Question
In: Statistics and Probability
The research department of an insurance company conducted a survey of the cause of automobile accidents...
The research department of an insurance company conducted a survey of the cause of automobile accidents in the last calendar year. A random sample of 200 policies written on single persons revealed that 60 had been in at least one accident. A similar survey of 300 policies written on married persons revealed that 75 had been in at least one accident. At the .05 level of significance, is there a significant difference in the proportion of the population of single and married persons having at least one accident during the period of these surveys? The hypothesis statements for this problem are: H0: πS – πM = 0 H1: πS – πM ≠ 0
Solutions
orchestra answered 2 years agoRelated Solutions
The research department of an insurance company conducted a survey of the cause of automobile accidents...
The research department of an insurance company conducted a
survey of the cause of automobile accidents in the last calendar
year. A random sample of 200 policies written on single persons
revealed that 60 had been in at least one accident. A similar
survey of 300 policies written on married persons revealed that 75
had been in at least one accident. At the .05 level of
significance, is there a significant difference in the proportion
of the population of single...
A survey of 70 people conducted by a companys statistical research department concluded that 45 respondents...
A survey of 70 people conducted by a companys statistical
research department concluded that 45 respondents claimed they do
not trust their insurance companies. According to the company it
has become common place to believe that the majority of respondents
do not trust insurance companies. Test the validity of this claim
with a significance level of 0.05.
The claims department at Wise Insurance Company believes that younger drivers have more accidents and, therefore,...
The claims department at Wise Insurance Company believes that
younger drivers have more accidents and, therefore, should be
charged higher insurance rates. Investigating a sample of 1,200
Wise policyholders revealed the following breakdown on whether a
claim had been filed in the last 3 years and the age of the
policyholder. Is it reasonable to conclude that there is a
relationship between the age of the policyholder and whether or not
the person filed a claim? Use the .05 significance...
A study conducted by the research department of a pharmaceutical company claims that the annual spending...
A study conducted by the research department of a pharmaceutical
company claims that the annual spending (per person) for
prescription drugs for allergy relief,
μ1
, is greater than or equal to the annual spending (per person)
for non-prescription allergy relief medicine,
μ2
. A health insurance company conducted an independent study and
collected data from a random sample of
225
individuals for prescription allergy relief medicine. The sample
mean is found to be
17.5
dollars/year, with a sample standard...
A survey conducted by the American Automobile Association showed that a family of four spends an...
A survey conducted by the American Automobile Association showed
that a family of four spends an average of $215:60 per day while on
vacation. Suppose a sample of 64 families of four vacationing at
Niagara Falls resulted in a sample mean of $252:45 per day and a
sample standard deviation of $74:50. Using a 98% confidence
interval estimate, you wish to determine whether the average amount
a family of four will spend when vacationing at Niagara Falls per
day is...
A survey conducted by the American Automobile Association showed that a family of four spends an...
A survey conducted by the American Automobile Association showed
that a family of four spends an average of $215.60 per day while on
vacation. Suppose a sample of 64 families of four vacationing at
Niagara Falls resulted in a sample mean of $252.45 per day and a
sample standard deviation of $74.50.
a. Develop a 95% confidence interval estimate of the mean amount
spent per day by a family of four visiting Niagara Falls (to 2
decimals).
( , )...
A survey conducted by the American Automobile Association showed that a family of four spends an...
A survey conducted by the American Automobile Association showed
that a family of four spends an average of $315.60 per day when on
vacation. Assume that $315.60 is the population mean expenditure
per day for a family of four and that $85.00 is the population
standard deviation. Suppose that a random sample of 40 families
will be selected for further study.
What is the standard deviation of the sampling distribution of
xbar values where xbar is the mean expenditure per...
For a study conducted by the research department of a pharmaceutical company, 295 randomly selected individuals...
For a study conducted by the research department of a
pharmaceutical company, 295 randomly selected individuals were
asked to report the amount of money they spend annually on
prescription allergy relief medication. The sample mean was found
to be $17.60 with a standard deviation of $5.70. A random sample of
235 individuals was selected independently of the first sample.
These individuals reported their annual spending on
non-prescription allergy relief medication. The mean of the second
sample was found to be...
A market research company conducted a telephone survey of 255 Melbourne households to determine the proportion...
A market research company conducted a telephone survey of 255
Melbourne households to determine the proportion of households
seeking to purchase a new car in the next two years. An interval
estimate of 0.151 to 0.289 was calculated. Determine the level of
confidence (answer as percentage correct to two decimal
places).
A market research company conducted a telephone survey of 320 Melbourne households to determine the proportion...
A market research company conducted a telephone survey of 320
Melbourne households to determine the proportion of households
seeking to purchase a new car in the next two years. An interval
estimate of 0.134 to 0.196 was calculated. Determine the level of
confidence (answer as percentage correct to two decimal
places).
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