An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in California. Suppose that the mean income is found to be $32.6 for a random sample of 459 people. Assume the population standard deviation is known to be $9.3 . Construct the 80% confidence interval for the mean per capita income in thousands of dollars. Round your answers to one decimal place.
80% confidence interval for is
- z * / sqrt(n) < < + z * / sqrt(n)
32.6 - 1.2816 * 9.3 / sqrt(459) < < 32.6 + 1.2816 * 9.3 / sqrt(459)
32.0 < < 33.2
80% CI is ( 32.0 , 33.2 )
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