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 =

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Respuesta :

AlonsoDehner AlonsoDehner · Answer 1 · 2019-11-16T13:19:07+01:00

Answer:

-1.96 and 2.237

Step-by-step explanation:

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

[tex]H_0: \bar x = 126.96\\H_a: \bar x <126.96[/tex]

(Left tailed test)

Sample mean= 130.29

Mean difference [tex]= 130.29-126.96 = 3.33[/tex]

Population std dev is given so we use z test.

Z statistic = [tex]\frac{3.33}{\frac{8.6}{\sqrt{34} } } \\=\frac{3.33}{1.475} \\=2.237[/tex]

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

b) Test statistic = 2.237