© M. Innovera 2003

Review Questions

dependent samples, matched samples, matched pairs

matching , sequential samples, cross-over trials, "before/after" designs

There are no natural pairings in independent samples. Independent samples represent separate group.

µ_d

False. The confidence seeks to capture population mean difference µ_d (not sample mean difference x-bar_d)

0.50

... retain an incorrect null hypothesis.

7.l Placebo effect in Parkinson's disease patients. ...Test the difference for significance. [Understand all steps, not just the conclusion.]

H₀: µ_d = 0 vs. H₁: µ_d 0

Test statistic: SE = 0.181 / sqrt(6) = 0.0739; t_stat = (_d - µ₀) / sem = (-0.326 - 0 ) / 0.0739 = -4.41 with df = 6 - 1 = 5

The one-tailed P value is between 0.005 and 0.001 and the two-tailed P--value is between 0.01 and 0.002.

The evidence against H₀ is highly significant.

7.3 Oral contraceptive use and blood pressure

(A) Calculate a 95% confidence interval for the mean change in systolic blood pressure. SE = 2.6 / 30 = 0.475; 95% CI for µ_d = 1.4 ± (2.05)(0.475) = 1.4 ± 1.0 = (0.4 to 2.4)
(B) From information provided, can you determine the effect of this formulation in individuals? [Brief narrative response.] The confidence interval tells you nothing about the effect in individuals. It tells you only about the population mean. However, you could use the information about the mean and standard deviation provided in the question combined with Chebychev's principal to say that 75% of the individuals had a change between 1.4 ± 2(2.6) = 1.4 ± 5.2 = -3.8 to 6.6.

7.5 Water fluoridation

(A) Calculate the changes ...[and]... construct a stemplot of these differences ... interpret the stemplot. Below.

DELTA = AFTER - BEFORE. Positive values represent an increase in the cavity-free rates.

CITY BEFORE AFTER DELTA
----  ------- ----- -----
1    18.2 49.2   31.0
2    21.9 30.0    8.1
3     5.2 16.0   10.8
4    20.4 47.8   27.4
5     2.8   3.4    0.6
6    21.0 16.8   -4.2
7    11.3 10.7   -0.6
8     6.1   5.7   -0.4
9    25.0 23.0   -2.0
10    13.0 17.0    4.0
11    76.0 79.0    3.0
12    59.0 66.0    7.0
13    25.6 46.8   21.2
14    50.4 84.9   34.5
15    41.2 65.2   24.0
16    21.0 52.0   31.0

Stemplot

-0|0024 0|03478 1|0 2|147 3|114 ×10 (change per 100 children)
Location: median is 7.5 (underlined)
Spread: -4 to +34
Shape: distribution may(?) be bimodal; no apparent outliers.

(B) What percentage of cities showed an improvement in their cavity-free rate? Twelve (12) of the 16 cities (75%) show an increase in cavity-free rates.
(C) Estimate the mean change in the cavity-free rate in all similar cities with 95% confidence. SE_d = 13.62 / 16 = 3.405, df = 16 - 1 = 15, 95% CI for µ_d = 12.21 ± (2.131)(3.405) = 4.95 to 19.47.

7.7 Salivary cotinine

6|9
5|
4|19
3|09
2|
1|59
x10 (decline in cotinine)

The median decline was 39. Data spread from 15 to 69. All observations showed a decline. The data set is too small to make definitive statements about shape.

7.9 Benign prostate hyperplasia

(A) DELT_QoL = QoL_3MO - QOL_BASE = {1, 3, 2, 1, 3, 4, 2, -1, 0, 2}. Positive scores reflect an improvement:.

4|0
3|00
2|000
1|00
0|
-0|0
-1|0
×1

(B) Hypothesis test

H₀: µ_d = 0 vs. H₁: µ_d 0
t_stat = 1.70 / (1.494 / 10) = 3.60 with df = 10 - 1 = 9
Two-sided P-value is less than 0.01 and more than 0.005 (P = 0.0057)
The evidence against H₀ is highly significant.

7.11 NASA experiment Notice the large spread and two outliers (one high and one low).

-7|4 -6| -5| -4|9 -3|7 -2|5 -1|97654443 -0|865421 0|001356788 1|0114477 2|123569 3|3 4| 5| 6| 7| 8|5 ×1000

Unit 7 Key (Odd)

Review Questions