Question

The old machine filled 12 soda cans with an average of 11.9 oz of soda with a standard deviation of 0.3 ounces. The new machine still fills them (11 cans) with an average of 12.1 ounces, with a standard deviation of 0.3 oz. Is the new machine filling with significantly more than the old machine? (α=.05)

Answer 1

Mean = 11.90, S.D = 0.30

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_Old> u_New
Alternative hypothesis: u_Old < u_New

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the mean difference between sample means is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 0.12523
DF = 21

t = [ (x₁ - x₂) - d ] / SE

t = 0.799

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is thesize of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.

The observed difference in sample means produced a t statistic of 0.799.

Therefore, the P-value in this analysis is 0.217.

Interpret results. Since the P-value (0.217) is greater than the significance level (0.05), we cannot reject the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that machine filling with significantly more than the old machine.