TEST OF HYPOTHESIS ABOUT
mUSING A Z-SCORE
A nutritionist believes that 12-ounce boxes of cereal should contain an average of 1.2 ounces of bran. She suspects that a popular cereal has a different mean bran content. She carefully analyzes the contents of a random sample of twenty 12-ounce boxes of the cereal and finds that the mean bran content is 1.16 ounces. It is known that the standard deviation of the bran content of all such boxes of cereal is 0.08.A. Do the data provide sufficient evidence to conclude that the mean bran content of all 12-ounce boxes of this cereal differs from 1.2 ounces? Use a .05 level of significance.
B. State when this procedure is valid.
SOLUTION
A. The parameter of interest is
m, the mean bran content of all 12-ounce boxes of this cereal. It is desired to test a hypothesis about m.
Decision Rule: Accept Ha if the calculated p-value < .05.
p-value = 2(.5 - .4875) = .0250
Interpretation: At a .05 level of significance it can be concluded that the mean bran content of all 12-ounce boxes of this cereal is less than 1.2 ounces.
A graph of the appropriate z-distribution and the resulting p-value is below.
B. Since a random sample of n = 20 boxes of cereal was observed, and n = 20 < 30, the population of all bran contents must be normally distributed for this procedure to be valid. If a large random sample had been obtained, the population would not need to have a normal distribution.
COMMENTS ABOUT THE SOLUTION

The following comments will apply to all the test of hypotheses problems you solve in this course.
a; Give the decision rule; Give the formula for the test statistic; Calculate the value of the test statistic and the p-value; Interpret the results of your calculations in terms of the problem. This will include the variable, units, etc. of the problem.State the null hypothesis, H0; State the alternative hypothesis, H