the mean population A is approximately [7.12, 13.28]. This means that we're 95% confident that the true population A mean lies somewhere between 7.12 and 13.28.

Key concepts

Understanding Standard Deviation

Standard deviation is a measure reflective of the amount of variation or dispersion present in a set of data values. To put it simply, it tells us how much the individual data points differ from the mean, or average, of the data set. A low standard deviation indicates that the data points tend to be close to the mean, while a high standard deviation suggests that the data points are spread out over a larger range of values.

For instance, in the exercise scenario, each group—A, B, and C—has its own standard deviation. Group A's standard deviation is 2.864, which can be interpreted as the average distance of each observation in group A from the group's mean (10.2). When computing confidence intervals, this value helps us assess the precision of the sample mean as an estimate of the population mean. In general, the smaller the standard deviation, the more precise our estimate is likely to be.

ANOVA and Its Role in Statistical Analysis

ANOVA, which stands for Analysis of Variance, is a statistical method used to compare means between three or more groups to determine if at least one group mean is statistically different from the others. It's useful when testing hypotheses about whether group means differ within a population based on categorical variables.

The ANOVA produces an F-statistic, which is the ratio of the variance calculated between the groups to the variance within the groups. The P-value reported next to this F-statistic informs us whether the observed variance between groups is statistically significant. In our exercise, with a P-value of 0.002, we conclude that there's a significant difference between at least two group means. It is through the ANOVA that we also derive the mean square error (MS error), a component required to calculate the standard error.

Standard Error: Estimating the Margin of Error

Standard error measures the accuracy with which a sample represents a population. In other words, it quantifies the variability of the sample mean estimate of a population mean. The smaller the standard error, the more representative the sample mean is of the population mean.

In calculations, the formula for the standard error of the mean (SEM) commonly involves dividing the sample standard deviation by the square root of the sample size. However, in situations like our exercise where the population standard deviation is unknown, the standard error is derived from the Mean Square Error obtained from ANOVA output. This value provides an estimate of the population variance which, when divided by the sample size and square-rooted, gives us the SEM. Consequently, the SEM plays a pivotal role in determining the width of the confidence interval.

T-distribution and Its Significance

The T-distribution, or Student's t-distribution, is a probability distribution that is symmetrical and bell-shaped like the normal distribution but has heavier tails. This means there is a greater possibility of obtaining values far away from the mean in a t-distribution compared to a normal distribution. It becomes particularly important when dealing with small sample sizes or when the population standard deviation is unknown.

In the context of confidence intervals, we often use the T-distribution to determine the 'critical value' for the margin of error. The result is the T-value multiplied by the standard error to give the full range that is added and subtracted from the sample mean to form the interval, like in step 4 of the exercise. As the sample size increases, the T-distribution resembles the normal distribution more closely, which is why for larger sample sizes, it's common to use the normal (Z) distribution instead.