Draw conclusions based on significance levels and P-values.

This statement is false because a very high​ P-value is strong evidence that the null hypothesis is true.

This statement is true.

This statement is false because a very high​ P-value proves that the null hypothesis is false.

This statement is false because it is a low​ P-value that provides evidence that the null hypothesis is false.

This statement is true.

This statement is false because a very low​ P-value only shows strong evidence that the null hypothesis is false.

This statement is false because a very low​ P-value proves that the null hypothesis is true.

This statement is false because it is a very high​ P-value that proves that the null hypothesis is false.

This statement is false because a high​ P-value shows that the null hypothesis is false.

This statement is true.

This statement is false because a high​ P-value shows that the data is not consistent with the null​ hypothesis, and can only prove that the null hypothesis is false.

This statement is false because a high​ P-value shows that the data is consistent with the null​ hypothesis, but can never prove that the null hypothesis is true.

This statement is false because the null hypothesis is rejected whenever the​ P-value is below the value of α​, which may not necessarily be 0.05.

This statement is false because a​ P-value below 0.05 proves that the null hypothesis is​ false, which is a much stronger result than simply rejecting the null hypothesis.

This statement is true.

This statement is false because it is a​ P-value that is above 0.05 which is always considered sufficient evidence to reject a null hypothesis.

If there is no difference in​ effectiveness, the chance of seeing an observed difference this large or larger is 1.5​% by natural sampling variation.

There is a 1.5​% difference between the effectiveness of the new treatment and the effectiveness of the traditional ointment.

There is a 1.5​% chance that the new treatment is effective.

If there is a difference in​ effectiveness, the chance of seeing an observed difference this large or larger is 1.5​% by natural sampling variation.

The​ P-value 0.45 is the probability of getting results like the ones obtained in the study by natural variation​ alone, assuming that the proportion of commuters who wear seat belts has increased.

One minus the​ P-value, or 0.55​, is the probability that the proportion of commuters who wear seat belts has changed.

The​ P-value 0.45 is the probability that the proportion of commuters who wear seat belts has changed.

The​ P-value 0.45 is the probability of getting results like the ones obtained in the study by natural variation​ alone, assuming that the proportion of commuters who wear seat belts has not changed.

The​ P-value 0.45 is the probability that the proportion of commuters who wear seat belts has not changed.

His decision would have been the same for both α=0.005 and α=0.10.

His decision may have been different for α=0.10 but would have been the same for α=0.005.

His decision may have been different for α=0.005 but would have been the same for α=0.10.

His decision may have been different for both α=0.005 and α=0.10.

There is only a 0.6​% chance that 90.5​% is the actual percentage of vaccinated children.

There is only a 0.6​% chance that 90.5​% is not the actual percentage of children vaccinated.

There is only a 0.6​% chance that a sample proportion of 88.5​% is statistically different from the reported 90.5​% of children vaccinated

There is only a 0.6​% chance of seeing a sample proportion as low as 88.5​% vaccinated by natural sampling variation if 90.5​% have really been vaccinated.

None of these

2.8% more drug stores remove​ over-the-counter drugs from the shelves when the drugs are past the expiration date.

There is a​ 97.2% chance the drug stores remove more expired​ over-the-counter drugs.

97.2% more drug stores remove​ over-the-counter drugs from the shelves when the drugs are past the expiration date than food stores.