Question

Use the given confidence level and sample data to find a confidence interval for the population standard deviation sigma. Assume that a simple random sample has been selected from a population that has a normal distribution. Salaries of college graduates who took a statistics course in college 80​% ​confidence; nequals61​, x overbarequals​$56 comma 600​, sequals​$16 comma 432

Answer 1

Solution :

Given that,

= 56,600

s = 16432

n = 61

Degrees of freedom = df = n - 1 = 61 - 1 = 60

At 80% confidence level the t is ,

= 1 - 80% = 1 - 0.80 = 0.20

  / 2 = 0.20 / 2 = 0.10

t _/2,df = t_0.10,60 = 1.296

Margin of error = E = t_/2,df * (s /n)

= 1.296 * ( 16432 / 61)

= 2726.66

The 95% confidence interval estimate of the population mean is,

- E < < + E

56600 - 2726.66 < < 56600 + 2726.66

53873.34 < < 59326.66

(53873.34, 59326.66 )