Question

How much money does the average professional football fan spend on food at a single football game? That question was posed to 60 randomly selected football fans. The sampled results show thatthe sample mean was $70.00 and prior sampling indicated that the population standard deviation was $17.50. Use this information to create a 95 percent confidence interval for the the average professional football fan spend on food at a single football game.

Answer 1

To create a 95% confidence interval for the average amount of money spent on food by a professional football fan at a single game, we need to use the sample information provided and the population standard deviation. Given that we have a sample size of 60, a sample mean of $70.00, and a population standard deviation of $17.50, we can proceed with the following steps:

1. First, we need to find the critical value for a 95% confidence interval. Since the normal distribution is symmetric, we look for the z-score that corresponds to the upper half of 2.5% (as 95% confidence leaves out 2.5% on both ends). Using a standard normal distribution table or a calculator, we find that the z-score for 97.5 percentile is approximately 1.96.

2. The next step is to compute the standard error of the mean. This is the estimated standard deviation of the sample mean and can be calculated by dividing the population standard deviation by the square root of the sample size:
Standard Error = Population Standard Deviation / sqrt(Sample Size)

Here, it looks like this:
Standard Error = 17.50 / sqrt(60)

3. Multiply the z-score by the standard error to get the margin of error. This will tell us how much we should add and subtract from the sample mean to get our confidence interval.
Margin of Error = z-score * Standard Error

For our data, it is:
Margin of Error ≈ 1.96 * (17.50 / sqrt(60))

4. Subtract the margin of error from the sample mean to get the lower bound of the confidence interval and add it to the sample mean to get the upper bound of the confidence interval. This will give us the range in which we believe the true mean lies with 95% confidence.
Lower Bound = Sample Mean - Margin of Error
Upper Bound = Sample Mean + Margin of Error

Doing the calculations (or referring to a previous calculation) yields the following results:

- z-score ≈ 1.96
- Margin of Error ≈ 4.43
- Lower Bound of Confidence Interval ≈ $70.00 - $4.43 = $65.57
- Upper Bound of Confidence Interval ≈ $70.00 + $4.43 = $74.43

Thus, we are 95% confident that the average professional football fan spends between $65.57 and $74.43 on food at a single football game.

Answer 2

$70.00 +/- $4.43

= ( $65.57, $74.43)

Therefore at 95% confidence interval (a,b) = ( $65.57, $74.43)

Explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Sample Mean x = $70.00

Standard deviation r = $17.50

Number of samples n = 60

Confidence interval = 95%

z(at 95% confidence) = 1.96

Substituting the values we have;

$70.00+/-1.96($17.50/√60)

$70.00+/-1.96($2.259240285287)

$70.00+/-$4.428110959163

$70.00+/-$4.43

= ( $65.57, $74.43)

Therefore at 95% confidence interval (a,b) = ( $65.57, $74.43)