A random sample of 859 persons showed that 17.9% do not have any
health insurance. Based...
A random sample of 859 persons showed that 17.9% do not have any
health insurance. Based on this sample, the 95% confidence interval
for the proportion of all persons who do not have any health
insurance is:
A random sample of 45 life insurance policy holders showed that
the average premiums paid on their life insurance policies was $345
per year with a sample standard deviation of $65.
Construct a 90% confidence interval for the population mean.
Make a statement about this in context of the problem.
In a random sample of 100 persons with low incomes, 100 persons
with average incomes, and 100 persons with high incomes, there
were, respectively, 40, 50, and 60 who love biking.
(a) Use the 0.05 level of significance to test the null
hypothesis θ1 = θ2 = θ3 = 0.5 (that the proportion of persons
loving biking is 50% for all three income groups) against the
alternative hypothesis that the three θ’s are not all equal to 0.5.
In carrying...
part A:
Based on a random sample of 200 homeowner insurance claims, the
length of time it takes until the insurance adjustor first contacts
the homeowner follows a normal distribution with a mean of 4.5
hours and a standard deviation of 1.1 hours.
Create a 95% confidence interval for the population average amount
of time it takes for the adjustor to first contact the homeowner.
(Round your answers to 1 decimal place)
Confidence Interval: (___________), (__________) (round to one...
any state auto insurance company took a random sample
of 360 insurance claims paid out during a one year. The average
claim paid was $1,575 assume the name equals $238 find a 0.90
confidence interval for the mean claim round your answers to two
decimal places lower limit of the find a 0.99 confidence interval
for the mean claim payment round your answers to two decimal places
lower limit upper limit
In the US, people die because they do not have private health
insurance or their insurance does not cover the treatments they
need. A 2009 Harvard study published in the American Journal of
Public Health found more than 44,800 excess deaths annually in the
United States due to Americans lacking health insurance, equivalent
to one excess death every 12 minutes. More broadly, the total
number of people in the United States, whether insured or
uninsured, who die because of lack...
A simple random sample of 400 persons is taken to estimate the
percentage of Republicans in a large population. It turns out that
210 of the people in the sample are Republicans. True or False and
explain.
a. The sample percentage is 52.5%; the SE for the sample
percentage is 2.5%.
b. 52.5% ± 2.5% is a 75%-confidence interval for the population
percentage.
c. 52.5% ± 5% is a 95%-confidence interval for the sample
percentage.
d. 52.5% ± 5% is...
A simple random sample of 3,600 persons is taken, to estimate
the percentage of smokers in a certain large population. It turns
out that 1,217 people in the sample are smokers.
(a) Find the estimated percentage of smokers in the
population.
(b) Find the standard error of the estimated percentage of
smokers in (a).
(c) Find a 95% confidence interval of the percentage of smokers
in the population.
(d) Find the margin of error for your estimation of the
percentage...
q5. A random sample of 500 persons was taken in a
certain area if test is positive for the covid-19. let 60 persons
were found with positive test. obtain 98% confidence limit for the
proportion of the covid 19 positive cases
A random sample of 500 persons is questioned regarding
their political affiliation and opinion on a tax reform
bill. The observed contingency table is given below:
Favored
Indifferent
Opposed
Row total
Democrat
138
83
64
285
Republican
64
67
84
215
Column total
202
150
148
500
Which variable can be reasonably thought of as the dependent
variable, and which is the independent variable?
What is the probability that a person favored the tax reform
bill, given that the person...