Flashcards in Statistical Inference: practice questions Deck (18):

1

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The standard error of the mean

A. Measures the variability of the observations

B. Describes the accuracy with which each observation is measured

C. Measures how far the sample mean is likely to be from the population mean

D. Increases as sample size increases

E. Is greater than the estimated standard deviation of the population

### (C) The standard error of the mean is a measure of how far, on average, a sample mean will be from the true population mean. The standard error of the mean refers to the variability of the sample mean, while the standard deviation refers to the variability of individual observations, so (A) and (B) are false. Because the standard error of the mean equals the standard deviation of the observations divided by the square root of the sample size, the standard error of the mean is smaller than the standard deviation and increases with decreasing sample size.

2

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In a sample of 100 adults, mean triglyceride level was 130 mg/dl with a standard deviation of 120 mg/dl. If the sample size were increased to 1,000 we would expect:

A. The mean to decrease

B. The standard error of the mean to decrease

C. The standard deviation to decrease

D. The sample variance to increase

E. The 95% confidence interval for the mean to be wider

### (B) SEM is estimated as s/√n, where s = standard deviation and n = sample size, so it decreases with increasing sample size. The width of a 95% confidence interval is directly proportional to the SEM (width = 2 x 1.96 x SEM) so it also decreases with increasing sample size. The mean, standard deviation and variance describe fixed population parameters and therefore do not depend on the sample size.

3

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The average body fat percentage in a large population of male high school athletes is 15%, with a standard deviation of 6%. If the mean is estimated repeatedly from different samples of size 100, which of the following statements is NOT true about the sample means:

A. The distribution of the sample means is approximately Normal

B. The mean of the distribution of sample means is 15%

C. The standard deviation of the distribution of sample means is 6%

D. If a 95% confidence interval is estimated for each sample mean, 95% of the confidence intervals will include the value 15%.

### (C) The standard deviation of the sample means is the standard error of the mean, s/√n, which is .6%, not 6%. (A) and (B) are true by the Central Limit Theorem and (D) is true by the definition of a confidence interval.

4

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Researchers plan to estimate the mean LDL cholesterol by selecting a sample from the entire U.S. population. The precision of the estimate will depend on all of the following EXCEPT:

A. The number in the sample

B. The variance of LDL cholesterol in the population

C. The standard deviation of LDL cholesterol in the population

D. The mean LDL cholesterol in the population

### (D) The precision of the estimate is measured by the standard error of the mean, s/√n, which does not depend on the mean itself, but does depend on the standard deviation (s), variance (s2 ) and sample size (n).

5

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The 95% confidence limits for the mean estimated from a set of observations

A. Are limits between which, in the long run, 95% of observations fall

B. Describe the precision of the individual observations

C. Are limits within which the sample mean falls with probability .95

D. Are limits which would include the population mean for 95% of possible samples

E. Describe the variability of observations in the population

### (D) By definition of confidence interval. The confidence interval does not describe the variation in individual observations, so A, B and E are false. (C) is a common, subtle misinterpretation of the confidence interval so (D) is the best answer

6

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Many cases of gallstones can be explained in terms of established risk factors, especially obesity. However, gallstones develop in some women who are not obese, and the causes are unknown. Biochemical studies have shown that slow intestinal transit is associated with lithogenic bile. A study was conducted (Lancet 341:8, 1993) to test the hypothesis that intestinal transit is abnormally slow in normal-weight women with gallstones.

"The mean WGTT (in hours) was significantly longer in the women with gallstones than in the controls (82 versus 63 hrs.; mean difference 19 hrs., 95% CI: 2-37 hrs.). Stool output was also lower in the women with gallstones (74 [SD 54] versus 141 [SD 56] grams per 24 hrs., P = 0.015). There was no significant difference between cases and controls in body mass index, waist-hip circumference ratio, parity, plasma triglyceride concentration, or alcohol intake. Normal-weight women with gallstones tend to have slow intestinal transit and this feature could explain why they have gallstones."

The authors report, "The mean WGTT (in hours) was significantly longer in the women with gallstones than in the controls (82 versus 63 hrs.; mean difference 19 hrs., 95% CI: 2-37 hrs.).” Based on this, one would conclude that it is unlikely that the expected mean increase in WGTT (over that of the controls) in normal weight women with gallstones is

A. between 2-37 hrs.

B. greater than 19 hrs.

C. greater than 40 hrs.

D. greater than 2 hrs.

E. less than 37 hrs.

### (C) The 95% confidence interval for the difference in mean values of WGTT, when gallstone patients were compared to healthy controls, was 2 to 37 hours. Only answer C has no value within this confidence interval.

7

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Many cases of gallstones can be explained in terms of established risk factors, especially obesity. However, gallstones develop in some women who are not obese, and the causes are unknown. Biochemical studies have shown that slow intestinal transit is associated with lithogenic bile. A study was conducted (Lancet 341:8, 1993) to test the hypothesis that intestinal transit is abnormally slow in normal-weight women with gallstones.

"The mean WGTT (in hours) was significantly longer in the women with gallstones than in the controls (82 versus 63 hrs.; mean difference 19 hrs., 95% CI: 2-37 hrs.). Stool output was also lower in the women with gallstones (74 [SD 54] versus 141 [SD 56] grams per 24 hrs., P = 0.015). There was no significant difference between cases and controls in body mass index, waist-hip circumference ratio, parity, plasma triglyceride concentration, or alcohol intake. Normal-weight women with gallstones tend to have slow intestinal transit and this feature could explain why they have gallstones."

The authors report, “Stool output was also lower in the women with gallstones (74 [SD 54] versus 141 [SD 56] grams per 24 hrs., P = 0.015).” The reported p-value indicates that the difference in stool output between women with gallstones and control women was

A. not statistically significant

B. likely due to chance

C. unlikely if there is really no difference between the two groups

D. likely if the difference between the two groups is greater than zero

### (C) By the definition of p-value. P

8

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In the Heart Protection Study (Lancet 2002; 360: 7–22), “All-cause mortality was significantly reduced (1328 [12·9%] deaths among 10,269 allocated simvastatin versus 1507 [14·7%] among 10,267 allocated placebo). The corresponding risk ratio is 12.9 / 14.7 = 0.88 with a 95% confidence interval of 0.82 to 0.94. Based on this you can conclude that

A. There is no significant difference between simvastatin and placebo

B. Risk of death is likely to be between 6% and 18% lower in patients taking simvastatin

C. Risk of death is likely to be between 82% and 94% lower in patients taking simvastatin

D. Risk of death could be as much as 94% higher in patients taking placebo

### (B) The 95% confidence interval for the risk ratio indicates that a range of plausible values for mortality in the simvastatin group is from 82% to 94% of the mortality in the placebo group, for a risk reduction of (100%-94% =) 6% to (100% - 82% = )18%. Because the 95% confidence interval does not include the null value (1.0) it is statistically significant at the 5% level.

9

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The researchers conducted a statistical hypothesis test to compare mortality in the simvastatin group with mortality in the placebo group. Based on their reported 95% confidence interval for the relative risk (0.82 to 0.94), which of the following p-values could they have obtained?

A.

### (A) Because the 95% confidence interval does not include the null value (1.0) it is statistically significant at the 5% level, so the p-value must be less than 0.05.

10

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A statistical method for attempting to rule out chance (sampling variation) as an explanation for an observed difference.

A. Type II error

B. observed significance level (p-value)

C. significance level (alpha level)

D. Type I error

E. significance test

### E

11

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Failing to reject a null hypothesis when false (failing to conclude two treatments are different when in fact they are).

A. Type II error

B. observed significance level (p-value)

C. significance level (alpha level)

D. Type I error

E. significance test

### A

12

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Incorrectly concluding a difference exists (incorrectly rejecting a null hypothesis).

A. Type II error

B. observed significance level (p-value)

C. significance level (alpha level)

D. Type I error

E. significance test

### D

13

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Maximum probability of incorrectly rejecting a null hypothesis that the investigator is willing to accept

A. Type II error

B. observed significance level (p-value)

C. significance level (alpha level)

D. Type I error

E. significance test

### C

14

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Probability that the observed result (or one more extreme) would occur by chance if the null hypothesis is true.

A. Type II error

B. observed significance level (p-value)

C. significance level (alpha level)

D. Type I error

E. significance test

### B

15

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The purpose of a statistical significance test is to attempt to rule out which of the following possible explanations for differences in treatment results?

A. bias

B. treatment effects

C. chance (variation)

D. confounding

E. all of the above

### (C) Significance tests are used primarily to rule out chance as an explanation of the differences observed in outcome measures. Bias and confounding are addressed by proper study design, e.g. use of randomization and blinding. By ruling out chance, bias and confounding we hope to show that any differences are likely due to the treatment.

16

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A clinical trial is being planned in which a new drug (A) is to be compared to the drug in current use (B). Patients satisfying criteria for acceptance into the study will be randomly allocated into two groups --one group to receive drug A, the other group drug B. Patients in each group will have systolic blood pressure (SBP) measurements taken during a baseline period and after a prescribed period on the drug therapy. It is planned to determine the effectiveness of the new drug by comparing the difference in mean SBP changes (mean drop with drug A compared to mean drop with drug B) to determine if the difference in mean blood pressure changes with the two drugs is statistically significant.

The investigators planning this study are concerned that the study be large enough so that it is unlikely that any clinically important improvement that the new drug might offer will be missed. They agree that the study should have a 5% chance of concluding the new drug is superior (leads to a greater mean reduction in SBP) when in fact it is no more effective than the drug in current use. In addition, they desire an 80% chance of concluding the new drug is superior (getting a significant difference) when in fact the true reduction in mean SBP achieved by the new drug exceeds that of the current drug by at least 25%.

A. 5%

B. 95%

C. 20%

D. 25%

E. none of the above

16. The probability of a Type I error has been set at

17. The probability of a Type II error has been set at

18. The power of this study has been set at

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16. (A) The statement, “5% chance of concluding the new drug is superior…when in fact it is no more effective” implies that the probability of rejecting the null hypothesis when it is true (i.e. the probability of a type I error) is 0.05.

17. (C) The statement, “80% chance of concluding the new drug is superior…when in fact the true reduction…achieved by the new drug exceeds that of the current drug” implies that the probability of rejecting the null hypothesis when it is false (i.e., the power of the test) is 80%. Since power = 1 – P(type II error), the probability of a type II error is 20%.

18. E

17

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A report of a randomized, double-blind clinical trial comparing the ability of two drugs to reduce serum cholesterol concludes that "the difference in serum cholesterol reduction between the two drugs was not significant (p > 0.05)". This implies

A. randomization failed to insure comparability of patients receiving the two drugs.

B. that the probability that the difference is due to chance is greater than 95%.

C. that there is less than 5% chance of making a Type I error.

D. that the observed mean reduction in serum cholesterol was similar for the two drugs.

E. none of the above.

### (E) From the p-value statement, one cannot infer anything about randomization. "p>.05" implies that the probability that the difference is due to chance is greater than 5%. The Type I error was fixed at the start of the analysis probably at 5%. The observed difference can, in fact, be large even if not statistically significant if the variability in the treatments is large or the sample size is small.

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