Answer:
B)
No, we can only say there is 27% chance of seeing the observed effectiveness from natural sampling variation. There is no evidence the new formula is more effective but we cannot conclude equal effectiveness.
Step-by-step explanation:
Hello!
The company compared the old antiacid formula against the new one. The claim is that the new formula is more effective.
The hypotheses are
H₀: μ₁ ≤ μ₂
H₁: μ₁ > μ₂
Where the subfix 1 represents the new formula and the subfix 2 represents the old formula.
The statistical analysis threw a p-value of 0.27.
Remember if the p-value ≥ α, n the decision is to not reject the null hypothesis.
If p-value < α, the decision is to reject the null hypothesis.
Let's say α: 0.1 ⇒ you'd decide to not reject the null hypothesis.
Then there would not be enough evidence to say the new formula is better than the old one (μ₁ > μ₂) instead you'd conclude that the new formula is at most as effective as the old one (μ₁ ≤ μ₂). To know if it is equally effective as the old one or less effective a new test should be made.
In simple words, the p-value is the probability of obtaining the value of the statistic under the null hypothesis. In this case, there is a 27% of possibility of observing the effectiveness of the new antiacid formula from a sampling error than because the new antiacid formula is, in fact, effective.
I hope it helps!
