80 fans at a basketball game are randomly sampled and asked how much they spent on parking for the game. The mean of the sample was $12.41. Assuming a standard deviation of $1.18, find the 98% confidence interval for the sample mean. Use x=2.326. Round to the nearest cent.
To find the 98% confidence interval for the sample mean, we can use the formula:
CI = x̄ ± (z * σ/√n)
Where:
CI: Confidence Interval
x̄: Sample mean
z: Z-score corresponding to the desired confidence level (98% in this case)
σ: Standard deviation of the population
n: Sample size
Given:
x̄ = $12.41 (sample mean)
σ = $1.18 (standard deviation of the population)
n = 80 (sample size)
z = 2.326 (Z-score corresponding to 98% confidence level)
Let's substitute the values into the formula:
CI = 12.41 ± (2.326 * 1.18 / √80)
First, let's calculate (2.326 * 1.18 / √80):
(2.326 * 1.18) / √80 = 1.592848 / 2.828427 = 0.56349262
Now, substitute this value into the formula:
CI = 12.41 ± 0.56349262
To find the lower bound of the confidence interval, subtract the value we just calculated from the sample mean:
Lower bound = 12.41 - 0.56349262 = 11.84650738
To find the upper bound, add the value we just calculated to the sample mean:
Upper bound = 12.41 + 0.56349262 = 12.97349262
Therefore, the 98% confidence interval for the sample mean is approximately $11.85 to $12.97.