The FDA has decided to measure a storebrands 16ounce bags of
The FDA has decided to measure a store-brand’s 16-ounce bags of potato chips. The weights of 50 bags are :
. H0: 16 (one-tailed). Use a p-value approach to the t-test at the = 1% significance level. Does the FDA have enough evidence to issue a small fine and a warning to recalibrate their packaging machinery?
| 16.08 |
| 16.02 |
| 15.97 |
| 16.04 |
| 16.04 |
| 16 |
| 15.84 |
| 16.01 |
| 15.97 |
| 15.81 |
| 16.03 |
| 15.89 |
| 15.98 |
| 15.82 |
| 15.91 |
| 15.92 |
| 16.03 |
| 15.91 |
| 15.96 |
| 15.99 |
| 15.93 |
| 16.04 |
| 15.92 |
| 16.01 |
| 16.1 |
| 16.13 |
| 15.86 |
| 15.98 |
| 15.93 |
| 15.98 |
| 16.05 |
| 16.03 |
| 16.08 |
| 15.93 |
| 15.96 |
| 16.07 |
| 16.02 |
| 15.91 |
| 16.04 |
| 15.88 |
| 16.03 |
| 16.11 |
| 16.07 |
| 16.12 |
| 15.85 |
| 15.99 |
| 16.07 |
| 15.97 |
| 15.91 |
| 15.9 |
Solution
H0: 16
Ha: mu <=16
One tailed test.
sample size n = 50
Mean difference = -0.082
t statistic = -7.2005
Alpha = 0.01
df = 50-1=49
p value < .00001.
The result is significant at p < .01.
Since p < 0.01, reject null hypothesis.
the FDA has enough evidence to issue a small fine and a warning to recalibrate their packaging machinery
| 16.08 | |
| 16.02 | |
| 15.97 | |
| 16.04 | |
| 16.04 | |
| 16 | |
| 15.84 | |
| 16.01 | |
| 15.97 | |
| 15.81 | |
| 16.03 | |
| 15.89 | |
| 15.98 | |
| 15.82 | |
| 15.91 | |
| 15.92 | |
| 16.03 | |
| 15.91 | |
| 15.96 | |
| 15.99 | |
| 15.93 | |
| 16.04 | |
| 15.92 | |
| 16.01 | |
| 16.1 | |
| 16.13 | |
| 15.86 | |
| 15.98 | |
| 15.93 | |
| 15.98 | |
| 16.05 | |
| 16.03 | |
| 16.08 | |
| 15.93 | |
| 15.96 | |
| 16.07 | |
| 16.02 | |
| 15.91 | |
| 16.04 | |
| 15.88 | |
| 16.03 | |
| 16.11 | |
| 16.07 | |
| 16.12 | |
| 15.85 | |
| 15.99 | |
| 16.07 | |
| 15.97 | |
| 15.91 | |
| 15.9 | |
| 15.9818 | Mean |
| 0.080526 | sd |
| 0.011388 | Std error |


