Question

The U.S. Dairy Industry wants to estimate the mean yearly milk consumption. A sample of 20 people reveals the mean yearly consumption to be 72 gallons with a standard deviation of 15 gallons. Assume the population distribution is normal. (Use t Distribution Table.)

What is the best estimate of this value?

For a 99% confidence interval, what is the value of t? (Round your answer to 3 decimal places.)

Develop the 99% confidence interval for the population mean. (Round your answers to 3 decimal places.)

Answer 1

Solution :

Given that,

Best estimate = = 72

s = 15

n = 20

Degrees of freedom = df = n - 1 = 20 - 1 = 19

At 99% confidence level the t is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,19 = 2.861

t value = 2.861

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

= 2.861 * (15 / 20)

= 9.596

The 99% confidence interval estimate of the population mean is,

- E < < + E

72 - 9.596 < < 72 + 9.596

62.404 < < 81.596

(62.404 , 81.596)