BIOL/STAT 335 Sample Questions for Final Exam.
1. Eating out vegetarian. Independent random samples of 434 U.S. women and 747 U.S. men showed that 195 women and 276 men said that they
sometimes order a restaurant dish “without mean, fish, or fowl” (vegetarian). Let p1 and p2 be the probabilities for women and men, respectively.
Here p1-hat = 195/434, p1-tilde = 196/436, p2-hat = 276/747, p2-tilde = 277/749
Find a 95% confidence interval for (p1 – p2)
95% C.I. p1 – p2 = p1-tilde – p2-tilde +/- z_{.025}SE
= (196/436 – 277/749) +/- (1.96)sqrt[(196/436)(240/436)/436 + (277/749)(472/749)/749]
= .0797 +/- (1.96)(.0296)
= .0797 +/- .0581.
Or .022 < p1 – p2 < .138
Use this confidence interval to test at the 5% level of significance whether men and women eat out vegetarian the same percentage of times.
State the null hypothesis
H0: p1 – p2 = 0
State the alternative HA:
p1 – p2 not= 0
Reject H0 because the C.I. does not contain p1 – p2 = 0.
Conclusion of test Reject the Null
State your conclusion in context of the problem
There is sufficient evidence (P-value < .05) to conclude that men and women do not eat vegetarian in the same proportion.
2. Getting taller? Independent samples of 10 men aged 25-34 years and 15 men aged 45-54 years yields the following height data.
Welch’s formula yields 21.7
Age
n
Mean
SD
25-34
10
70.19
2.951
df = 21
45-54
15
68.58
3.543
Find the 95% confidence interval for the difference in height.
mu1 – mu2 = (y1-bar – y2-bar) +/- t_{21,.025}SE
= 1.61 +/- (2.080)sqrt[(2.951)^2/10 + (3.543)^2/15]
= 1.61 +/- 2.718
Or -1.108 < mu1 – mu2 < 4.328
Test at the 5% level of significance, assuming the populations have equal standard deviations, to decide whether men aged 25-34 years are, on the
average, taller than men aged 45-54 years.
State the null hypothesis mu1 = mu2
State the alternative mu1 > mu2
Do we need to assume normality of the population to use this test? YES
freedom df = 9 + 14 = 23
Pooled SD sp = sqrt[(9(2.951)^2 + 14(3.543)^2)/(9+14)] = 3.324
Degrees of
SE = sp sqrt[1/10 + 1/15] = 3.324sqrt[.1667] = 1.357 Test statistic ts = {(y1-bar – y2-bar) – (mu1 – mu2)}/SE = 1.61/1.357 = 1.186
P-value (be as specific as possible) P-value = Pr(t > ts) = Pr(t > 1.186) = .1243 (using technology), or 0.10 < P-value < 0.20 (using Table 4)
Conclusion of test Retain the Null
Detailed conclusion (be specific in context of the problem): There is insufficient evidence (0.10 < P-value < 0.20) to conclude that men are getting
taller.
3. Glaucoma and corneal thickness. A researcher measured corneal thickness (in microns) of eight patients who had glaucoma in one eye but not in the
other.
Patient
1
484
Normal
Glaucoma 488
Difference
-4
2
478
478
0
3
492
480
+12
4
444
426
+18
5
436
440
-4
6
398
410
-12
7
464
458
+6
8
476
460
+16
Mean
459
455
4
SD
31.31
27.69
10.74
Welch’s formula yields 13.79. At the 10% level of significance, use the t test to decide whether corneal thickness is greater in normal eyes than in eyes with
glaucoma.
State the null hypothesis mu1 = mu2
The two samples of data are
paired
State the alternative mu1 > mu2
Degrees of freedom df = n – 1 = 7
Do we need to assume normality of the population to use this test? YES The table from the book to use for this problem in on page Table 4 on page 618
Test statistic ts =(yd-bar – mud)/SEd = 4/(10.74/sqrt[8]) = 1.053
P-value (be as specific as possible) P-value = Pr(t > ts) = Pr(t > 1.053) = .16357 (using technology), or 0.10 < P-value < 0.20 (using Table 4)
Conclusion of test Retain the Null
Detailed conclusion (be specific in context of the problem): There is insufficient evidence (0.10 < P-value < 0.20) to conclude that corneal thickness is
less in eye with glaucoma.
4. Glaucoma and corneal thickness again. Repeat problem above but use the sign test instead.
A researcher measured corneal thickness (in microns) of eight patients who had glaucoma in one eye but not in the other.
Patient
Normal
Glaucoma
Difference
1
484
488
-4
2
478
478
0
3
492
480
+12
4
444
426
+18
5
436
440
-4
6
398
410
-12
7
464
458
+6
8
476
460
+16
Mean
459
455
4
SD
31.31
27.69
10.74
Welch’s formula yields 13.79. At the 10% level of significance, use the sign test to decide whether corneal thickness is greater in normal eyes than in eyes
with glaucoma.
State the null hypothesis mu1 = mu2
State the alternative mu1 > mu2
Do we need to assume normality of the population to use this test? NO Test statistic Bs = N+ = 4 (with nd = 7)
The table from the book to use for this problem in on page Table 7 on page 625. P-value from the table (be as specific as possible) Off chart: P-value > .10
Calculate the exact P-value P-value = 7C4(.5)^4(1-.5)^3 + 7C5(.5)^5(1-.5)^2 + 7C6(.5)^6(1-.5)^1 + 7C7(.5)^7(1-.5)^0 = (35 + 21 + 7 + 1)/2^7 = 64/128
= .50
Conclusion of test Retain the Null
Detailed conclusion (be specific in context of the problem): There is insufficient evidence (0.10 < P-value < 0.20)
to conclude that corneal thickness is less in eye with glaucoma.
5. Glaucoma and corneal thickness again. Repeat problem above but use the Wilcoxon Signed-Rank test instead.
A researcher measured corneal thickness (in microns) of eight patients who had glaucoma in one eye but not in the other.
Patient
1
484
Normal
Glaucoma 488
Difference
Signed Rank
-4
-1.5
2
478
478
0
XX
3
492
480
+12
+4.5
4
444
426
+18
+7
5
436
440
-4
-1.5
6
398
410
-12
-4.5
7
464
458
+6
+3
8
476
460
+16
+6
Mean
459
455
4
SD
31.31
27.69
10.74
Welch’s formula yields 13.79. At the 10% level of significance, use the Wilcoxon Signed-Rank test to decide whether corneal thickness is greater in
normal eyes than in eyes with glaucoma.
State the null hypothesis mu1 = mu2
State the alternative mu1 > mu2
Do we need to assume normality of the population to use this test? NO Test statistic W– = 7.5, W+ = 20.5, Ws = W+ = 20.5 (with nd = 7)
The table from the book to use for this problem in on page Table 8 on page 626. P-value from the table (be as specific as possible) Off chart: P-value > .10
Conclusion of test Retain the Null
Detailed conclusion (be specific in context of the problem): There is insufficient evidence (0.10 < P-value) to
conclude that corneal thickness is less in eye with glaucoma.
6. Smoking and low birthweight babies. In a study done in 1971 on smoking among pregnant women and low birthweight, the following data was
collected.
TOTAL
Low birthweight
Normal birthweight
Smokers
185 (143.81)
1891 (1932.19)
2076
Non-smokers
134 (175.19)
2395 (2353.81)
2529
TOTAL
319
4286
4605
Expected counts shown above in parentheses
Perform a one-sided chi-square test at the 1% level of significance to determine whether low birthweights are associated with smoking.
State the null hypothesis Smoking and birthweight are independent State the alternative Smoking and birthweight are related
Do we need to assume normality of the population to use this test? NO (variable is categorical) Degrees of freedom df = (r-1)(c-1) = 1
Test statistic Chi-Sq = (185 – 143.81)^2/143.81 + …… + (2395 – 2353.81)^2/2535.81 = 11.798 + .8781 + 9.685 + 0.7208 = 23.081
P-value (be as specific as possible) (Off chart) P-value < 0.0001/2 = .00005 Conclusion of test Reject the Null
Detailed conclusion (be specific in context of the problem): There is strong evidence (P-value < .00005) to conclude that smjoking and birthweight are
related.
7. A study was conducted to determine whether dietary lipid content is important in the population growth of Chesapeake Bay copepods. Independent
random samples of copepods were placed in containers with lipid-rich diatoms, bacteria, and leafy macroalgae. There were 12 containers total with four
replicates per diet. Five egg bearing females were placed in each container. After 14 days, the number of copepods in each container were as follows.
Sample
Treatment
Mean
SD
A
Diatoms
426
467
438
497
457.00
31.74
B
Bacteria
303
301
293
328
306.25
15.13
C
Macroalgae
277
324
302
272
293.75
24.06
Calculate the pooled variance. sp^2 = {3(31.74)^2 + 3(15.13)^2 + 3(24.06)^2}/{3 + 3 + 3} = 605.11
Calculate the grand mean {4(457.00) + 4(306.25) + 4(293.75)}/12 = 352.3333
Complete the ANOVA table and carry out the F test at  = 5%.
Source
df
SS
MS
F
Between treatments
2
66043
33021.5
Within treatments
9
5446
605.11
54.57
TOTAL
11
71489
Null hypothesis muA = muB =muC Alternative hypothesis Not all equal (at least one mu is different)
= 54.58
Table from the book to use for this problem is on page Table 10 on page 629
Circle the correct decision:
growth.
Degrees of freedom 2 and 9
Test statistic Fs
P-value (be as specific as possible) Off chart: P-value < .0001
Reject Null Detailed conclusion (be specific in context of the problem): Not all treatments result in the same mean
8. Use the Bonferroni method to construct a 95% confidence interval for the difference between the treatment means for treatments B and C and use this
interval to state whether or not you feel they are indeed equal. Show all work. Here df = 9, sp^2 = 605.11, SEp = sqrt[sp^2/nB + sp^2/nC] =
sqrt[605.11/4 + 605.11/4] = 17.39
Bonferroni Interval muB – muC = (yB-bar – yC-bar) +/- t_{9,.025/(3C2)}SE = 12.5 +/- (2.933)(17.39) = 12.5 +/- 51.01
Or
-38.51 < muB – muC < 63.51.
Note that this C.I. consistent with the null hypothesis that muB = muC at alpha = 5%, since the 95% Bonferroni confidence interval contains muB –
muC = 0.
9. Age and body fat. Eighteen randomly selected adults were measured for percentage of body fat, using dual-photon absorptiometry. The following table
shows the results of the measurements and the ages of the adults.
Subject
1
2
3
4
5
6
x = Age
23
23
27
27
39
41
y = %Fat
9.5
27.9
7.8
17.8
31.4
25.9
Subject
7
8
9
10
11
12
x = Age
45
49
50
53
53
54
y = %Fat
27.4
25.2
31.1
34.7
42.0
29.1
Subject
13
14
15
16
17
18
x = Age
56
57
58
58
60
61
y = %Fat
32.5
30.3
33.0
33.8
41.1
34.5
For this sample we have
x-bar = 46.33, y-bar = 28.61, sx = 13.22, sy = 9.14, r = .7918 and Σ (x – x-bar)(y – y-bar) = 1626.524,. SS(resid) = 529.736
Σ (x – x-bar)2 = 2971.063, Σ (y – y-bar)2 = 1420.173
Find an equation of the major axis Slope : m = sy/sx = 9.14/13/22 = .6914, Point: (x-bar,y-bar) = (46.33,28.61) Eqn: y – y-bar = b1(x – x-bar) or y –
28.61 = .6914(x – 46.33) or y = -3.42 + .6914x
Find an equation of the regression line
Slope : b1 = r(sy/sx) = (.7918)(9.14/13/22) = .5480, Point: (x-bar,y-bar) = (46.33,28.61) Eqn: y – y-bar = b1(x – x-bar) or y – 28.61 = .5480(x – 46.33)
or y = 3.221 + .5480x
Estimate the %Fat of a person whose age is 58 years. y-hat = 3.221 +(.5480)(58) = 35.00%
Estimate the SD of the %Fat of a person aged 58 years s_{e} = sy(sqrt[1-r^2]sqrt[(n-1)/(2-1)] = sqrt[SS(resid)/(n-2)] = 5.754
Give a 95% confidence interval for the slope of the regression line beta1 = b1 +/- t_{16,.025}SE = .5480 +/- (2.120)(.1056) = .5480 +/- .2239 or .3241 <
beta1 < .7719
Use linear regression to test the hypothesis that age has no effect on percentage body fat against the alternative that percentage body fat increases with age.
Let alpha = 1%. H0: beta1 = 0, HA: beta1 > 0, alpha = .01,
Test statistic ts = (b1 – beta1)/SE = (.5480 – 0)/.1056 = 5.19, P-value Off chart: P-value < .0005; Decision:
Reject Null
Detailed conclusion (be specific in context of the problem): There is strong evidence (P-value < .0005) to conclude that body fat increases with age.
10. The data in the table below are taken from a study on Parkinson’s disease, a disease that, among other things, affects a person’s ability to speak. Eight of
the people in this study had received one of the most common operations to treat the disease, and the other fourteen received no operation. Subsequently,
each patient was given a test to assess their speaking ability, and these scores are reported in the table (lower scores indicate more problems with speaking).
Use alpha = 0.05
Use the Wilcoxon-Mann-Whitney test to decide whether there is a difference between the treated patients and the untreated patients.
Note that Welch’s equation (7.1) yields df = 18.8.
State the null hypothesis The population distributions are the same State the alternative The population distributions are different
Do we need to assume normality of the population to use this test? NO Test statistic Here K1 = 90.5 and K2 = 21.5, so Us = 90.5 with n = 14 and n’ = 8
(check: K1 + K2 = 90.5 + 21.5 = 112, n*n’ = (14)(8) = 112)
The table from the book to use for this problem in on page Table 6 on page 622 P-value from the table (be as specific as possible) .01 < P-value < .02
Conclusion of test (circle one) Reject the Null or Retain the Null
Detailed conclusion (be specific in context of the problem): There is sufficient
evidence (0.01 < P-value < 0.02) to conclude that the operation does improve a patient’s ability to speak.
Table
Number w/o
operation that are
smaller
Pts. who had
operation
(ORDERED)
7
1.7
9.5
2.0
12,12
12,12
12.5
13.5
n’ = 8
K1 = 90.5
2.5,2.5
2.6,2.6
2.7
3.0
Pts. who didn’t have
operation
(ORDERED)
1.2
1.3,1.3,1.3
1.5,1.5
1.6
Number w/operation
that are smaller
0
0,0,0
0,0
0
1.8,1.8
2.0
2.2
2.3
1,1
1.5
2
2
2.7
3.0
6.5
7.5
n=14
K2 = 21.5
How would you assess normality for the above data? Use the Anderson-Darling test or a normal probability plot
MORE SAMPLE QUESTIONS
11. Six patients with renal disease underwent plasmapheresis. Urinary protein excretion (grams of protein per gram of creatinine) was measured for each
patient before and after plasmapheresis. The data are given in the table below. Assume that the relevant population distribution(s) is (are) normally
distributed. Note that the Welch’s Formula (Equation 7.1) yields 8.8 for these data.
Patient
1
2
3
4
5
6
Mean
SD
Before
20.3
6.3
7.6
6.1
5.9
4.0
8.37
5.96
After
11.8
2.1
3.0
2.6
4.0
4.2
4.62
3.61
Difference
8.5
4.2
4.6
3.5
1.9
-0.2
3.75
2.92
a. Use these data to obtain a 98% confidence interval for the difference of the average protein excretion before and after plasmapheresis. Show all work.
muA – muB = 3,75 +/- t_{5,.01}(2.92/sqrt[6]) = 3.76 +/- (3.365) (2.92/sqrt[6]) = 3.75 +/- 4.01
b. Use these data and  = 2% to investigate whether or not plasmapheresis affects urinary protein excretion in patients with renal disease.
Since muA – muB = 0 is included in the interval, we retain H0. There is insufficient evidence (P-value > .02) to conclude that there is a difference.
12. Use the sign test to test the same hypothesis.
Calculated Test Statistic
Bs = 5
Either [a] Use table from the book to use for this problem is on page 625 (Table 7)
nondirectional P-value > .20
P-value from the table (be as specific as possible). This test is
or [b] Calculate the exact P-value = 2(Pr{Y >= 5}) = 2(6C5(.5)^5(1-.5)^1 + 6C6(.5)^6}) = 14(.5)^6 = .2185
the Null
Reject the Null
Circle the correct decision:
Fail to reject
13. Two drugs, zidovudine and didanosine, were tested for the effectiveness in preventing progression of HIV disease in children. In a double-blind clinical
trial, 276 children with HIV were given zidovudine, 281 were given didanosine, and 274 were given zidovudine plus didanosine. The following table shows
the survival data for the three groups.
Died
Survived
Total
Zidovudine
Didanosine
17 (11.29)
259 (264.71)
276
7 (11.50)
274 (269.50)
281
Zidovudine &
Didanosine
10 (11.21)
264 (262.79)
274
Total
34
797
831
Given that a patient received zidovudine, use these data to estimate the probability that (s)he died. 17/276
Given that a patient survived, use these data to estimate the probability that (s)he received (only) didanosine. 274/797
Use  = 10% and these data to conduct a test of the null hypothesis that survival and treatment are independent.
Null Hypothesis Survival and Treatment are independent Alternative Hypothesis Survival depends on Treatment
Calculated Test Statistic (show all work) Chi-Sq = 2.885 + 1.759 + 0.131 + 0.123 + 0.075 + 0.006 = 4.978
Table from the book to use for this problem is on page 627 (Table 9)
Degrees of Freedom 2
P-value (be as specific as possible) .05 < P-value < .10 (=alpha), reject H0
Circle the correct decision: Fail to reject the Null
Reject the Null
conclude that the survival rate is related to the treatment.
Detailed conclusion: There is sufficient evidence (.05 < P-value < .10) to
14. In a study of the mutual effects of the air pollutants ozone and sulfur dioxide, Blue Lake snap beans were grown in open-top field chambers. Some
chambers were fumigated repeatedly with sulfur dioxide. The air in some chambers was carbon-filtered to remove ambient ozone. In all, there were 4
treatments – labeled A (ozone absent, SO2 absent), B (ozone absent, SO2 present), C (ozone present, SO2 absent) and D (ozone present, SO2 present) – and
there were three chambers per treatment combination (allocated at random). After one month of treatment, total yield (mg) of bean pods was recorded for
each chamber, with results shown in the following table. The resulting data yielded SS(Between) = 54.68 and SS(Within) = 27.51.
Chamber
1
2
3
Mean
SD
A
15.20
18.50
13.90
15.87
2.37
Treatment
C
11.50
13.00
15.70
13.40
2.13
B
14.90
15.50
12.10
14.17
1.81
D
9.50
10.60
9.90
10.00
0.56
Complete the ANOVA table and carry out the F test at  = 5%.
Source
df
SS
MS
F
5.30
Between treatments
3
54.68
18.23
Within treatments
8
27.51
3.44
TOTAL
11
82.19
s_p = sqrt[3.44] = 1.8547
Null hypothesis muA = muB = muC = muD Alternative hypothesis At least one mean is not equal to others.
Degrees of freedom 3 and 8
Test statistic Fs = 5.30
Table from the book to use for this problem is on page 630
Circle the correct decision:
Reject Null
P-value (be as specific as possible) .02 < P-value < .05 (= alpha), Reject H0.
Fail to Reject Null
Use the Bonferroni method to construct a 95% confidence interval for the difference between the treatment means for treatments C and D and use this
interval to state whether or not you feel they are indeed equal. Show all work. Bonferroni Interval muC – muD = 3.40 +/- t_{8,.025/(4C2)}s_p{sqrt[1/3 +
1/3]} = 3.40 +/- 3.479(1.8547)sqrt[1/3+1/3] = 3.40 +/- 5.27. Since the interval contains muC = muD = 0, we retain equality.
15. Ten treatments were compared for their effect on the liver of mice. There were 13 animals in each treatment group. The ANOVA gave MS(within) =
0.5842. The mean liver weights are given in the table below.
Group
1
2
3
4
5
6
7
8
9
10
Mean
2.59
2.28
2.34
2.40
2.59
2.84
2.29
2.45
2.76
2.37
n
13
13
13
13
13
13
13
13
13
13
Use the Bonferroni method with using  = 5% to determine whether any pairs of means are significantly different.
The sorted group means are as follows
Treatment
4
2
7
Mean
2.07
2.28
2.29
3
2.34
10
2.37
5
2.40
8
2.45
1
2.59
9
2.76
6
2.84
The most extreme difference among groups occurs between groups 4 and 6, with means 2.84 and 2.07, respectively.
We have df = (10)(13-1) =120. We use df = 100, spooled = sqrt[0.5842] = 0.7643, and the Bonferroni multiplier 2.871.
The Bonferroni confidence interval for the difference of means between these two groups is
(2.84 – 2.07) +/- (2.871)(spooled )sqrt[1/13 + 1/13] = 0.77 +/- (2.871)(0.7643) sqrt[1/13 + 1/13] = 0.77 +/- 0.8607.
This confidence interval contains 0.
Since even the largest difference among the groups is not significant, there is no significant difference among the group means.
16. In a study of the dietary treatment of anemia in cattle, researchers randomly divided 144 cows into four treatment groups. Group A was a control group,
and groups B, C and D received different regimens of dietary supplementation with selenium. After a year of treatment, blood samples were drawn and
assayed for selenium. The accompanying table shows the mean selenium concentrations (g/dl).
The resulting data yielded a grand mean (y-bar-bar) of 4.17 and spooled = sqrt[MS(within)] = sqrt[ 2.071] = 1.439
Group
A
B
C
D
Mean
0.8
5.4
6.2
5.0
n
36
36
36
36
It turns out that the test of A = B = C = D is rejected using  = 5%. Use the Bonferroni method to compare all pairs of means at the 5% level.
Summarize your findings using the “underline method”
C
B
D
A
______
______
There are k = 6 pairs to compare and df = (4)(36 – 1) = 140. From Table 11, we find that the Bonferroni multiplier is 2.676.
For each pair, the SE = (spooled )sqrt[1/36 + 1/36] = 0.3392. We obtain +/- (2.676)(0.3392) = +/-0.9077.
The largest difference between means is that between C and A. We start with that pair.
C vs A
C vs D
B vs A
C vs B
B vs D
D vs A
muC – muA =
muC – muD =
muB – muA =
muC – muB =
muB – muD =
muD – muA =
5.4 +/- 0.9077.
1.2 +/- 0.9077.
4.6 +/- 0.9077.
0.8 +/- 0.9077.
0.4 +/- 0.9077.
4.2 +/- 0.9077.
We reject equality of muC and muA.
We reject equality of muC and muD.
We reject equality of muB and muA.
We do NOT reject equality of muC and muB.
We do NOT reject equality of muB and muD.
We reject equality of muD and muA.
This is a summary of the underline method expressed in plain English. All diets are superior to A. While diet C produces the highest mean
selenium concentration among cows in our sample, there is not enough evidence to claim C is the best. This is because we cannot conclude that C is
superior to B as the C vs B interval contains zero. Similarly, we cannot conclude that B is superior to D as the B vs D interval contains zero.
However, C is superior to D.
17. In a study of crop losses due to air pollution, plots of Blue Lake snap beans were grown in open-top field chambers, which were fumigated with various
concentrations of sulfur dioxide (x, measured in ppm). After a month of fumigation, the plants were harvested and the total yield of bean pods (y, measured
in kg) was recorded for each chamber. The results are summarized by n = 12, x-bar = .120, y-bar = 1.117, sx = .117, sy = .312
and the partial Minitab output:
Regression Analysis: Yield versus SulfurDioxide
Predictor
Constant
SulfurDio
Coef
1.38810
-2.2619
S = 0.1719
SE Coef
0.07264
0.4421
R-Sq = 72.4%
R-Sq(adj) = 69.6%
[a] Calculate the equation of the linear regression of yield on SO2 concentration. Y = 1.38810 – 2.2619X [b] Give a measure of the variability of the yield
measurements (with units) sy = .312 kg per chamber [c] Give a measure of the variability of the yield measurements about the regression line (with
units) s_{e} = 0.1719 [d] Give a 95% confidence interval for 1 = -2.2619 +/- t_{10,.025}SE_{b1} = -2.2619 +/- 2.228(0.4421) = -2.2619 +/- .9850 [e] Give
a biological a biological interpretation of b1: For each increase of 1 ppm SO2, there is a decrease in yeld of 2.2619 kg per chamber.
[f] Perform a statistical test to decide whether the population correlation coefficient (), equivalently the slope of the regression line (1), differs from zero.
Null hypothesis rho = 0 (beta1 = 0) Alternative rho not equal 0 (beta1 not equal 0) Calculated Test Statistic (show all work) (b1 – 0)/SE_{b1} = 2.2619/0.4421 = -5.12
Degrees of Freedom n-2 = 10 Table from the book to use for this problem is on page 618 P-value (be as specific as possible) P-value < 2(.0005) = .001
Circle the correct decision:
Fail to reject the Null
Reject the Null
Assumptions we must make for this test . Each data pair (x,y) is independent of all other observations. For each X, the distribution of Y is normal. If
for each X, we denote the population SD of Y by sigma_{Y|X}, then sigma_{Y|X} does not depend on X (that is, sigma_{Y|X} is constant for all X).
If for each X we denote the mean of Y by mu_{Y|X}, then mu_{Y|X} is a linear function of X (that is, mu_{Y|X} = beat0 + (beta1)X).
18. Consider the following scores of six students on two tests in a statistics course.
Student
1
2
3
4
5
6
7
Score on Score on
first test second test
40
50
70
80
80
100
70
----AV = 70
SD = 20
80
65
85
70
85
80
95
----AV = 80
SD = 10
zx
zy
(zx)(zy)
-1.5
-1.0
0.0
+0.5
+0.5
+1.5
0.0
0.0
-1.5
+0.5
-1.0
+0.5
0.0
+1.5
0.0
+1.5
0.0
-0.5
+0.25
0.0
0.0
sum = 0
sum = 0
sum = 1.25
b1 = rsy/sx = (1.25/6)(10/20) = .1042, b0 = y-bar – b1(x-bar) = 80 – (.1042)(70) = 72.71
Find the correlation coefficient r = sum(zx)(zy)/(n-1) = +1.25/6 = .208333 Find an equation of the regression line Y = .1042X + 72.71
19. To estimate forearm lengths of men, the following results were obtained for about 1000 men:
Average height = 69 inches, SD for height = 2.5 inches,
Average forearm length = 18 inches, SD = 1 inch;
Correlation coefficient = .80.
Estimate the length of the forearms of a man whose height is
[a] unknown
18 + 1 Here we use y-bar and s_y.
[b] 69 inches 18 + .6. Here we use y-bar(the regression line goes thru the point (x-bar,.y-bar))and s_{e} = sqrt[1-r^2](s_y) = sqrt[1 - .8^2](1) = 0.6
[c] 73 inches 19.28 + .6. Here we use the regression line and s_{e}. 1. x = 73, 2. z_x = (73-69)/2.5 = 1.6, 3. z_y = r z_x = (.8)(1.6) = 1.28, 4. y = 18 +
(1.28)(1) = 18.28.
[d] 66 inches 17.04 + .6