Question

A simple random sample of 36 bottles of Helo, a healthy juice mix, has a mean volume of 12.19 oz. Assume that the standard deviation of all bottles of Helo is 0.11 oz. Use a 0.01 significance level to test the claim that bottles of Helo have volumes with a mean of 12 oz, as stated on the label. What is your final conclusion. Could anyone show me how to do this?

 

Answer 1

SAMPLE STATISTICS: mean volume (x bar)=12.19 fluid oz, size (n)=36 bottles.

POPULATION STATISTIC: standard deviation (σ)=0.11 fluid oz

CLAIM: Mean volume is 12 fluid oz

COUNTERCLAIM: Mean volume is not 12 fluid oz

SIGNIFICANCE LEVEL: ɑ=0.01 (99% confidence level)

Null hypothesis, H₀: µ=12 (claim)

Alternative hypothesis, H₁: µ≠12 (counterclaim)

This is a 2-tail test, because “not equals” means less than (left tail) or greater than (right tail), so the significance level is split between the two tails giving us the required confidence interval of 99%.

TEST STATISTIC: Z=(x bar-µ)/(σ/√n)=(12.19-12)/(0.11/√36)=10.37, corresponding to a P-value of 0.000. ɑ/2=0.005 and sample P-value 0.000<0.005, which means that H₀ is rejected, and H₁ is accepted to be true. The conclusion is that the stated claim is false, and the mean volume of Helo is not 12 oz.

Another way of confirming this result is to use the 2-tail critical value for the 0.01 significance level, which corresponds to |Z|=2.575. 10.37>2.575 meaning that the sample Z-score is more extreme (way more extreme!) than the critical value, and H₀ is rejected and H₁ is accepted.