Approximately 20 of 1000 fish in a pond are affected with a skin .docx

Approximately 20% of 1000 fish in a pond are affected with a
skin disease. A random sample of 20 fish are selected. What is
the mean of the sampling distribution for the proportion of your
sample that is infected? What is the standard deviation of the
sampling distribution for the proportion of your sample that is
infected?
Chapter 4, Section 3, Exercise 075
Match the p-values with the appropriate conclusion:
(a) The evidence against the null hypothesis is significant, but
only at the 10% level.
(b) The evidence against the null and in favor of the alternative
is very strong.
(c) There is not enough evidence to reject the null hypothesis,
even at the 10% level.
(d) The result is significant at a 5% level but not at a 1% level.
Sleep or Caffeine for Memory?
The consumption of caffeine to benefit alertness is a common
activity practiced by 90% of adults in North America. Often
caffeine is used in order to replace the need for sleep. One

recent study1 compares students' ability to recall memorized
information after either the consumption of caffeine or a brief
sleep. A random sample of 35 adults (between the ages of 18-39
) were randomly divided into three groups and verbally given a
list of 24 words to memorize. During a break, one of the groups
takes a nap for an hour and a half, another group is kept awake
and then given a caffeine pill an hour prior to testing, and the
third group is given a placebo. The response variable of interest
is the number of words participants are able to recall following
the break. The summary statistics for the three groups are in the
table below. We are interested in testing whether there is
evidence of a difference in average recall ability between any
two of the treatments. Thus we have three possible tests
between different pairs of groups: Sleep vs Caffeine, Sleep vs
Placebo, and Caffeine vs Placebo.
Group
Sample size
Mean
Standard Deviation
Sleep
12
15.25
3.3
Caffeine
12
12.25
3.5
Placebo
11
13.70
3.0
1 Mednick, Cai, Kanady, and Drummond, "Comparing the
benefits of caffeine, naps and placebo on verbal, motor and
perceptual memory", Behavioural Brain Research, 193 (2008),

79-86.
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(a) In the test comparing the sleep group to the caffeine group,
the p-value is 0.003.
What is the conclusion of the test?
H0.
In the sample , which group had better recall ability?
According to the test results, do you think sleep is really better
than caffeine for recall ability?
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(b) In the test comparing the sleep group to the placebo group,
the p-value is 0.06.
What is the conclusion of the test, using a 5% significance
level?
H0.
What is the conclusion of the test, if we use a 10% significance
level?
H0.
How strong is the evidence of a difference in mean recall ability
between these two treatments?
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(c) In the test comparing the caffeine group to the placebo
group, the p-value is 0.22.
What is the conclusion of the test?
H0.
In the sample , which group had better recall ability?
According to the test results, would we be justified in
concluding that caffeine impairs recall ability?
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(d) According to this study, what should you do before an exam

that asks you to recall information?
Take a nap.
Have a placebo.
Have some coffee.
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Arsenic in Chicken
A restaurant chain is measuring the levels of arsenic in chicken
from its suppliers. The question is whether there is evidence
that the mean level of arsenic is greater than 80 ppb, so we are
testing H0: =80 vs Ha: >80, where represents the average level
of arsenic in all chicken from a certain supplier. It takes money
and time to test for arsenic so samples are often small. Suppose
n=6 chickens from one supplier are tested, and the level of
arsenic (in ppb) are
63,71,82,92,93,139.
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(a) What is the sample mean for the data?
Round your answer to the nearest integer.
x=
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the tolerance is +/-2%

(b) Translate the original sample data by the appropriate amount
to create a new data set in which the null hypothesis is true.
How do the sample size and standard deviation of this new data
set compare to the sample size and standard deviation of the
original data set?
They are different.
The sample size is the same but the standard deviation is
different.
The standard deviation is the same but the sample size is
different.
They are the same.
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(c) Write the six new data values from part (b) on six cards.

Sample from these cards with replacement to generate one
randomization sample. (Select a card at random, record the
value, put it back, select another at random, until you have a
sample size of 6 , to match the original sample size.) Give the
sample mean.
x=
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Hypotheses for a statistical test are given, followed by several
possible confidence intervals for different samples. In each
case, use the confidence interval to state a conclusion of the test
for that sample, and give the significance level used. In
addition, in each case for which the results are significant, state
which group (1 or 2) has the larger mean.
Hypotheses: H0: 1=2 vs Ha:1≠2
(a) 95% confidence interval for : 1-2: 0.16 to 0.58
Conclusion: H0
Significance level:
Group with the larger mean:
(b) 99% confidence interval for 1-2: -2.5 to 5.5

Conclusion: H0
Significance level:
(c) 90% confidence interval for 1-2: -10.8 to -3.3
Conclusion: H0
Significance level:
Are you "In a Relationship"?
A new study1 shows that relationship status on Facebook
matters to couples. The study included 58college-age
heterosexual couples who had been in a relationship for an
average of 19 months. In 45 of the 58 couples, both partners
reported being in a relationship on Facebook. In 31 of the 58
couples, both partners showed their dating partner in their
Facebook profile picture. Men were somewhat more likely to
include their partner in the picture than vice versa. However,
the study states: "Females' indication that they are in a
relationship was not as important to their male partners
compared with how females felt about male partners indicating
they are in a relationship." Using a population of college-age
heterosexual couples who have been in a relationhip for an
average of 19 months:
(a) A 95% confidence interval for the proportion with both
partners reporting being in a relationship on Facebook is about
0.66 to 0.88. What is the conclusion in a hypothesis test to see
if the proportion is different from 0.5? What significance level
is being used?

Conclusion: H0
Significance level:
(b) A 95% confidence interval for the proportion with both
partners showing their dating partner in their Facebook profile
picture is about 0.40 to 0.66 . What is the conclusion in a
hypothesis test to see if the proportion is different from 0.5?
What significance level is being used?
Conclusion: H0
Significance level:
1 Roan, Shari, "The true meaning of Facebook's 'in a
relationship'", Los Angeles Times, February 23, 2012, reporting
on a study in Cyberpsychology, Behavior, and Social
Networking.
Find the specified areas for a normal density.
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(a) The area below 77 on a N(75,10) distribution
Round your answer to three decimal places.
area=
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(b) The area above 26 on a N(20,6) distribution
area
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(c) The area between 11 and 14 on a N(12.2, 1.6) distribution
area=
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Find endpoint(s) on the given normal density curve with the
given property.
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(a) The area to the left of the endpoint on a N(5,4) curve is
about 0.10.
Round your answer to two decimal places.
endpoint
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(b) The area to the right of the endpoint on a N(500,24) curve is
about 0.05.
endpoint=
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Choose the graph for the middle 80% for a standard normal
distribution.

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Choose the graph for the the middle 80% for a standard normal
distribution converted to a N(100,15) distribution.

Random Samples of College Degree Proportions
The distribution of sample proportions of US adults with a
college degree for random samples of size n=500 is
N(0.275,0.02). How often will such samples have a proportion,
p, that is more than 0.300?
Round your answer to one decimal place.
% of samples of 500 US adults will contain more
than 30.0% with at least a bachelor’s degree.
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the absolute tolerance is +/-0.1
Curving Grades on an Exam
A statistics instructor designed an exam so that the grades
would be roughly normally distributed with mean =75 and
standard deviation δ=11. Unfortunately, a fire alarm with ten
minutes to go in the exam made it difficult for some students to
finish. When the instructor graded the exams, he found they
were roughly normally distributed, but the mean grade was 60
and the standard deviation was 18. To be fair, he decides to
‘‘curve” the scores to match the desired N(75,11) distribution.

To do this, he standardizes the actual scores to z-scores using
the N(60,18) distribution and then ‘‘unstandardizes” those z-
scores to shift to N(75,11).
What is the new grade assigned for a student whose original
score was 45?
new score=
How about a student who originally scores an 87?
new score
Find the indicated confidence interval. Assume the standard
error comes from a bootstrap distribution that is approximately
normally distributed.
A 95% confidence interval for a proportion p if the sample has
n=200 with p=0.34, and the standard error is SE=0.03
Round your answers to three decimal places.
The 95% confidence interval is to .
Find the indicated confidence interval. Assume the standard
error comes from a bootstrap distribution that is approximately
normally distributed.
A 90% confidence interval for a mean if the sample has n=80
with x=22.9 and s=5.8, and the standard error is SE=0.65

Where Is the Best Seat on the Plane?
A survey of 1000 air travelers1 found that 60% prefer a window
seat. The sample size is large enough to use the normal
distribution, and a bootstrap distribution shows that the
standard error is SE=0.015.
Use a normal distribution to find a 90% confidence interval for
the proportion of air travelers who prefer a window seat.
1Willingham, A., ‘‘And the best seat on a plane is... 6A!,”
HLNtv.com, April 25, 2012.
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Smoke-Free Legislation and Asthma
Hospital admissions for asthma in children younger than 15
years was studied1 in Scotland both before and after
comprehensive smoke-free legislation was passed in March
2006. Monthly records were kept of the annualized percent
change in asthma admissions, both before and after the
legislation was passed. For the sample studied, before the
legislation, admissions for asthma were increasing at a mean
rate of 5.2% per year. The standard error for this estimate is
0.7% per year. After the legislation, admissions were decreasing
at a mean rate of 18.2% per year, with a standard error for this
mean of 1.79%. In both cases, the sample size is large enough to
use a normal distribution.
1Mackay, D., et. al., ‘‘Smoke-free Legislation and
Hospitalizations for Childhood Asthma,” The New England
Journal of Medicine, September 16, 2010; 363(12):1139-45.
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(a) Find a 95% confidence interval for the mean annual percent
rate of change in childhood asthma hospital admissions in

Scotland before the smoke-free legislation.
Round your answers to one decimal place.
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(b) Find a 95% confidence interval for the same quantity after
the legislation.
Round your answers to one decimal place.
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(c) Is this an experiment or an observational study?
Experiment
Observational study
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(d) The evidence is quite compelling. Can we conclude cause
and effect?
Yes

No
Penalty Shots in World Cup Soccer
A study1 of 138 penalty shots in World Cup Finals games
between 1982 and 1994 found that the goalkeeper correctly
guessed the direction of the kick only 41% of the time. The
article notes that this is ‘‘slightly worse than random chance.”
We use these data as a sample of all World Cup penalty shots
ever. Test at a 5% significance level to see whether there is
evidence that the percent guessed correctly is less than 50%.
The sample size is large enough to use the normal distribution.
The standard error from a randomization distribution under the
null hypothesis is SE=0.043.
1St.John, A., ‘‘Physics of a World Cup Penalty-Kick Shootout -
2010 World Cup Penalty Kicks,” Popular Mechanics, June 14,
2010.
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State the null and alternative hypotheses.
What is the test statistic?
z=
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What is the p-value?
p-value=
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What is the conclusion?
Reject H0 and find evidence that the proportion guessed
correctly is not less than half.
Reject H0 and find evidence that the proportion guessed
correctly is less than half.
Do not reject H0 and do not find evidence that the proportion
guessed correctly is less than half.
Do not reject H0and find evidence that the proportion guessed
correctly is less than half.
How Often Do You Use Cash?

In a survey1 of 1000 American adults conducted in April 2012,
43% reported having gone through an entire week without
paying for anything in cash. Test to see if this sample provides
evidence that the proportion of all American adults going a
week without paying cash is greater than 40%. Use the fact that
a randomization distribution is approximately normally
distributed with a standard error of SE=0.016. Show all details
of the test and use a 5% significance level.
1‘‘43% Have Gone Through a Week Without Paying Cash,”
Rasmussen Reports, April 11, 2011.
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State the null and alternative hypotheses.
What is the test statistic?
z=

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What is the p-value?
p-value=
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What is the conclusion?
Do not reject H0 and find evidence that the proportion is greater
than 40%.
Do not reject H0 and do not find evidence that the proportion is
greater than 40%.
Reject H0 and find evidence that the proportion is not greater
than 40%.
Reject H0 and find evidence that the proportion is greater than
40%.
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Approximately 20 of 1000 fish in a pond are affected with a skin .docx

Approximately 20 of 1000 fish in a pond are affected with a skin .docx

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