Final answer:
To construct a 92% confidence interval for the population mean spending of a sample, we use a Z-value of approximately 1.753 and calculate the margin of error. Adding and subtracting this from the sample mean of $1838.53, we get a confidence interval of $1831.00 to $1846.06, within which the true population mean likely resides.
Step-by-step explanation:
To construct a confidence interval for the population mean with a known standard deviation using a specific confidence level, we employ the formula for a confidence interval for a normal distribution:
Confidence interval = µ ± (Z* × (σ/sqrt(n)))
where:
- µ is the sample mean,
- Z* is the Z-value corresponding to the confidence level from the standard normal distribution,
- σ is the population standard deviation, and
- n is the sample size.
Given the sample mean (µ) is $1838.53, the sample standard deviation (σ) is $215.76, the sample size (n) is 2530, and the required confidence level is 92%, we must first find the Z-value that corresponds to a 92% confidence level. Consulting a standard normal (Z) distribution table or calculator, the Z-value is approximately 1.753.
Next, calculate the margin of error (E) using the Z-value and the standard deviation:
E = Z* × (σ/sqrt(n))
E = 1.753 × ($215.76/sqrt(2530))
E ≈ 1.753 × ($215.76/50.2987)
E ≈ 1.753 × 4.2916
E ≈ 7.5284
Finally, add and subtract this margin of error from the sample mean to find the confidence interval:
Lower limit = µ - E = $1838.53 - 7.5284 ≈ $1831.00
Upper limit = µ + E = $1838.53 + 7.5284 ≈ $1846.06
Thus, we are 92% confident that the true population mean is between $1831.00 and $1846.06.