Based on a sample x 1, x 2, …, x 12 of 12 values from a population that is presumed normal, Genevieve tested H 0 :  = 20 versus H 1 :   20 at the 5% level of significance. She got a t statistic of 2.23, and Minitab reported the p ‑ value as Genevieve properly rejected H 0. But Genevieve is a little concerned. Why? (a)She is worried that she might have made a type II error. (b)She is concerned about whether the 95% confidence interval might contain the value 20. (c)The p ‑ value is close to 0.05, and she worries that she might have a type I error. (d)She wonders if using a test at the 10% level of significance might have given her an even smaller p ‑ value.

Since Genevieve rejected H 0, type II error is not in play. Thus choice (a) is incorrect. As for (b), we know for sure that the comparison value 20 is outside a 95% confidence interval, as H 0 was rejected. Thus (b) is also incorrect. The best answer is (c). The hypothesis test with a small sample size just barely rejected H 0. It is certainly possible that a Type I error is involved.

Choice (d) is completely misguided. After all, the p- value is computed from the data and its calculation has nothing to do with the level of significance. Keep in mind that the p-value can be compared to a specified significance level for the simple purpose of telling whether H 0 would have been rejected at that level. For example, using α = 0.05 and getting then p = from the data would correspond to rejecting H 0.

Based on a sample x 1, x 2, …, x 19,208 of 19,208 observations from a population that is presumed normal, Karl tested H 0 :  = 140 versus H 1 :   140. He got a t statistic of ‑ 2.18, and Minitab reported the p ‑ value as Karl properly rejected H 0. But Karl is a little concerned. Why? (a)He is considering whether a sample size of around 40,000 might give him a smaller p ‑ value. (b)He is concerned that his results would not be found significant at the 0.01 level of significance. (c)He is concerned that his printed t table did not have a line for 19,207 degrees of freedom. (d)He is worried that his significant result might not be useful.

The best answer is (d). These results are useless! Observe that t = ‑ 2.18 =, which leads to  ‑ It seems that the true mean μ is only about 1.57% of a standard deviation away from the target 140. No one will notice or care!

Statement (a) happens to be true, but it should not be the cause of any concern. Yes, a bigger sample size is going to get a smaller p-value, but the results with a sample size over 19,000 are useless. It would be madness to run up the expenses for a sample of 40,000! Statement (b) is also true, but not the proper cause of any concern. Yes, it’s awkward to run such a large experiment and not get significance at the 1% level. It’s already clear that the results are useless, so this kind of concern is not helpful.

As for (c), the value of t 0.025; 19,207 is very, very close to Minitab gives this as

Impossible … or … could happen ? Alicia tested H 0 :  = 405 against H 1 :   405 and rejected H 0 with a p ‑ value of Alicia was unaware of the true value of  and committed a Type I error. Yes, this could happen. If we reject H 0 we might be making a Type I error.

Impossible … or … could happen ? Celeste tested H 0 : p = 0.70 versus H 1 : p  0.70, where p is the population probability that a randomly-selected subject will like Citrus ‑ Ola orange juice better that Tropicana. She had 85 subjects, and she ended up rejecting H 0 with  = Lou worked the next day with 105 subjects (all 85 that Celeste had, plus an additional 20), and Lou accepted H 0 with  = This could happen. It’s not expected, however. Usually, enlarging the sample size makes it even easier to reject H 0.

Impossible … or … could happen ? In an experiment with two groups of subjects, it happened that s x = standard deviation of x ‑ group = 4.0 s y = standard deviation of y ‑ group = 7.1 s p = pooled standard deviation = 7.8 This is impossible, as s p must be between s x and s y.

Impossible … or … could happen ? In an experiment with two groups of subjects, it happened that s x = standard deviation of x ‑ group = 4.0 s y = standard deviation of y ‑ group = 7.1 s z = standard deviation of all subjects combined = 7.8 This can happen. In fact, the standard deviation of combined groups is usually larger than the two separate standard deviations.

True... or... false ? Alvin always uses  = 0.05 in his hypothesis testing problems. In the long run, Alvin will reject about 5% of all the hypotheses he tests. False! Alvin’s rejection rate depends on what hypotheses he chooses to test. True: In the long run, Alvin will reject about 5% of the TRUE hypotheses that he tests.

True... or... false ? If you perform 40 independent hypothesis tests, each at the significance level 0.05, and if all 40 null hypotheses are true, you will commit exactly two Type I errors. False. The number of Type I errors is a binomial random variable. The expected number of errors is two.

True... or... false ? The test of H 0 :  =  0 versus H 1 :    0 at level 0.05 with data x 1, x 2, …, x 50 will have a larger type II error probability than the test of H 0 :  =  0 versus H 1 :    0 at level 0.05 with data x 1, x 2, …, x 100. True. All else equal, the major benefit to enlarging the sample size is reducing the Type II error probability.

True... or... false ? Steve had a binomial random variable X based on n = 84 trials. He tested H 0 : p = 0.60 versus H 1 : p  0.60 with observed value x = 65, and this resulted in a p ‑ value of The probability is 99.9% that H 0 is false. This is false. This is a very serious misinterpretation of the p-value. Remember that p = P[ these data | H 0 true ].

True... or... false ? You are testing H 0 : p = 0.70 versus H 1 : p  0.70 at significance level  = A sample of size n = 200 will allow you to claim a smaller Type I error probability than a sample of size n = 150. This is false. The probability of Type I error is 0.05, exactly the value you specify. From Museum of Bad Art, Somerville, Massachusetts.