Question
In: Statistics and Probability
A survey was taken of randomly selected Americans, age 65 and older, which found that 409...
A survey was taken of randomly selected Americans, age 65 and older, which found that 409 of 1001
men and 537 of 1064 women suffered from some form of arthritis.
a) Are the assumptions and conditions necessary for inference satisfied? Why?
b) Create a 95% confidence interval for the difference in the proportions of senior men and women who have this disease.
c) What is the proportion of American men and women age 65 and older who suffer from arthritis between?
d) Does this suggest that arthritis is more likely to afflict women than men? Why?
Solutions
Expert Solution
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orchestra answered 2 years agoRelated Solutions
A survey was taken of randomly selected Americans, age 65 and older, which found that 420...
A survey was taken of randomly selected Americans, age 65 and
older, which found that 420 of 1012 men and 531 of 1061 women
suffered from some form of arthritis.
a) Let p1 be the sample proportion of senior women suffering
from some form of arthritis, and let p2 be the sample proportion
of senior men suffering from some form of arthritis. Create a 95%
confidence interval for the difference in the proportions of senior
men and women who have...
The Centers for Disease Control and Prevention reported a survey of randomly selected Americans age 65...
The Centers for Disease Control and Prevention reported a
survey of randomly selected Americans age 65 and older, which found
that 411 of 1012 mean and 535 of 1062 women suffered from some form
of arthritis.
Test the claim that a different proportion of
senior men and women who have this disease at the α = 0.05.
The Centers for Disease Control and Prevention reported a survey of randomly selected Americans age 65...
The Centers for Disease Control and Prevention reported a survey
of randomly selected Americans age 65 and older, which found 411 of
1012 men and 535 of 1062 women suffered from some sort of
arthritis. Find a 90% confidence interval for the difference in the
proportion of senior men and senior women who have this
disease.
13) Answer the following:
a) Is this a “proportions of success” or a “means” problem?
b) What calculator function are you using?
c) What...
2) A survey of 25 randomly selected customers found that their average age was 31.84 years...
2) A survey of 25 randomly selected customers found that their
average age was 31.84 years with a standard deviation of 9.84
years. a. What is the standard error of the mean?
b. How would the standard error have changed if the sample size
had been 100 instead of 25 (assuming that the mean and standard
deviations were the same)?
c. How many degrees of freedom would the t-statistic have for
this set of data?
d. What would the critical...
At a local store, 65 female employees were randomly selected and it was found that their...
At a local store, 65 female employees were randomly selected and
it was found that their mean monthly income was $625 with a
standard deviation of $121.50. Seventy-five male employees were
also randomly selected and their mean monthly income was found to
be $667 with a standard deviation of $168.70. Test the hypothesis
that male employees have a higher monthly income than female
employees. Use α = 0.01. Use any method, however, follow the
PHANTOMS acronym.
P - Parameter Statement...
At a local store, 65 female employees were randomly selected and it was found that their...
At a local store, 65 female employees were randomly selected and
it was found that their mean monthly
income was $625 with a standard deviation of $121.50. Seventy-five
male employees were also randomly
selected and their mean monthly income was found to be $667 with a
standard deviation of $168.70. Test the
hypothesis that male employees have a higher monthly income than
female employees. Use α = 0.01.
Use any method, however, follow the PHANTOMS acronym.
P - Parameter Statement...
At a local college, 65 female students were randomly selected and it was found that their...
At a local college, 65 female students were randomly selected
and it was found that their mean monthly income
was $609. Seventy-five male students were also randomly selected
and their mean monthly income was found to
be $651. Test the claim that male students have a higher monthly
income than female students. Assume the
population standard deviation for the females is $121.50 and for
the males $131. Use α = 0.01
A researcher has selected a random sample of 120 older residents (age 65+) in Nebraska and...
A researcher has selected a random sample of 120 older residents
(age 65+) in Nebraska and asked them how many times they’ve been
victimized by crime in the past year. He found that the senior
citizens averaged 2.8 victimizations last year. The researcher also
has information on the amount of victimization experienced in the
entire state. On average, the population of Nebraskans experienced
2.4 victimizations last year, with a standard deviation of 2.7. The
researcher wonders whether senior citizens are...
A survey of 2,306 adult Americans aged 18 or older found that 417 have donated blood...
A survey of 2,306 adult Americans aged 18 or older found that
417 have donated blood in the past two years. The point estimate
p-hat = .181, and the sample size n = 2,306. Given the value of
LaTeX: \sigmaσσp-hat found in the above question, and with the
knowledge that zLaTeX: _{\frac{\alpha}{2}}α 2α 2=1.96 for LaTeX:
\alphaαα=.05, Construct a 95% confidence interval for the
population proportion of adult Americans aged 18 or older who have
donated blood in the past...
In a survey of 356 randomly selected gun owners, it was found that 82 of them...
In a survey of 356 randomly selected gun owners, it was found
that 82 of them said they owned a gun primarily for protection.
Find the margin of error and 95% confidence interval for the
percentage of all gun owners who would say that they own a
gun primarily for protection. Round all answers to 3 decimal
places.
Margin of Error (as a percentage): %
Confidence Interval: % to %
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