Question

Starting salaries of 64 college graduates who have taken a statistics course have a mean of $43,500 with a standard deviation of $6,800. Find an 80% confidence interval for ? μ . (NOTE: Do not use commas or dollar signs in your answers. Round each bound to three decimal places.) Find upper and lower bounds.

Answer 1

Solution

Given that,

= 62.5

= 6800

n = 64

At 80% confidence level the z is ,

= 1 - 80% = 1 - 0.80 = 0.20

/ 2 = 0.20 / 2 = 0.10

Z/2 = Z 0.10 = 1.282

Margin of error = E = Z/2* (/n)

= 1.282 * (6800 / 64)

= 1089.312

At 80% confidence interval estimate of the population mean is,

- E < < + E

43500 - 1089.312 < < 43500 + 1089.312

42410..681 < < 44589.329

Lower bound = 42410.681

Lower bound= 44589.329