Chapter 7 (Key, Odd)

(7.1) REVIEW QUESTIONS
(A) dependent samples
(B) matching
(C) An independent independent sample.
(D) There are no natural pairings in independent samples. Independent samples represent separate group.
(E) Exploratory Data Analysis
(F) difference
(G) True. If X is bigger than Y, the difference will be positive.
(H) stem-and-leaf, boxplot, histogram, etc.
(I) Parameter: �_d
(J) Estimator: _d
(K) 17
(L) a = .01. Thus use the t percentile corresponding to: 1 - a/2 = 1 - (.01/2) = .995.
(M) False. The confidence seeks to capture the population mean.
(N) Choice (b) is true since �_d either does or does not fall into the interval. There is only one �_d but there are many possible confidence intervals that can be derived from the population.. Notice that the "95% confidence" refers to our confidence in whether the interval captures �_d.
(O) (point estimate) � (tcoefficient) (standard error)
(P) H₀: �_d = 0
(Q) Again, df = n - 1, where n represent the number of pairs.
(R) Several methods may be available. A t table will allow you to derive an approximate p value by convert the test statistics to the approximate area under the curve in the tail (or tails) of the distribution. Computer programs such as SPSS and StaTable.exe convert the t_stat to a precise two-sided p value automatically.
(S) II
(T) 0.50
(U) This probability is greater than 50%
(V) D = "a difference worth detecting" F = "the function indicating the area under the curve to the right of the specified value" s = the population standard deviation.
(W) 80% - 90% is considered adequate for most purposes.

(7.3) OC-BP
(A) Calculation: 5.2 � (t_9,1-_.05/2)(4.88/sqrt(10)) = 5.2 � (2.26)(1.54) = 5.2 � 3.5 = (1.7, 8.7).
Interpretation: We are 95% confident the population mean difference in systolic blood pressure lies between 1.7 and 8.7 mm Hg. (Comments: Any interpretation of the confidence must acknowledges that the population mean difference may or may not be in the interval. Another way to view this confidence interval is to say that 95% of similarly constructed intervals from the same population will capture the population mean difference, and 5% will not.)

(B) H₀: �_d = 0 vs. H₁: �_d not = 0
a = .05
t_stat = 5.2 / (4.88/sqrt(10)) = 3.37; df = 9
[Draw the t distribution and determine the area under the curve]: the area in one tail < .005. Thus, the two-sided p < .01. (Precise p via StaTable = .0083.)
Reject H₀

(7.5) FLUORIDE

(A) Data (changes in cavity-free rates per 100 children) are:

CITY BEFORE AFTER DELTA (AFTER - BEFORE)
----  ------- ----- ---------------------
1    18.2 49.2   31.0
2    21.9 30.0    8.1
3     5.2 16.0   10.8
4    20.4 4(7.8   2(7.4
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 1(7.0    5.0
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

Notice that positive values represent increases in the cavity-free rates within cities.

Plot drawn with SPSS (version 10):

Frequency Stem & Leaf 4.00 -0 . 0024 5.00 0 . 03478 1.00 1 . 0 3.00 2 . 147 3.00 3 . 114

Stem width: 10.00 Each leaf: 1 case(s)

Notice that there are two zeros on the stem, one to receive negative values between 0 and -9, and one to receive positive values between 0 and +9.

Interpretation:

The central location of this distribution is approx. 10. (Comment: the quickest way to find the "middle" (median) of the distribution is to identify the value with a depth of (n+1) /2, by counting from the top or bottom of the stem-and-leaf plto. In this instance, n = 16, so the median has a depth of 8.5. This puts it between 7 and 8.)
Data range (spread) from -4 to +34.
The shape is (possibly) bi-modal. (Either that, or right-skewed.)
Four (4) of the 16 cities (25%) failed to show an improvement.

(B) H₀: �_d = 0 vs. H₁: �_d not = 0
a = .05
t_stat = 12.21 / 3.404 = 3.59; df = 16 - 1 = 15; �p < .005; p < .01
Reject H₀ (There was a significant increase in cavity-free rates.)

(7.7) BPH-SAMP
(A) These are paired samples because the sequential measurements represent natural couplings of measurements within observations.
(B) Data (recall that lower numbers represents improvements):

ID QoLTX QoL3Mo DELTA (QoL3Mo - QoL3TX) -- ----- ------ ----------------------- 1 2 1 -1 2 4 1 -3 3 3 1 -2 4 4 3 -1 5 5 2 -3 6 6 2 -4 7 4 2 -2 8 4 5 1 9 3 3 0 10 3 1 -2

(C) _d= -17 / 10 = -1.7
SS_d = (-1 - -1.7)� + (-3 - -1.7)� + . . . + (-2 - -1.7)�
    =     0.7�     +    1.3�      + . . . +    0.3�
    =     0.49     +    1.69      + . . . +    0.09
    =     20.1
s_d = (20.1 / [10 - 1]) = 1.494
sem_d = 1.494 / sqrt(10) = 0.473

(D) H₀: �_d = 0 vs. H₁: �_d 0
a = .05
t_stat = -1.70 / 0.473 = -3.60
df = 10 - 1 = 9
Draw the t distribution, place the tstat on the curve, and shade the area corresponding to half the p value. We find t_9,.995 = 3.25. Therefore, the region to the left of the t_stat < .005 and the two-sided p value < (2)(.005), which is less than .01. (Precise p value via StaTable = .0057.)
Reject H₀

(7.11) DELTAPOW

1 - b = F(-1.96 + [|1|sqrt(25)/10]) = F(-1.96 + 0.5) = F(-1.46) = .07054. [from the z table]

Interpretation: The test did NOT have adequate power to show a difference of 1 unit (the non-significant finding are likely due to a type II error).