A local juice manufacturer distributes juice in bottles labeled 32 ounces. A gov

Question

A local juice manufacturer distributes juice in bottles labeled 32 ounces. A government agency thinks that the company is cheating its customers. The agency selects 25 of these bottles, measures their contents, and obtains a sample mean of 31.7 ounces with a standard deviation of 0.70 ounce. Use a 0.05 significance level to test the agency's claim that the company is cheating its customers.

A. The appropriate null hypothesis for this situation is: ____ (select a letter option: a, b, c, or d.)

B. The appropriate alternative hypothesis for this situation is: ____ (select a letter option: a, b, c, d, e, f, g or h.)

C. The statistic of the test is = ____ and the p-value is ____ (round answers to the nearest thousandth. Do not enter the 0 before the decimal point).

D. There ____(is /is not) enough statistical evidence, at the 1% level, that the company is cheating its customers.

Explanation / Answer

Here the choice of options are not visible. i am providing the answers to ghe given questions.

a) The average juice in the bottle =32 ounces

b) The average juice is less than 3 ounces (cheating in the sense providing less than the labelled)

c)statistic of the test t- value = -2.1429 and the P-value is .0212

d)There is not enough statistical evidence , at the 1% level, that the company is cheating tis customers.

Note : In the problem description, you mentioned it 5% level, but problem (d) it is 1% , so conclusion will change for 5% and 1%

Calcuations given below for 5%

Data Null Hypothesis                m= 32 Level of Significance 0.05 Sample Size 25 Sample Mean 31.7 Sample Standard Deviation 0.7 Intermediate Calculations Standard Error of the Mean 0.14 Degrees of Freedom 24 t Test Statistic -2.142857143 Lower-Tail Test Lower Critical Value -1.710882067 p-Value 0.021236064

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