Classwork:
Given and
, we have
(1)
Using the composition found in (1), we have
(2)
(3)
Using the composition found in (3),
(4)
Homework:
Now if , we would have
(1)
For (2), we could explicitly find then evaluate it at <em>x</em> = -1 like we did in the classwork section, but we don't need to.
(2)
(3) We can demonstrate that both methods work here:
• by using the result from (1),
• by evaluating the inner function at <em>x</em> = 2 first,
Answer:
The p-value is less than the significance level, so we will reject the null hypothesis, and conclude that at 5% significance level, the proportion for women that ran 5 km is more than the proportion of male that ran the 5 km. Thus, women are more likely to run the 5 km race.
Step-by-step explanation:
We are given;
Number of males selected; n₁ = 150
Number of females selected; n₂ = 190
Proportion of the males; p₁ = 0.6
Proportion of females; p₂ = 0.5
Let's define the hypotheses;
Null hypothesis; H0: p₁ ≥ p₂
Alternative hypothesis; Ha: p₁ < p₂
Now, the z score formula for this is;
z = (p₁ - p₂)/√(p^(1 - p^))(1/n₁ + 1/n₂))
Where;
p^ = (p₁ + p₂)/2
p^ = (0.6 + 0.5)/2
p^ = 0.55
Thus;
z = (0.6 - 0.5)/√(0.55(1 - 0.55))((1/150) + (1/190))
z = 0.1/0.0543
z = 1.84
From online p-value from z-score calculator attached, using z = 1.84, significance level = 0.05, one tailed hypothesis, we have;
P-value = 0.033
The p-value is less than the significance level, so we will reject the null hypothesis, and conclude that at 5% significance level, the proportion for women that ran 5 km is more than the proportion of male that ran the 5 km. Thus, women are more likely to run the 5 km race.
Although if we check at 0.01 significance level, we will not have the same answer as the p-value will be greater.
