Seconds Speed
144 37.5
149.2 36.193029
148.2 36.437247
149 36.241611
149.6 36.096257
146.8 36.784741
148.6 36.339166
149.8 36.048064
149 36.241611
148.2 36.437247
147.8 36.535859
147.2 36.684783
147 36.734694
149.8 36.048064
148.2 36.437247
The above data is the column of seconds and speed .calculate independent sample t test for equal variance. Using alpha 0.05.
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.
To calculate an independent sample t-test for equal variance, we need to follow these steps:
Step 1: State the null hypothesis (H0) and the alternative hypothesis (HA).
H0: The means of the two groups are equal.
HA: The means of the two groups are not equal.
Step 2: Set a significance level (alpha).
In this case, alpha (α) is given as 0.05.
Step 3: Calculate the mean and standard deviation for each group.
Group 1 (Seconds):
Mean1 = (144 + 149.2 + 148.2 + 149 + 149.6 + 146.8 + 148.6 + 149.8 + 149 + 148.2 + 147.8 + 147 + 147) / 15
= 148.253333
Standard deviation1 = sqrt(((144-148.253333)^2 + (149.2-148.253333)^2 + (148.2-148.253333)^2 + ... + (147-148.253333)^2) / 14)
= sqrt(0.802222/14)
= 0.162726
Group 2 (Speed):
Mean2 = (37.5 + 36.193029 + 36.437247 + 36.241611 + 36.096257 + 36.784741 + 36.339166 + 36.048064 + 36.241611 + 36.437247 + 36.535859 + 36.684783 + 36.734694 + 36.048064 + 36.437247) / 15
= 36.478059
Standard deviation2 = sqrt(((37.5-36.478059)^2 + (36.193029-36.478059)^2 + (36.437247-36.478059)^2 + ... + (36.437247-36.478059)^2) / 14)
= sqrt(0.065444/14)
= 0.041773
Step 4: Calculate the test statistic (t-value):
t = (Mean1 - Mean2) / sqrt((Standard deviation1^2 / n1) + (Standard deviation2^2 / n2))
= (148.253333 - 36.478059) / sqrt((0.162726^2 / 15) + (0.041773^2 / 15))
= 111.7752745 / sqrt(0.0170738 + 0.001097)
= 111.7752745 / sqrt(0.0181708)
≈ 111.7752745 / 0.134877
≈ 828.487
Step 5: Determine the degrees of freedom (df).
df = n1 + n2 - 2
= 15 + 15 - 2
= 28
Step 6: Find the critical value based on alpha and degrees of freedom.
For alpha = 0.05 and 2-tailed test:
The critical value (t-critical) is approximately ±2.048 (lookup from t-distribution table).
Step 7: Make a decision.
If |t| > t-critical, reject the null hypothesis (H0).
If |t| <= t-critical, fail to reject the null hypothesis (H0).
In this case, |t| = 828.487 and t-critical = 2.048.
Since |t| > t-critical, we reject the null hypothesis (H0).
Therefore, there is enough evidence to suggest that the means of the two groups (seconds and speed) are not equal.