Running heats. In Olympic running events, preliminary heats are determined by random draw, so we should expect that the abilities of runners in the various heats to be about the same, on average. Here...

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Test: µ1 6= µ2
Assumption is that both the samples are drawn independently and randomly.
Next, there are 2 ways to test this, one assuming equal population variance for 2
populations (pooled variance method), and the other is without this assumption.
Select the appropriate one, as different professors prefer different ones.
Heat 2:
51.02, 51.29, 51.5, 52.04, 52.1, 52.53, 56.01
We calculate sample mean and std deviation from given data.
Sample Mean, x̄ =
∑
(x)
n
= 366.497 = 52.355714
Sample Variance, s2 =
∑
(x− x̄)2
n− 1 =
17.201371
6 = 2.866895
Sample std dev, s =
√
s2 =
√
2.866895 ≈ 1.693191
Heat 5:
51.2, 51.37, 51.53, 51.9, 52.85, 52.87, 54.58
We calculate sample mean and std deviation from given data.
Sample Mean, x̄ =
∑
(x)
n
= 366.37 = 52.328571
Sample Variance, s2 =
∑
(x− x̄)2
n− 1 =
8.647886
6 = 1.441314
Sample std dev, s =
√
s2 =
√
1.441314 ≈ 1.200547
Test: µ1 6= µ2 using equal variance assumption
Hypothesis test:
H0 : µ1 − µ2 = 0 (Null Hypothesis)
Ha : µ1 − µ2 6= 0 (Alternative Hypothesis, also called H1)
This is two tailed test.
sample 1 sample 2
x1 = 52.355714 x2 = 52.328571
s1 = 1.693191 s2 = 1.200547
n1 = 7 n2 = 7
Significance Level, α = 0.05 (If no value...