Download Sample Exam 3 Solutions - University of Arizona Math

Sample Exam 3 Solutions - Math 263-009 - Fall 13 -Kennedy
1. A test of significance was conducted with a random sample of 25 subjects; the p-value was 0.028 and the significance level was 5%. Mark
each of the following statements either T (true) or F (false):
(a) There is evidence against the null hypothesis. TRUE
(b) The null hypothesis is definitely false. FALSE
(c) The p-value was calculated under the assumption that the alternate hypothesis was true. FALSE
(d) The probability of a value of the test statistic at least as extreme
as that observed in this study, assuming the null hypothesis is
true, is 0.028. TRUE
(e) If the test is repeated by taking a new random sample of 25 subjects, the same p-value will be obtained. FALSE
2. Mark each of the following statements as T (true) or F (false). In each
scenario, everything remains fixed except the quantity that is increasing
or decreasing.
(a) The margin of error for a 95% confidence interval for the mean µ
increases as the sample size increases. FALSE
(b) The margin of error for a confidence interval for the mean µ, based
on a specified sample size n, increases as the confidence level decreases. FALSE
(c) The margin of error for a 95% confidence interval for the mean µ
decreases as the population standard deviation decreases. TRUE
(d) The sample size required to obtain a confidence interval of specified margin of error increases as the confidence level increases.
TRUE
3. The area under the normal curve to the right of a test statistic z = 1.812
is 0.035. Mark each of the following statements as T (true) or F (false).
Assume µ0 is a constant.
(a) If the alternative hypothesis is of the form Ha : µ > µ0 , the data
are statistically significant at significance level α = 0.05. TRUE
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(b) If the alternative hypothesis is of the form Ha : µ > µ0 , the data
are statistically significant at significance level α = 0.10. TRUE
(c) If the alternative hypothesis is of the form Ha : µ 6= µ0 , the data
are statistically significant at significance level α = 0.05. FALSE
(d) If the alternative hypothesis is of the form Ha : µ 6= µ0 , the data
are statistically significant at significance level α = 0.10. TRUE
4. The heights of a random sample of 400 male high school sophomores
are measured. The sample mean is x = 66.2 inches. The heights of
male high school sophomores follow a normal distribution with standard deviation σ = 4.1 inches. What is a 95% confidence interval for
the mean, µ?
We know the population mean, so we use a z statistic. For a 95%
confidence interval z ∗ = 1.960. So the interval is
4.1
66.2 ± 1.960 · √
= (65.80, 66.60)
400
5. (Continuation of previous problem): Suppose the heights of 100 male
sophomores were measured, rather than 400. Which of the following
statements is true?
(a) The margin of error for the 95% confidence interval would increase.
6. To assess the accuracy of a laboratory scale, an object known to weigh
exactly 1 gram is weighed n times and the mean scale weight, x, is computed. The scale readings are normally distributed with an unknown
mean of µ and a standard deviation of σ = 0.01gm. How large should
n be so that a 95% confidence interval for µ has a margin of error of
±0.0001gm ?
We know the population standard deviation, so we use a z statistic.
For 95% confidence interval z ∗ = 1.96. Since the margin of error must
be 0.0001, we have
0.01
1.96 · √ = 0.0001
n
This gives n = 38, 416.
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7. The US Environmental Agency (EPA) defines the Action Level for lead
in tap water as 15 parts per billion (ppb). Above this level, further
investigation and public education campaigns are required. A random
sample of 10,341 water specimens from the Washington DC area found
that 7,496 samples were below the Action Level. Find a 95% confidence
interval for the proportion of all possible water specimens that exceed
the Action Level in the DC area.
Note that the problem asked how many exceed the standard, so we take
p to be the proportion that exceed. There are 10, 341 − 7496 = 2845
samples that exceeded the Action level, so p̂ = 2845/10341 = 0.2751.
The standard error is
r
0.2751(1 − 0.2751)
= 0.0044,
10341
The 95% confidence interval is
0.2751 ± 1.96 · 0.044 = (0.266, 0.284) = (26.6%, 28.4%)
8. To determine the pollution level in a river or lake, samples of water are
tested for biological oxygen demand (BOD); higher values generally
mean more pollution. An aquaculture farm takes water from a stream
and returns it after the water has circulated through fish tanks. Ten
samples of upstream water have a sample mean BOD of 6.654 mg/l with
a sample standard deviation 0.461 mg/l; ten samples of downstream
water have a sample mean BOD of 8.687 mg/l with a sample standard
deviation 0.547 mg/l. Assume BOD levels are normally distributed.
Is the downstream BOD level significantly higher than the upstream
BOD level? Use the following steps:
Let Population 1 be the downstream water and Population 2 the upstream water. The BOD level is a quantitative variable - we are comparing two population means.
(a) Null hypothesis: H0 : µ1 = µ2
(b) Alternate hypothesis: Ha : µ1 > µ2
(c) We know that x1 = 8.687, x2 = 6.654. s1 = 0.547, s2 = 0.461,
and n1 = n2 = 10, so we have a t-statistic with df = 9 and
8.687 − 6.654
= 8.98.
t= q
0.5472
0.4612
+
10
10
3
(d) Since this t-value is large, the p-value is essentially 0. (The calculator gives approximately 4.3 × 10−6
(e) Since the p-value is small, we reject the null hypothesis.
(f) We conclude that the BOD levels downstream are significantly
higher than upstream.
9. A researcher compares two cholesterol-reducing medications. The researcher finds 22 adult volunteers with high cholesterol. The volunteers
are randomly divided into two groups of 11 people each. Group 1 takes
Medication 1, and Group 2 takes Medication 2. The mean cholesterol
reduction in Group 1 was 60 mg/dL. The mean cholesterol reduction
in Group 2 was 40 mg/dL. Assume that the standard deviation of both
populations is 5 mg/dL and that both populations are normal.
(a) What are the two populations in this study? One is people with
high cholestrol who take medication 1, the other is people with
high cholestrol who take medication 2.
(b) Find a 98% confidence interval for the difference in cholesterol
reduction between groups 1 and 2. Give at least 3 decimal places.
We know the population standard deviations so we use a z statistic. For a 98% confidence interval z ∗ = 2.33. So the interval
is
r
52
52
(60 − 40) ± 2.33
+
= 20 ± 4.97 = (15.03, 24.97)
11 11
(c) Interpret your confidence interval by finishing the following sentence. Do NOT use any variables in your answer; instead, you
should give your answer in terms that someone who did not know
statistics would understand.
There is a 98% chance that... the true reduction in cholestrol level
is in the confidence interval given above.
10. Burning fossil fuels lead to acid rain that can turn lakes acidic; this
acidity damages fish populations and disrupts plant life. Acidity is
measured by pH, where a pH of 6 or below is considered acidic. A
pH above 6 is nonacidic. A study of 15 high mountain lakes in the
Southern Alps of Italy and Switzerland found the average pH of 6.6
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and standard deviation 0.672. Does this data provide evidence that
high mountain lakes are nonacidic? Use a 1% significance level and the
following steps to decide.
(a) What is the null hypothesis? H0 : µ = 6
(b) What is the alternate hypothesis? Ha : µ > 6
(c) What is the test statistic? (Make clear if this is t or z.) We do not
know the population standard deviation, so we use the t statistic.
t=
6.6 − 6
√ = 3.46
0.672/ 15
(d) Find the p-value. There are 14 degrees of freedom, so we find
p = tcdf (3.46, 100, 14) = 0.0019. Or if we use the table we find
the p-value is between 0.0025 and 0.001
(e) Give a conclusion about the null hypothesis. We reject the null
hypothesis.
(f) Explain your conclusion in words about lakes. The data provides
strong evidence that high mountain lakes in the Southern Alps
have an average pH above 6, so they are nonacidic.
11. The clinical trial of Reclast (a new osteoporosis drug) included a test
of the side effects of 3862 patients taking the drug, 209 quit because of
side effects. Of 3852 patients taking the placebo, 187 quit because of
side effects.
Is there a significant difference in the proportions of the two groups
that quit?
We take population one to be the people taking the new drug, population two to be the people taking the placebo.
(a) Null hypothesis: H0 : p1 = p2
(b) Alternate hypothesis: Ha : p1 6= p2
(c)
SE =
s
0.0513(1 − 0.0513)
5
1
1
+
3862 3852
= 0.00502
So
z=
p̂1 − p̂2
= 1.11
SE
(d) The p-value is p = P (|Z| > 1.11) = 2 · 0.1335 = 0.267 = 26.7%.
(e) Since the p-value is large, we do not have evidence to reject the
null hypothesis. We do not have evidence that there is a difference
in the proportions of people that have side effects, i.e., there is no
evidence (from this study) that Reclast produces side effects more
or less often than a placebo.
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