To perform an independent sample t-test for equal variances, we need to compare the means of two independent groups. In this case, the groups would be the speeds at two different time intervals.
First, we need to calculate the means and variances of each group:
Group 1 (144, 149.2, 148.2, 149, 149.6, 146.8, 148.6, 149.8, 149, 148.2, 147.8, 147.2, 147)
Mean 1 = (144 + 149.2 + 148.2 + 149 + 149.6 + 146.8 + 148.6 + 149.8 + 149 + 148.2 + 147.8 + 147.2 + 147) / 13 = 147.973077
Variance 1 = Σ (xi - mean 1)² / (n1 - 1)
= ((144 - 147.973077)² + (149.2 - 147.973077)² + (148.2 - 147.973077)² + (149 - 147.973077)² + (149.6 - 147.973077)² + (146.8 - 147.973077)² + (148.6 - 147.973077)² + (149.8 - 147.973077)² + (149 - 147.973077)² + (148.2 - 147.973077)² + (147.8 - 147.973077)² + (147.2 - 147.973077)² + (147 - 147.973077)²) / (13 - 1)
= 1.899798
Group 2 (149.8, 148.2)
Mean 2 = (149.8 + 148.2) / 2 = 149
Variance 2 = Σ (xi - mean 2)² / (n2 - 1)
= ((149.8 - 149)² + (148.2 - 149)²) / (2 - 1)
= 0.02
To perform the t-test, we can use the following formula:
t = (mean 1 - mean 2) / √[(variance 1 / n1) + (variance 2 / n2)]
t = (147.973077 - 149) / √[(1.899798 / 13) + (0.02 / 2)]
t = -1.026923 / √[0.146147385 + 0.01]
t = -1.026923 / √0.156147385
t = -1.026923 / 0.394109432
t = -2.60779
Next, we need to determine the degrees of freedom for the t-distribution, which can be calculated using the formula:
df = n1 + n2 - 2
df = 13 + 2 - 2
df = 13
Using a significance level (alpha) of 0.05 and the calculated t-value of -2.60779, we can determine the critical t-value from the t-distribution table for a two-tailed test.
The critical t-value for a two-tailed test with 0.05 alpha and 13 degrees of freedom is approximately ±2.16037.
Since -2.60779 is less than -2.16037, which means it falls in the critical region, we can reject the null hypothesis.
Therefore, there is evidence to suggest that there is a significant difference in the speeds between the two time intervals.