Question 1203976: Three assembly lines are used to produce a certain component for an airliner. To examine the production rate, a random sample of six hourly periods is chosen for each assembly line and the number of components produced during these periods for each line is recorded. The output from a statistical software package is:
Summary
Groups Sample Size Sum Average Variance
Line A 6 250 41.66667 0.266667
Line B 6 260 43.33333 0.666667
Line C 6 249 41.5 0.7
ANOVA
Source of Variation SS df MS F p-value
Between Groups 12.33333 2 6.166667 11.32653 0.001005
Within Groups 8.166667 15 0.544444
Total 20.5 17
Compute 99% confidence intervals that estimate the difference between each pair of means. (Negative amount should be indicated by a minus sign. Round your answers to 2 decimal places.) Which pairs of means are statistically different?
Answer by asinus(45)   (Show Source):
You can put this solution on YOUR website! To compute the 99% confidence intervals for the difference between each pair of means and determine which pairs are statistically different, we follow these steps:
---
### **Step 1: Data Summary**
From the summary table, the means, variances, and sample sizes for each group are:
- \( \bar{X}_A = 41.67, \, \bar{X}_B = 43.33, \, \bar{X}_C = 41.50 \)
- \( s_A^2 = 0.27, \, s_B^2 = 0.67, \, s_C^2 = 0.70 \)
- \( n_A = n_B = n_C = 6 \)
From the ANOVA table:
- \( MS_{\text{Within}} = 0.544444 \) (pooled variance estimate).
---
### **Step 2: Pooled Standard Error for the Difference Between Means**
The pooled standard error for the difference between two group means is given by:
\[
SE = \sqrt{MS_{\text{Within}} \cdot \left( \frac{1}{n_1} + \frac{1}{n_2} \right)}
\]
Substitute \( MS_{\text{Within}} = 0.544444 \) and \( n_1 = n_2 = 6 \):
\[
SE = \sqrt{0.544444 \cdot \left( \frac{1}{6} + \frac{1}{6} \right)} = \sqrt{0.544444 \cdot \frac{2}{6}} = \sqrt{0.181481} \approx 0.426.
\]
---
### **Step 3: Critical Value for 99% Confidence Interval**
For a 99% confidence level and \( df = 15 \) (degrees of freedom from the within-group variation), the critical \( t \)-value is:
\[
t_{0.005, 15} \approx 2.947.
\]
---
### **Step 4: Compute Confidence Intervals for Each Pair of Means**
The confidence interval for the difference between two means is:
\[
(\bar{X}_1 - \bar{X}_2) \pm t \cdot SE
\]
#### **Pair 1: \( \bar{X}_A - \bar{X}_B \)**
\[
\bar{X}_A - \bar{X}_B = 41.67 - 43.33 = -1.66
\]
\[
CI = -1.66 \pm 2.947 \cdot 0.426
\]
\[
CI = -1.66 \pm 1.255
\]
\[
CI = [-2.92, -0.41]
\]
#### **Pair 2: \( \bar{X}_A - \bar{X}_C \)**
\[
\bar{X}_A - \bar{X}_C = 41.67 - 41.50 = 0.17
\]
\[
CI = 0.17 \pm 2.947 \cdot 0.426
\]
\[
CI = 0.17 \pm 1.255
\]
\[
CI = [-1.09, 1.42]
\]
#### **Pair 3: \( \bar{X}_B - \bar{X}_C \)**
\[
\bar{X}_B - \bar{X}_C = 43.33 - 41.50 = 1.83
\]
\[
CI = 1.83 \pm 2.947 \cdot 0.426
\]
\[
CI = 1.83 \pm 1.255
\]
\[
CI = [0.58, 3.08]
\]
---
### **Step 5: Determine Which Pairs are Statistically Different**
For a pair of means to be statistically different, the confidence interval must **not include 0**:
1. **\( \bar{X}_A - \bar{X}_B \):** CI = \([-2.92, -0.41]\). This does **not include 0**, so \( \bar{X}_A \) and \( \bar{X}_B \) are statistically different.
2. **\( \bar{X}_A - \bar{X}_C \):** CI = \([-1.09, 1.42]\). This includes 0, so \( \bar{X}_A \) and \( \bar{X}_C \) are **not** statistically different.
3. **\( \bar{X}_B - \bar{X}_C \):** CI = \([0.58, 3.08]\). This does **not include 0**, so \( \bar{X}_B \) and \( \bar{X}_C \) are statistically different.
---
### **Final Results:**
#### Confidence Intervals:
- \( \bar{X}_A - \bar{X}_B \): \([-2.92, -0.41]\)
- \( \bar{X}_A - \bar{X}_C \): \([-1.09, 1.42]\)
- \( \bar{X}_B - \bar{X}_C \): \([0.58, 3.08]\)
#### Statistically Different Pairs:
- \( \bar{X}_A \) and \( \bar{X}_B \)
- \( \bar{X}_B \) and \( \bar{X}_C \)
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