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MrMuchimi
3 years ago
8

During the 2000 season, the home team won 138 out of 240 regular season National Football League games. (15 points) a) Construct

a 95% confidence interval for the winning proportion of the home team during this season. b) At the 0.01 significance level, is there strong evidence of a home field advantage (they win more than half of the games) in professional football? State hypotheses, calculate the test statistic and p-value, and make a conclusion in context
Mathematics
1 answer:
jeka943 years ago
3 0

Answer:

Step-by-step 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.

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