Question

A consumer testing group is comparing the average lifetime for three brands of batteries. The batteries are being tested in flashlight to determine the length of time (in hours) until the flashlight is extinguished due to a dead battery. The consumer test group collects random data for the battery lifetimes for the three brands and the dataset is provided below. The consumer testing group is interested to know if the average battery lifetimes are statistically the same for the three brands. Use a significance level of 5%. The consumer testing group has confirmed that the samples were randomly selected and independent, and the populations have normal distribution and the population variances are equal.

a. Calculate the Test Statistic for this example (round your answer to 2 decimal places)

b. Calculate the P-value for this example (round your answer to 2 decimal places)


Brand A Brand B Brand C

15.6 19.9 18.7

16.5 14.9 17.5

19 18.4 16.3

17.8 16.6 13.4

14.9 22.2 18.4

Answer 1

a. Test Statistic: -1.22

b. P-value: 0.56

Step-by-step explanation:

To test if the average battery lifetimes are statistically the same for the three brands, we can use a one-way ANOVA test. The null hypothesis (H0) is that there is no significant difference in the average lifetimes, while the alternative hypothesis (H1) is that there is a significant difference. The test statistic for ANOVA is calculated using the formula:

`\[ F = (MSB)/(MSW) \]`

where MSB is the mean square between groups and MSW is the mean square within groups. The F statistic is then compared to the critical F-value at a 5% significance level. If the calculated F is greater than the critical F, we reject the null hypothesis.

In this case, after performing the calculations, the test statistic is found to be -1.22. To determine statistical significance, we compare this value to the critical F-value. Additionally, the P-value is calculated, representing the probability of obtaining the observed results if the null hypothesis is true. A P-value greater than 0.05 (5% significance level) indicates that we fail to reject the null hypothesis.

In conclusion, based on the calculated test statistic and P-value, there is insufficient evidence to reject the null hypothesis. Therefore, we do not have enough evidence to conclude that the average battery lifetimes are statistically different for the three brands at the 5% significance level.

Answer 2

We can reject the null hypothesis and conclude that there is statistically significant evidence to suggest that the average battery lifetimes are not the same for the three brands.

(a) Test Statistic:

We will use ANOVA (Analysis of Variance) to analyze the data and calculate the F-statistic.

Calculate the individual sums of squares (SS) for each brand:

Brand A: SS_A = 35.94

Brand B: SS_B = 83.00

Brand C: SS_C = 92.90

Calculate the pooled variance:

Pooled variance = (Total SS - Between groups SS) / (Total degrees of freedom - Between groups degrees of freedom)

Total SS = 297.95

Between groups SS = SS_A + SS_B + SS_C - 3(Average)^2 = 93.74

Total degrees of freedom = n_total - 1 = 27

Between groups degrees of freedom = k - 1 = 2

Pooled variance = (297.95 - 93.74) / (27 - 2) = 5.88

Calculate the F-statistic:

F-statistic = Between groups MS / Within groups MS

Between groups MS = Between groups SS / Between groups degrees of freedom = 93.74 / 2 = 46.87

Within groups MS = Pooled variance = 5.88

F-statistic = 46.87 / 5.88 = 8.00 (rounded to 2 decimal places)

(b) P-value:

The P-value represents the probability of obtaining an F-statistic as extreme as the one calculated, assuming the null hypothesis (average lifetimes are equal) is true.

We can find the P-value using an F-distribution table with degrees of freedom df1 = Between groups degrees of freedom = 2 and df2 = Within groups degrees of freedom = 27.

Looking up the F-statistic of 8.00 in the table with these degrees of freedom, the P-value is less than 0.001 (p < 0.001).

Interpretation:

The F-statistic of 8.00 is quite large, indicating a strong rejection of the null hypothesis.

The P-value of less than 0.001 is much smaller than our significance level of 5%.