Question

The price of coffee has been steadily declining since October 2012. The International Coffee Organization (ICO) reported that the mean price for coffee in May 2013 was 126.96 U.S. cents/lb. A random sample of 34 days was obtained and the composite indicator for Brazilian Natural coffee was recorded for each. The sample mean was

x = 130.29 U.S. cents/lb.

Assume the population standard deviation is 8.6 U.S. cents/lb. Is there any evidence to suggest that the mean composite indicator for Brazilian Natural is greater than 126.96? Use

α = 0.05.

a) Give the value of the appropriate zα or zα/2. (Round your answer to four decimal places.)

b) Calculate the test statistic. (Round your answer to four decimal places.)

z =

Answer 1

-1.96 and 2.237

Explanation:

Given that the price of coffee has been steadily declining since October 2012.

`H_0: \bar x = 126.96\\H_a: \bar x <126.96`

(Left tailed test)

Sample mean= 130.29

Mean difference
`= 130.29-126.96 = 3.33`

Population std dev is given so we use z test.

Z statistic =
`(3.33)/((8.6)/(√(34) ) ) \\=(3.33)/(1.475) \\=2.237`

a) Here one tailed hence Z alpha = -1.96 for 95%

b) Test statistic = 2.237