Question

A store sells "16-ounce" boxes of Captain Crisp cereal. A random sample of 12 boxes was taken and weighed. The average weight of cereal was 15.93 ounces with the sample standard deviation 0.135 ounces. The company that makes Captain Crisp cereal claims that the average weight of cereal in a box is at least 16 ounces. Assume the weight of cereal in a box is normally distributed. We wish to test H 0 : μ 16 vs H 1 : μく16 Find the value of the test statistic; the p-value (approximately) for this test; the critical value(s) at α = 0.10 level of significance: the decision at α 0.10 level of significance iv)

Answer 1

Data:    

n = 12   

μ = 16   

s = 0.135   

x-bar = 15.93   

Hypotheses:    

Ho: μ ≥ 16   

Ha: μ < 16   

Decision Rule:    

α = 0.1   

Degrees of freedom = 12 - 1 = 11

Critical t- score = -1.363430318   

Reject Ho if t < -1.363430318   

Test Statistic:    

SE = s/√n = 0.135/√12 = 0.038971143

t = (x-bar - μ)/SE = (15.93 - 16)/0.0389711431702997 = -1.796200837

p- value = 0.049973656   

Decision (in terms of the hypotheses):    

Since -1.796200837 < -1.363430318 we reject Ho and accept Ha

Conclusion (in terms of the problem):    

There is sufficient evidence that μ < 16 ounces. The company's claim is not valid.

Answers:

(i) -1.7962

(ii) 0.05

(c) -1.3634

(d) Reject Ho. The company's claim is not valid.