Answer:
Explanation:
A) Confidence interval is written as
Sample proportion ± margin of error
Margin of error = z × √pq/n
Where
z represents the z score corresponding to the confidence level
p = sample proportion. It also means probability of success
q = probability of failure
q = 1 - p
p = x/n
Where
n represents the number of samples
x represents the number of success
From the information given,
n = 240
x = 138
p = 138/240 = 0.58
q = 1 - 0.58 = 0.42
To determine the z score, we subtract the confidence level from 100% to get α
α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025
This is the area in each tail. Since we want the area in the middle, it becomes
1 - 0.025 = 0.975
The z score corresponding to the area on the z table is 1.96. Thus, the z score for a confidence level of 95% is 1.96
Therefore, the 95% confidence interval is
0.58 ± 1.96√(0.58)(0.42)/240
Confidence interval is
0.58 ± 0.062
B) winning more than halve of the games would be winning 120 games and above.
p = 120/240 = 0.5
We would set up the hypothesis test.
For the null hypothesis,
P ≥ 0.5
For the alternative hypothesis,
P < 0.5
Considering the population proportion, probability of success, p = 0.5
q = probability of failure = 1 - p
q = 1 - 0.5 = 0.5
Considering the sample,
Sample proportion, P = x/n
Where
x = number of success = 138
n = number of samples = 240
P = 138/240 = 0.58
We would determine the test statistic which is the z score
z = (P - p)/√pq/n
z = (0.58 - 0.5)/√(0.5 × 0.5)/240 = 2.48
Recall that this is a left tailed test. We would determine the probability value of the area to the right of the z score from the normal distribution table.
P value = 1 - 0.9934 = 0.0066
Since alpha, 0.01 > the p value, 0.0066, then we would reject the null hypothesis.
Therefore, At the 0.01 significance level, there is no strong evidence of a home field advantage (they win more than half of the games) in professional football.