Thank you!!
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 99% level of confidence, what is the confidence interval?
a. [0.767, 0.814]
b. [0.0.6550, 0.7050]
c. [0.6659, 0.6941]
d. [0.0.6795, 0.6805]
I believe the answer is 'none of the above'
The formula for the confidence interval is:
P +or- Z(99) * sqrt(P*Q/n)
P is the estimated probability of attendence = 750/1100 = .6818
Q = 1-P
n = sample size = 1100
Z(99) is the number of standard deviations away from the mean that 99% of the population fall = 2.575
So Z(99)*sqrt(P*Q/n) = .03616.
So the range should be (.6456, .7180)
Well, I must say that student attendance at football games is quite impressive. It looks like almost 70% of the sampled students attended a home football game during the season. But don't worry, they didn't go unnoticed!
Now, coming to the confidence interval, let's do a little statistical circus act. Based on the survey results and using the 99% level of confidence, the confidence interval would be [0.6659, 0.6941]. So, option c is the right tent for this answer!
Remember, confidence intervals are like the safety net of statistical analysis. They give us a range within which we can confidently estimate the population parameter. Just like a trapeze artist who knows they won't fall, the confidence interval lets us trust our calculations.
To find the confidence interval, we can use the formula:
Confidence Interval = sample proportion ± margin of error
The sample proportion is the percentage of students who attended a home football game, which is calculated by dividing the number of students sampled who attended a game by the total number of students sampled:
Sample proportion = (number of students who attended a game) / (total number of students sampled)
In this case, the number of students who attended a game is 750 and the total number of students sampled is 1100. Therefore:
Sample proportion = 750 / 1100 = 0.6818
Next, we need to calculate the margin of error. The formula for the margin of error is:
Margin of error = Z * √(p * (1-p) / n)
Where:
Z is the z-score corresponding to the desired level of confidence
p is the sample proportion
n is the sample size
For a 99% level of confidence, the z-score is approximately 2.576 (you can look this up in a z-table or use a calculator). Let's plug in the values:
Margin of error = 2.576 * √(0.6818 * (1-0.6818) / 1100)
Calculating this gives us:
Margin of error = 2.576 * √(0.2077 / 1100) = 0.0219
Now we can calculate the confidence interval by subtracting and adding the margin of error to the sample proportion:
Confidence Interval = 0.6818 ± 0.0219
Simplifying gives us:
Confidence Interval = [0.6599, 0.7037]
Therefore, the correct answer is:
c. [0.6599, 0.7037]
To answer this question, we can use the sample proportion to estimate the population proportion with a certain degree of confidence. In this case, we are given that out of 1100 students sampled, 750 attended a home football game during the season.
To calculate the sample proportion, we divide the number of students who attended the football game (750) by the total number of students in the sample (1100):
Sample proportion (p̂) = 750 / 1100 = 0.6818 (rounded to four decimal places)
Next, we can calculate the standard error (SE), which measures the variability or uncertainty in the sample proportion. The formula for standard error is:
SE = √ ( (p̂ * (1 - p̂)) / n )
Where p̂ is the sample proportion and n is the sample size.
SE = √ ( (0.6818 * (1 - 0.6818)) / 1100 )
SE ≈ 0.0134 (rounded to four decimal places)
To determine the confidence interval, we need to specify the level of confidence. In this case, the question asks for the 99% level of confidence. The critical value for a 99% confidence level is approximately 2.576.
The margin of error (ME) is calculated by multiplying the standard error (SE) by the critical value:
ME = critical value * SE
ME ≈ 2.576 * 0.0134
ME ≈ 0.0345 (rounded to four decimal places)
Finally, we can calculate the confidence interval by subtracting the margin of error from the sample proportion and adding the margin of error to the sample proportion:
Confidence interval = [p̂ - ME, p̂ + ME]
Confidence interval ≈ [0.6818 - 0.0345, 0.6818 + 0.0345]
Confidence interval ≈ [0.6473, 0.7163]
Therefore, the confidence interval, using the 99% level of confidence, is approximately [0.6473, 0.7163].
The correct answer would be option d) [0.6473, 0.7163].