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%.