Download Confidence Intervals, Null Hypothesis - Study Guide | STAT 541 and more Exams Biostatistics in PDF only on Docsity!
6.4 Problems
6-1. Suppose that a random sample of n = 51 children is selected from the population of newborn infants in Mexico. The probability that a child in this population weighs at most 2500 grams is
presumed to be π = 0.15. Calculate the probability that six or fewer of the infants weigh at most
2500 grams, using… (a) the exact binomial distribution, (b) the normal approximation to the binomial distribution (with continuity correction ). [[Technically, an assumption under which this method may be used is violated here. Why?]]
(c) Suppose we wish to test the null hypothesis H 0 : π = 0.15 versus the alternative HA : π ≠ 0.15,
and that in this random sample of n = 51 children, we find six whose weights are under 2500 grams. Calculate the associated p -value (with continuity correction ) and 95% confidence
interval , and use each to arrive at a decision whether or not to reject H 0 at the α =.
significance level.
6-2. A new “smart pill” is tested on n = 36 individuals randomly sampled from a certain population
whose IQ scores are known to be normally distributed, with mean μ = 100 and standard
deviation σ = 27. After treatment, the sample mean IQ score is calculated to be x = 109.9, and
a two-sided test of the null hypothesis H 0 : μ = 100 versus the alternative hypothesis HA : μ ≠ 100
is performed, to see if there is any statistically significant difference from the mean IQ score of the original population. Using this information, answer the following. (a) Calculate the p -value of the sample. (b) Fill in the following table, concluding with the decision either to reject or not reject the null
hypothesis H 0 at the given significance level α.
Significance
Level α
Confidence
Level 1 − α
Confidence Interval Decision about H 0
. . . (c) Extend these observations to more general circumstances. Namely, as the significance level decreases, what happens to the ability to reject a null hypothesis? Explain why this is so, in terms of the p -value and generated confidence intervals.
6-3. Consider the distribution of serum cholesterol levels for all 20- to 74-year-old males living in the United States. The mean of this population is 211 mg/dL, and the standard deviation is 46. mg/dL. In a study of a subpopulation of such males who smoke and are hypertensive, it is assumed (not unreasonably) that the distribution of serum cholesterol levels is normally
distributed, with unknown mean μ, but with the same standard deviation σ as the original
population. (a) Formulate the null hypothesis and complementary alternative hypothesis , for testing whether
the unknown mean serum cholesterol level μ of the subpopulation of hypertensive male
smokers is equal to the known mean serum cholesterol level of 211 mg/dL of the general population of 20- to 74-year-old males. (b) In the study, a random sample of size n = 12 hypertensive smokers was selected, and found to have a sample mean cholesterol level of x = 217 mg/dL. Construct a 95% confidence interval for the true mean cholesterol level of this subpopulation.
(c) Calculate the p -value of this sample, at the α = .05 significance level.
(d) Based on your answers in parts (b) and (c), is the null hypothesis rejected in favor of the
alternative hypothesis, at the α = .05 significance level? Interpret your conclusion : What
exactly has been demonstrated, based on the empirical evidence? (e) Determine the 95% acceptance region and complementary rejection region for the null hypothesis. Is this consistent with your findings in part (d)? Why?
6-4. Consider a random sample of ten children selected from a population of infants receiving antacids that contain aluminum, in order to treat peptic or digestive disorders. The distribution of plasma
aluminum levels is known to be approximately normal; however its mean μ and standard deviation σ
are not known. The mean aluminum level for the sample of n = 10 infants is found to be x = 37.
μg/l and the sample standard deviation is s = 7.13 μg/l. Furthermore, the mean plasma aluminum
level for the population of infants not receiving antacids is known to be only 4.13 μg/l.
(a) Formulate the null hypothesis and complementary alternative hypothesis , for a two-sided test of whether the mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids. (b) Construct a 95% confidence interval for the true mean plasma aluminum level of the population of infants receiving antacids.
(c) Calculate the p -value of this sample (as best as possible), at the α = .05 significance level.
(d) Based on your answers in parts (b) and (c), is the null hypothesis rejected in favor of the
alternative hypothesis, at the α = .05 significance level? Interpret your conclusion : What
exactly has been demonstrated, based on the empirical evidence? (e) With the knowledge that significantly elevated plasma aluminum levels are toxic to human beings, reformulate the null hypothesis and complementary alternative hypothesis , for the appropriate one-sided test of the mean plasma aluminum levels. With the same sample data as above, how does the new p -value compare with that found in part (c), and what is the resulting conclusion and interpretation?
6-7. Recall that, for any random variable X , the population mean and population variance are, respectively,
Mean( X ) = μ X = E [ X ] Var( X ) = σ X^2 = E ⎡⎣(^ X − μ X )^2 ⎤⎦.
Likewise, for a random variable Y ,
Mean( Y ) = μ Y = E [ Y ] Var( Y ) = σ Y^2 = E ⎡⎣(^ Y − μ Y )^2 ⎤⎦.
In addition, suppose each population value of X corresponds to one and only one population value of Y. Then the population covariance between X and Y is defined by
Cov( X , Y ) = (^) σ (^) XY = (^) E (^) [ ( X − μ (^) X )( Y − μ Y )],
which measures the extent to which they vary with respect to one another.
These population “expected values” can be estimated by the sample mean , sample variance , and sample covariance , based on data values ( x 1 , x 2 , …, xn ) and ( y 1 , y 2 , …, yn ), respectively:
Mean( X ) ≈ x =
n (^) 1
n i i
x
∑ Var( X )^ ≈^ sx
n − 1
2 1
n i i
x x
∑ −
Mean( Y ) ≈ y =
n (^) 1 Var( Y )^ ≈^ s
n i i
y
∑ y
n − 1
2 1
n i i
y y
∑ −
Cov( X , Y ) ≈ sxy = 1 n − 1 1 (^ )(^ )
n i i i
x x y y
∑ −^ −.
It can be shown (using previous properties of mathematical expectation and some elementary algebra), that for any two random variables X and Y ,
- a) Mean ( X + Y ) = Mean( X ) + Mean( Y )
b) Var ( X + Y ) = Var( X ) + Var( Y ) + 2 Cov( X , Y )
2. a) Mean ( X − Y ) = Mean( X ) − Mean( Y )
b) Var ( X − Y ) = Var( X ) + Var( Y ) − 2 Cov( X , Y )
and that these properties hold for the sample mean , sample variance , and sample covariance as well.
For the following ordered data sets X and Y , form the new data sets X + Y and X − Y. Calculate all of their sample means, variances, and associated covariance, and verify that formulas 1 and 2 hold. (In R, use mean, var, and cov.)
X 0 6 12 18
Y 3 3 5 9
Recall that two events A and B are said to be independent if P ( A ∩ B ) = P ( A ) P ( B ). In the same spirit, two random variables X and Y are independent if their joint probability distribution P ( X ≤ x ∩ Y ≤ y ) = P ( X ≤ x ) P ( Y ≤ y ) for all x , y. It can be shown mathematically, that if X and Y are independent, then their covariance Cov( X , Y ) = 0. Repeat the above calculations for the following ordered data sets X and Y , paying special attention to
formula 2 b. What conclusion can you draw about Var( X − Y ) versus Var( X ) + Var( Y ),
if X and Y are independent? (This property is crucial in § 6.2.1.)
X 0 6 12 18
Y 3 9 3 5
6-8. The arrival time of my usual morning bus ( B ) is normally distributed, with a mean ETA at 8:00 AM, and a standard deviation of 4 minutes. My arrival time ( A ) at the bus stop is also normally distributed, with a mean ETA at 7:50 AM, and a standard deviation of 3 minutes.
(a) With what probability can I expect to catch the bus? ( Hint : What is the distribution of the random variable X = A – B , and what must be true about X in the event that I catch the bus?)
(b) How much earlier should I arrive, if I expect to catch the bus with 99% probability?
6-9. In this problem, assume that population cholesterol level is normally distributed.
(a) Consider a small clinical trial, designed to measure the efficacy of a new cholesterol- lowering drug against a placebo. A group of six high-cholesterol patients is randomized to either a treatment arm or a control arm, resulting in two numerically balanced samples of
n 1 = n 2 = 3 patients each, in order to test the null hypothesis H 0 : μ 1 = μ 2 vs. the alternative
HA : μ 1 ≠ μ 2. Suppose that the following data are obtained:
Placebo Drug 220 180 240 200 290 220
Obtain the 95% confidence interval for μ 1 − μ 2 , and the p -value of the data (use R if you wish),
and use each to decide whether or not to reject H 0 at the α = .05 significance level. Conclusion?
(b) Now imagine that the same drug is tested using another pilot study, with a different design. Serum cholesterol levels of n = 3 patients are measured at the beginning of the study, then re- measured after a six month treatment period on the drug, in order to test the null hypothesis
H 0 : μ 1 = μ 2 versus the alternative HA : μ 1 ≠ μ 2. Suppose that the following data are obtained:
Baseline End of Study 220 180 240 200 290 220
Obtain the 95% confidence interval for μ 1 − μ 2 , and the p -value of the data (use R if you wish),
and use each to decide whether or not to reject H 0 at the α = .05 significance level. Conclusion?
(c) Compare and contrast these two study designs and their results.
6-12. A formal hypothesis test for two-sample means using the confidence interval for μ 1 − μ 2 is
generally NOT equivalent to an informal side-by-side comparison of the individual
confidence intervals for μ 1 and μ 2 for detecting overlap between them.
(a) Suppose that two population random variables X 1 and X 2 are normally distributed, each
with standard deviation σ = 50. We wish to test the null hypothesis H 0 : μ 1 = μ 2 versus the
alternative H 0 : μ 1 ≠ μ 2 , at the α = .05 significance level. Two independent, random
samples are selected, each of size n = 100 , and it is found that the corresponding means are x 1 (^) = 215 and x 2 (^) = 200 , respectively. Show that even though the two individual 95%
confidence intervals for μ 1 and μ 2 overlap, the formal 95% confidence interval for the mean
difference μ 1 − μ 2 does not contain the value 0, and hence the null hypothesis can be
rejected. (See middle figure below.)
(b) In general, suppose that X 1 ~ N ( μ 1 , σ ) and X 2 ~ N ( μ 2 , σ ), with equal σ (for simplicity).
In order to test the null hypothesis H 0 : μ 1 = μ 2 versus the two-sided alternative H 0 : μ 1 ≠μ 2
at the α significance level, two random samples are selected, each of the same size n (for
simplicity), resulting in corresponding means x 1 and x 2 , respectively. Let CI μ 1 and CI μ 2
be the respective 100
1 2 / 2
x x d z (^) α σ n
(1 − α)% confidence intervals, and let =. (Note that
the denominator is simply the margin of error for the confidence intervals.) Also let CI (^) μ 1 −μ 2
be the 100 (1 − α)%confidence interval for the true mean difference μ 1 − μ 2. Prove:
y If d < 2 , then 0 ∈ CI μ 1 −μ 2 (i.e., “accept” H 0 ), and CI μ 1 ∩ CIμ 2 ≠ ∅ (i.e., overlap).
x 1
x 2
x 1 (^) − x 2
| 0
y If 2 < d < 2 , then 0 ∉^ CI μ 1 −μ 2 (i.e., reject H 0 ), butCI^ μ 1 ∩^ CIμ 2 ≠ ∅(i.e., overlap)!
y If d > 2 , then 0 ∉ CI μ 1 −μ 2 (i.e., reject H 0 ), and CI μ 1 ∩ CIμ 2 = ∅ (i.e., no overlap).
x 2
x 1 (^) − x 2
| 0 x 1
x 1
x 2
x 1 (^) − x 2
| 0
6-13. Z -tests and Chi-squared Tests
(a) Test of Independence. Imagine that a marketing research study surveys a random sample of n = 2000 consumers about their responses regarding two brands ( A and B ) of a certain product, with the following observed results.
Do You Like Brand B****? Yes No
Yes 335 915
Do You Like
Brand
A
?^ 1250
No 165 585 750
First consider the null hypothesis H 0 : π A | B = π A | BC , that is, in this consumer population,
“The probability of liking A , given that B is liked, is equal to probability of liking A , given that B is not liked.”
⇔ “There is no association between liking A and liking B .”
⇔ “Liking A and liking B are independent of each other.” (Why? See Exercise on page 3-14 of section 3.2.)
Calculate the point estimate πˆ A | B − πˆ A | BC. Determine the Z -score of this sample
(and thus whether or not H 0 is rejected at α = .05). Conclusion?
Now consider the null hypothesis H 0 : π B | A = π B | AC , that is, in this consumer population,
“The probability of liking B , given that A is liked, is equal to probability of liking B , given that A is not liked.”
⇔^ “There is no association between liking^ B^ and liking^ A .”
⇔ “Liking B and liking A are independent of each other.”
Calculate the point estimate πˆ B | A − πˆ B | AC. Determine the Z -score of this sample
(and thus whether or not H 0 is rejected at α = .05). How does it compare with the
previous Z -score? Conclusion?
Compute the Chi-squared score. How does it compare with the preceding Z -scores? Conclusion?
6-14. Consider the following 2 × 2 contingency table taken from a retrospective case-control study that investigates the proportion of diabetes sufferers among acute myocardial infarction (heart attack) victims in the Navajo population residing in the United States. MI Yes No Total
Yes 46 25 71
Diabetes^ No^98 119
Total 144 144 288
(a) Conduct a Chi-squared Test for the null hypothesis H 0 : π Diabetes | MI = π Diabetes | No MI versus the
alternative HA : π Diabetes | MI ≠ π Diabetes | No MI. Determine whether or not we can reject the null
hypothesis at the α = .01 significance level. Interpret your conclusion: At the α =.
significance level, what exactly has been demonstrated about the proportion of diabetics among the two categories of heart disease in this population?
(b) In the study design above, the 144 victims of myocardial infarction ( cases ) and the 144 individuals free of heart disease ( controls ) were actually age- and gender-matched. The members of each case-control pair were then asked whether they had ever been diagnosed with diabetes. Of the 46 individuals who had experienced MI and who were diabetic, it turned out that 9 were paired with diabetics and 37 with non-diabetics. Of the 98 individuals who had experienced MI but who were not diabetic, it turned out that 16 were paired with diabetics and 82 with non-diabetics. Therefore, each cell in the resulting 2 × 2 contingency table below corresponds to the combination of responses for age- and gender- matched case-control pairs, rather than individuals. MI Diabetes No Diabetes Totals
Diabetes 9 16 25
No MI^ No Diabetes
Totals 46 98 144
Conduct a McNemar Test for the null hypothesis H 0 : “The number of ‘diabetic, MI case’ - ‘non-diabetic, non-MI control’ pairs, is equal to the number of ‘non-diabetic, MI case’ - ‘diabetic, non-MI control’ pairs, who have been matched on age and gender,” or more succinctly, H 0 : “There is no association between diabetes and myocardial infarction in the Navajo population, adjusting for age and gender.” Determine whether or not we can reject the
null hypothesis at the α = .01 significance level. Interpret your conclusion: At the α =.
significance level, what exactly has been demonstrated about the association between diabetes and myocardial infarction in this population?
(c) Why does the McNemar Test only consider discordant case-control pairs? Hint : What, if anything, would a concordant pair (i.e., either both individuals in a ‘MI case - No MI control’ pair are diabetic, or both are non-diabetic) reveal about a diabetes-MI association, and why?
6-15. The following data are taken from a study that attempts to determine whether the use of electronic fetal monitoring (“exposure”) during labor affects the frequency of caesarian section deliveries (“disease”). Of the 5824 infants included in the study, 2850 were electronically monitored during labor and 2974 were not. Results are displayed in the 2 × 2 contingency table below.
Caesarian Delivery
Yes No Totals
Yes 358 2492
EFM Exposure
No 229 2745 2974
Totals 587 5237 5824
(a) Calculate a point estimate OR m for the population odds ratio OR , and interpret.
(b) Compute a 95% confidence interval for the population odds ratio OR.
(c) Based on your answer in part (b), show that the null hypothesis H 0 : OR = 1 can be rejected in
favor of the alternative HA : OR ≠ 1, at the α = .05 significance level. Interpret this
conclusion: What exactly has been demonstrated about the association between electronic fetal monitoring and caesarian section delivery? Be precise.
(d) Does this imply that electronic monitoring somehow causes a caesarian delivery? Can the association possibly be explained any other way? If so, how?
(d) To compute a 95% confidence interval for the summary odds ratio OR summary, we must first verify that the sample sizes in the two studies are large enough to ensure that the method used is valid.
Step 1: Verify that the expected number of observations of the ( i, j )th^ cell in the first table, plus the expected number of observations of the corresponding ( i, j )th^ cell in the second table, is greater than or equal to 5, for i = 1, 2 and j = 1, 2. Recall that the expected number of the ( i, j )th^ cell is given by Ei j = R (^) i C (^) j/ n.
Step 2: By its definition, the quantity L computed in part (b) is a weighted mean of log-odds ratios, and already represents a point estimate of ln( OR summary). The estimated standard error of L is given by
m^1 s.e.( L ) w 1 + w 2
Step 3: From these two values in Step 2, construct a 95% confidence interval for ln( OR summary), and exponentiate it to derive a 95% confidence interval for OR summary itself.
(e) Also compute the value of the Chi-squared test statistic for OR summary given at the end of § 6.2.3.
2
(f) Use the confidence interval in (d), and/or the χ 1 statistic in (e), to perform a Test of
Association of the null hypothesis H (^) 0 : OR summary = 1, versus the alternative HA : OR summary ≠ 1,
at the α = .05 significance level. Interpret your conclusion: What exactly has been
demonstrated about the association between the number of term pregnancies and the odds of developing epithelial ovarian cancer? Be precise.
6-17. In a random sample of n = 1200 consumers who are surveyed about their ice cream flavor preferences, 416 indicate that they prefer vanilla, 419 prefer chocolate, and 365 prefer strawberry.
(a) Conduct a Chi-squared “Goodness-of-Fit” Test of the null hypothesis of equal proportions
H 0 : π Vanilla = πChocolate =πStrawberry of flavor preferences, at the α = .05 significance level.
Vanilla Chocolate Strawberry
416 419 365
(b) Suppose that the sample of n = 1200 consumers is equally divided between males and females, yielding the results shown below. Conduct a Chi-squared Test of the null
hypothesis that flavor preference is not associated with gender, at the α = .05 level.
Vanilla Chocolate Strawberry Totals
Males 200 190 210 600
Females 216 229 155 600
Totals 416 419 365 1200
6-18. In the late 1980s, the pharmaceutical company Upjohn received approval from the Food and Drug Administration to market RogaineTM, a 2% minoxidil solution, for the treatment of androgenetic alopecia (male pattern hair loss). Upjohn’s advertising campaign for Rogaine included the results of a double-blind randomized clinical trial, conducted with 1431 patients in 27 centers across the United States. The results of this study at the end of four months are summarized in the 2 × 5 contingency table below, where the two row categories represent the treatment arm and control arm respectively, and each column represents a response category, the degree of hair growth reported. [Source: Ronald L. Iman, A Data-Based Approach to Statistics, Duxbury Press]
Degree of Hair Growth No Growth
New Vellus
Minimal Growth
Moderate Dense Growth Growth Total
Rogaine 301 172 178 58 5 714 Placebo 423 150 114 29 1 717 Total 724 322 292 87 6 1431
(a) Conduct a Chi-squared Test of the null hypothesis H 0 : πRogaine = πPlacebo versus the
alternative hypothesis HA : πRogaine ≠ π Placeboacross the five hair growth categories (That is,
H 0 : πNo Growth | Rogaine = πNo Growth | Placebo and πNew Vellus | Rogaine =πNew Vellus | Placebo and …
and .) Infer whether or not we can reject the null
hypothesis at the α = .01 significance level. Interpret in context: At the α = .01 significance
level, what exactly has been demonstrated about the efficacy of Rogaine versus placebo?
π Dense Growth | Rogaine =πDense Growth | Placebo
(b) Form a 2 × 2 contingency table by combining the last four columns into a single column
labeled Growth. Conduct a Chi-squared Test for the null hypothesis H 0 : π Rogaine = πPlacebo
versus the alternative H A : πRogaine ≠ πPlacebobetween the resulting No Growth versus Growth
binary response categories. (That is, H 0 : πGrowth | Rogaine = πGrowth | Placebo.) Infer whether or
not we can reject the null hypothesis at the α = .01 significance level. Interpret in context:
At the α = .01 significance level, what exactly has been demonstrated about the efficacy of
Rogaine versus placebo?
(c) Calculate the p -value using a two-sample Z -test of the null hypothesis in part (b), and show that the square of the corresponding z -score is equal to the Chi-squared test statistic found in
(b). Verify that the same conclusion about H 0 is reached, at the α = .01 significance level.
