Four brands of flashlight batteries are to be compared by testing each brand in five
flashlights. Twenty flashlights are randomly selected and divided randomly into four groups of
five flashlights each. Then each group of flashlights uses a different brand of battery. The
lifetimes of the batteries, to the nearest hour, are as follows.
BRAND
A
BRAND
B
BRAND
C
BRAND
D
42 28 24 20
30 36 36 32
39 31 28 38
28 32 28 28
29 27 33 25
Preliminary data analyses indicate that the independent samples come from normal populations
with equal standard deviations. At the 5% significance level, does there appear to be a
difference in mean lifetime among the four brands of batteries?

To determine whether there is a significant difference in the mean lifetime among the four brands of batteries, we can perform a **one-way ANOVA (Analysis of Variance)**. This method compares the means of multiple groups (in this case, the four battery brands) to see if at least one brand's mean differs significantly from the others.

### Steps for One-Way ANOVA:

1. **State the hypotheses**:
   - Null hypothesis \( H_0 \): There is no difference in the mean lifetimes among the four brands.
   - Alternative hypothesis \( H_1 \): At least one brand has a different mean lifetime.

2. **Level of significance**:
   The level of significance \( \alpha \) is 0.05 (5%).

3. **Calculate the sample means and variances** for each brand of battery:
   - Brand A: \( 42, 30, 39, 28, 29 \)
   - Brand B: \( 28, 36, 31, 32, 27 \)
   - Brand C: \( 24, 36, 28, 28, 33 \)
   - Brand D: \( 20, 32, 38, 28, 25 \)

4. **Compute the total variation** (sum of squares) and partition it into:
   - **Between-group variation**: How much the group means differ from the overall mean.
   - **Within-group variation**: The variability within each group.

5. **Calculate the F-statistic**:
   The F-statistic compares the between-group variance to the within-group variance.

6. **Compare the F-statistic to the critical F-value** from the F-distribution table, or use the p-value method to determine if \( H_0 \) should be rejected.

Let's calculate this step by step. I’ll compute the ANOVA values for you.The results of the one-way ANOVA are as follows:

- **F-statistic**: 0.739
- **p-value**: 0.544

### Conclusion:
At the 5% significance level (\( \alpha = 0.05 \)):
- Since the p-value (0.544) is greater than \( \alpha = 0.05 \), we **fail to reject the null hypothesis**.
- This means that there is no statistically significant difference in the mean lifetimes of the four brands of batteries.

Would you like more details on the calculations or the interpretation of the ANOVA? 

Here are five related questions you could explore:
1. How is the F-statistic in ANOVA calculated?
2. What assumptions must be met to use one-way ANOVA?
3. Can you perform a post-hoc test if the ANOVA result was significant?
4. How can you compute the effect size for an ANOVA?
5. What would happen if the sample sizes were different for each group?

**Tip:** Always check assumptions like normality and equal variances before applying ANOVA to ensure valid results.

Four brands of flashlight batteries are to be compared by testing each brand in five flashlights. Twenty flashlights are randomly selected and divided randomly into four groups of five flashlights each. Then each group of flashlights uses a different brand of battery. The lifetimes of the batteries, to the nearest hour, are as follows: BRAND A: 42, 30, 39, 28, 29 BRAND B: 28, 36, 31, 32, 27 BRAND C: 24, 36, 28, 28, 33 BRAND D: 20, 32, 38, 28, 25. Preliminary data analyses indicate that the independent samples come from normal populations with equal standard deviations. At the 5% significance level, does there appear to be a difference in mean lifetime among the four brands of batteries?

Learn how to use one-way ANOVA to compare the mean lifetimes of four different brands of flashlight batteries. This problem covers hypothesis testing, calculating F-statistics, and interpreting p-values at a 5% significance level. Suitable for college-level statistics students.

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