since genevieve rejected h 0 , type ii error is not in play. thus choice (a) is incorrect

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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 0.048. 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.

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Page 1: Since Genevieve rejected  H 0  ,  type II error is not in play.   Thus choice (a) is incorrect

Based on a sample x1, x2, … , x12 of 12 values from a population that is presumed normal, Genevieve tested H0:  = 20 versus H1:   20 at the 5% level of significance. She got a t statistic of 2.23, and Minitab reported the p‑value as 0.048. Genevieve properly rejected H0. 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.

Page 2: Since Genevieve rejected  H 0  ,  type II error is not in play.   Thus choice (a) is incorrect

Since Genevieve rejected H0 , 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 H0 was rejected. Thus (b) is also incorrect.

The best answer is (c). The hypothesis test with a small sample size just barely rejected H0 . It is certainly possible that a Type I error is involved.

Page 3: Since Genevieve rejected  H 0  ,  type II error is not in play.   Thus choice (a) is incorrect

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 H0 would have been rejected at that level. For example, using α = 0.05 and getting then p = 0.0381 from the data would correspond to rejecting H0 .

Page 4: Since Genevieve rejected  H 0  ,  type II error is not in play.   Thus choice (a) is incorrect

Based on a sample x1, x2, … , x19,208 of 19,208 observations from a population that is presumed normal, Karl tested H0:  = 140 versus H1:   140. He got a t statistic of ‑2.18, and Minitab reported the p‑value as 0.029. Karl properly rejected H0. 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.  

Page 5: Since Genevieve rejected  H 0  ,  type II error is not in play.   Thus choice (a) is incorrect

The best answer is (d). These results are useless!

Observe that t = ‑2.18 =  , which

leads to  ‑0.0157.

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!

14019,208 xs

140xs

Page 6: Since Genevieve rejected  H 0  ,  type II error is not in play.   Thus choice (a) is incorrect

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.

Page 7: Since Genevieve rejected  H 0  ,  type II error is not in play.   Thus choice (a) is incorrect

As for (c), the value of t0.025; 19,207 is very, very close to 1.96. Minitab gives this as 1.96009.

Page 8: Since Genevieve rejected  H 0  ,  type II error is not in play.   Thus choice (a) is incorrect

Impossible … or … could happen ?

Alicia tested H0:  = 405 against H1:   405 and rejected H0 with a p‑value of 0.0278. Alicia was unaware of the true value of and committed a Type I error.

Yes, this could happen. If we reject H0 we might be making a Type I error.

Page 9: Since Genevieve rejected  H 0  ,  type II error is not in play.   Thus choice (a) is incorrect

Impossible … or … could happen ?

Celeste tested H0: p = 0.70 versus H1: 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 H0 with  = 0.05. Lou worked the next day with 105 subjects (all 85 that Celeste had, plus an additional 20), and Lou accepted H0 with  = 0.05.

This could happen. It’s not expected, however. Usually, enlarging the sample size makes it even easier to reject H0.

Page 10: Since Genevieve rejected  H 0  ,  type II error is not in play.   Thus choice (a) is incorrect

Impossible … or … could happen ?

In an experiment with two groups of subjects, it happened that

sx = standard deviation of x‑group = 4.0 sy = standard deviation of y‑group = 7.1sp = pooled standard deviation = 7.8

This is impossible, as sp must be between sx and sy .

Page 11: Since Genevieve rejected  H 0  ,  type II error is not in play.   Thus choice (a) is incorrect

Impossible … or … could happen ?

In an experiment with two groups of subjects, it happened that

sx = standard deviation of x‑group = 4.0 sy = standard deviation of y‑group = 7.1sz = 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.

Page 12: Since Genevieve rejected  H 0  ,  type II error is not in play.   Thus choice (a) is incorrect

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.

Page 13: Since Genevieve rejected  H 0  ,  type II error is not in play.   Thus choice (a) is incorrect

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.

Page 14: Since Genevieve rejected  H 0  ,  type II error is not in play.   Thus choice (a) is incorrect

True . . . or . . . false ?

The test of H0:  = 0 versus H1:   0 at level 0.05 with data x1, x2, … , x50 will have a larger type II error probability than the test of H0:  = 0 versus H1:   0 at level 0.05 with data x1, x2, … , x100 .

True. All else equal, the major benefit to enlarging the sample size is reducing the Type II error probability.

Page 15: Since Genevieve rejected  H 0  ,  type II error is not in play.   Thus choice (a) is incorrect

True . . . or . . . false ?

Steve had a binomial random variable X based on n = 84 trials. He tested H0: p = 0.60 versus H1: p  0.60 with observed value x = 65, and this resulted in a p‑value of 0.001. The probability is 99.9% that H0 is false.

This is false. This is a very serious misinterpretation of the p-value. Remember that p = P[ these data | H0 true ].

Page 16: Since Genevieve rejected  H 0  ,  type II error is not in play.   Thus choice (a) is incorrect

True . . . or . . . false ?

You are testing H0: p = 0.70 versus H1: p 0.70 at significance level  = 0.05. 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.