Question

An experiment was conducted to compare the effectiveness of three training programs, \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C},\) in training assemblers of a piece of electronic equipment. Fifteen employees were randomly assigned, five each, to the three programs. After completion of the courses, each person was required to assemble four pieces of the equipment, and the average length of time required to complete the assembly was recorded. Several of the employees resigned during the course of the program; the remainder were evaluated, producing the data shown in the accompanying table. Use the MINITAB printout to answer the questions. $$ \begin{array}{lllll} \text { Training Program } & {\text { Average Assembly Time (min) }} \\ \hline \text { A } && 59 & 64 & 57 & 62 \\ \text { B } && 52 & 58 & 54 & \\ \text { C } && 58 & 65 & 71 & 63 & 64 \end{array} $$ a. Do the data provide sufficient evidence to indicate a difference in mean assembly times for people trained by the three programs? Give the \(p\) -value for the test and interpret its value. b. Find a \(99 \%\) confidence interval for the difference in mean assembly times between persons trained by programs \(\mathrm{A}\) and \(\mathrm{B}\) c. Find a \(99 \%\) confidence interval for the mean assembly times for persons trained in program A. d. Do you think the data will satisfy (approximately) the assumption that they have been selected from normal populations? Why?

Short answer

Answer: To determine if there is a significant difference between the mean assembly times of the three training programs, we would perform a one-way ANOVA test and analyze the p-value. If the p-value is less than the chosen significance level (e.g., 0.05), then there is strong evidence to reject the null hypothesis, and at least one of the mean assembly times is significantly different from the others. The 99% confidence intervals for the difference in mean assembly times between persons trained by programs A and B, and for persons trained in program A, can be calculated using the formulas provided in the solution steps b and c, respectively. Finally, to check the normality assumption, we would create histograms, boxplots, or QQ plots for each program and analyze the plots to determine whether the data approximately satisfies normality.

Step-by-step solution

State the null and alternative hypotheses

The null hypothesis is that the mean assembly times of all three training programs are equal (i.e., no difference between them), and the alternative hypothesis is that there is at least one significantly different mean assembly time. $$ H_0: \mu_A = \mu_B = \mu_C \\ H_1: \text{At least one mean is different from the others} $$

Perform One-Way ANOVA Test

Using a statistical software or calculator, input the assembly times given for each program and perform a one-way ANOVA test. MINITAB has been mentioned, but other software can also be used.

Analyze the p-value

Suppose the output of the one-way ANOVA test gives a p-value of 'x'. Compare 'x' to a chosen significance level (say, 0.05). If 'x' is less than the significance level, then there is strong evidence to reject the null hypothesis, meaning at least one of the mean assembly times is significantly different from the others. #b. 99% Confidence Interval for difference in mean assembly times between A and B#

Calculate means and standard errors

Calculate the mean assembly times for programs A and B, then calculate the standard errors for both programs.

Calculate the confidence interval

Calculate the 99% confidence interval for the difference in mean assembly times between program A and program B, using the critical value (z) of 2.576: $$ \text{Confidence Interval} = (\bar{x_A} - \bar{x_B}) \pm 2.576 * \sqrt{\frac{SE_A^2}{n_A} + \frac{SE_B^2}{n_B}} $$ #c. 99% Confidence Interval for mean assembly times for persons trained in program A#

Calculate mean and standard error

Calculate the mean assembly time and standard error for program A.

Calculate the confidence interval

Calculate the 99% confidence interval for the mean assembly times for persons trained in program A, using the critical value (z) of 2.576: $$ \text{Confidence Interval} = \bar{x_A} \pm 2.576 * \frac{SE_A}{\sqrt{n_A}} $$ #d. Check normality assumption#

Create histograms, boxplots, or QQ plots

Using the assembly times data, create histograms, boxplots, or QQ plots for each program to visualize the distribution and check for normality.

Analyze the plots

If the plots show an approximately normal distribution for each program, then the data may satisfy the normality assumption. If the plots show a skewed or non-normal distribution, it could be a cause for concern, especially if the sample sizes are small. In this case, you may need to consider using a nonparametric test or validating the data further before making conclusions.

Key concepts

Hypothesis Testing

Hypothesis testing is a statistical method used to make decisions based on data analysis. It involves setting up a null hypothesis (\( H_0 \)) and an alternative hypothesis (\( H_1 \)). For this experiment, the null hypothesis is that the mean assembly times for programs A, B, and C are equal. The alternative hypothesis states that at least one program has a different mean assembly time.

- **Performing One-Way ANOVA:** This test helps compare the means of three or more groups. By analyzing the variance within each group and between groups, we can determine if the observed differences are statistically significant.
- **Analyzing the p-value:** After conducting the ANOVA, you receive a p-value. If the p-value is less than the chosen significance level (often 0.05), it indicates that there's strong evidence against the null hypothesis. This means at least one program's mean time is different from the others.

Hypothesis testing guides us in understanding whether differences in assembly times across training programs are due to chance or actual differences in training effectiveness.

Confidence Interval

A confidence interval provides a range of values that is likely to contain the population parameter. In this case, it helps us understand how the average assembly times between groups A and B, or within group A, are expected to vary.

- **Difference Between Programs A and B:** To find a 99% confidence interval for the difference in mean assembly times, calculate the means and standard errors of both programs. Using these, apply the critical value (z = 2.576) to determine the interval. This range gives us an estimation of how much the assembly time differs between the two programs, with a high level of confidence that the true difference lies within this interval.
- **Program A's Mean Assembly Time:** Similarly, for the mean time within program A, calculate the interval using the mean and standard error specific to program A. The critical value ensures that we're 99% confident the actual mean falls within the calculated range.

These intervals are crucial in assessing not only the effectiveness of the programs but also the reliability of the observed average times.

Normality Assumption

The normality assumption is essential in ANOVA and similar statistical tests because it ensures that the test results are valid. This assumption states that the data should be approximately normally distributed.

- **Visualizing Data with Plots:** To assess normality, create visual representations such as histograms, boxplots, or QQ plots. These plots provide a quick way to see if the data follows a normal distribution.
- **Analyzing the Plots:** If the plots depict the data as bell-shaped or symmetrically distributed, the normality assumption is likely met. However, if the plots show skewed distributions, this might be an issue, especially with small sample sizes. In that case, consider a nonparametric test or further validate the data.

Ensuring normality in your data helps in drawing accurate conclusions from statistical analyses, like ANOVA, thereby enhancing the reliability of your results.