A survey of an urban university (population of 25,450) showed that 750 of 1100 students sampled attended a home football game during the season. What inferences can be made about student attendance at football games?

Using the 90% level of confidence, what is the confidence interval?

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To determine the confidence interval for student attendance at football games, we need to make some assumptions and perform a calculation using the provided data.

Assumptions:
1. The sample of 1100 students is randomly selected from the population of 25,450.
2. The sampled students are representative of the entire student population.
3. The number of students attending a home football game during the season is a binomial distribution.

Calculation:

1. Calculate the sample proportion (p̂):
p̂ = Number of students attended / Total sample size
= 750 / 1100
= 0.6818 (rounded to four decimal places)

2. Calculate the standard error (SE):
SE = √[(p̂ * (1-p̂)) / n]
= √[(0.6818 * (1-0.6818)) / 1100]
≈ 0.0148 (rounded to four decimal places)

3. Determine the critical value for a 90% confidence interval. Since the sample size is large (1100) and the data assumed to follow a normal distribution, we can use a z-table or calculator to find this value. The critical value for a 90% confidence interval is approximately 1.645.

4. Calculate the margin of error (ME):
ME = Critical value * SE
= 1.645 * 0.0148
≈ 0.0244 (rounded to four decimal places)

5. Calculate the lower and upper bounds of the confidence interval:
Lower bound = p̂ - ME
= 0.6818 - 0.0244
≈ 0.6574 (rounded to four decimal places)

Upper bound = p̂ + ME
= 0.6818 + 0.0244
≈ 0.7062 (rounded to four decimal places)

Therefore, using a 90% level of confidence, the confidence interval for student attendance at football games is approximately 0.6574 to 0.7062. This means we can be 90% confident that the true proportion of students attending football games falls within this interval.