"An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in Texas. Suppose that the mean income is found to be $20.3 for a random sample of 2333 people. Assume the population standard deviation is known to be $11.3. Construct the 98% confidence interval for the mean per capita income in thousands of dollars. Round your answers to one decimal place."

($19.8, $20.8)

Step-by-step explanation:

Confidence interval for mean : , where = Sample mean, n= sample size, = population standard deviation, z* = two tailed critical z-value.

Given:   = $20.3

n= 2333

= $11.3

For  98% confidence,  z* = 2.326

Then, the98% confidence interval for population mean will be:

Hence,  the 98% confidence interval for the mean per capita income in thousands of dollars = (19.8, 20.8)

i think the answer is d im not sure

step-by-step explanation:

your answer is a