Question
In: Statistics and Probability
A car insurance company has determined that the mean annual car insurance cost for a family...
A car insurance company has determined that the mean annual car insurance cost for a family in the town of Watlington is $1716. A researcher wants to perform a hypothesis test to determine whether the mean insurance cost for a family in the town of Putford is higher than this. The mean insurance cost for a random sample of 32 families in Putford was $1761. At the 10% significance level, do the data provide sufficient evidence to conclude that the mean insurance cost for a family in Putford is higher than $1716, the mean cost for a family in Watlington? Assume that the population standard deviation is $ 35.50. What is your conclusion?
Solutions
orchestra answered 1 year agoRelated Solutions
A car insurance company has determined that the mean annual car insurance cost for a family...
A car insurance company has determined that the mean annual car
insurance cost for a family in the town of Watlington is $1716. A
researcher wants to perform a hypothesis test to determine whether
the mean insurance cost for a family in the town of Putford is
higher than this. The mean insurance cost for a random sample of 32
families in Putford was $1761. At the 10% significance level, do
the data provide sufficient evidence to conclude that the...
A car insurance company has determined that 7% of all drivers were involved in a car...
A car insurance company has determined that 7% of all drivers
were involved in a car accident last year. Among the 11 drivers
living on one particular street, 3 were involved in a car accident
last year. If 11 drivers are randomly selected, what is the
probability of getting 3 or more who were involved in a car
accident last year? Group of answer choices 0.0317 0.0370 0.9683
0.481
A car insurance company has determined that 10% of drivers were involved in a car accident last year.
A car insurance company has determined that 10% of drivers were involved in a car accident last year. If 3 drivers are randomly selected, what is the probability of getting: a) one driver involved in a car accident last year? P(x = 1) = b) fewer than two drivers who were involved in a car accident last year? P(x2)=
A car insurance company has determined that 9% of all drivers were involved in a car accident last year.
A car insurance company has determined that 9% of all drivers were involved in a car accident last year. Among the 11 drivers living on one particular street, 3 were involved in a car accident last year. If 11 drivers are randomly selected, what is the probability of getting 3 or more who were involved in a car accident last year?
The number of successful claims on life insurance policies for an insurance company has been determined...
The number of successful claims on life insurance policies for
an insurance company has been determined to follow a Poisson
process with an average rate of 5 claims per week. If the amount of
money paid on each policy is uniformly distributed between $2,000
and $10,000, what is the mean of the total amount of money that the
company pays out in a four-week period?
Suppose that the mean annual cost of automobile insurance in NC is $989 with standard deviation...
Suppose that the mean annual cost of automobile insurance in NC
is $989 with standard deviation $225. What is the probability that
the average cost for a sample of 50 NC automobile insurance
policies is within $35 of the average annual cost for all policies?
(4 decimal places). Find answer on statcrunch
A researcher for a car insurance company wishes to estimate themean annual premium that men...
A researcher for a car insurance company wishes to estimate the
mean annual premium that men aged 20-24 years pay for their car
insurance. A random sample of 50 men aged 20-24 years shows a mean
of $711. Assume that the distribution of annual premiums has a
standard deviation of $185.a) Find a 94% confidence interval for the mean annual premium
for all men aged 20-24 years.b) What is the error of the estimate?c) Find the 99% confidence interval for...
Your insurance company has coverage for three types of cars. The annual cost for each type...
Your insurance company has coverage for three types of cars. The
annual cost for each type of car can be modeled using Gaussian
(Normal) distribution, with the following parameters:
Car Type 1: Mean=$520 and Standard Deviation=$110
Car Type 2: Mean=$720 and Standard Deviation=$170
Car Type 3: Mean=$470 and Standard Deviation=$80
Use the Random number generator and simulate 1000-long columns,
for each of the three cases. Example: for the Car Type 1, use
Number of variables=1, Number of random numbers=1000,
Distribution=Normal,...
Your insurance company has converage for three types of cars. The annual cost for each type...
Your insurance company has converage for three types of cars.
The annual cost for each type of car can be modeled using Gaussian
(Normal) distribution, with the following parameters:
Car Type 1: Mean=$520 and Standard Deviation=$110
Car Type 2: Mean=$720 and Standard Deviation=$170
Car Type 3: Mean=$470 and Standard Deviation=$80
Use the Random number generator and simulate 1000-long columns,
for each of the three cases. Example: for the Car Type 1, use
Number of variables=1, Number of random numbers=1000,
Distribution=Normal,...
Your insurance company has converged for three types of cars.The annual cost for each type...
Your insurance company has converged for three types of cars.
The annual cost for each type of cars can be modeled using Gaussian
(Normal) distribution, with the following parameters: (Discussions
allowed!)Car type 1 Mean=$520 and Standard Deviation=$110Car type 2 Mean=$720 and Standard Deviation=$170Car type 3 Mean=$470 and Standard Deviation=$80Use Random number generator and simulate 1000 long columns, for
each of the three cases. Example: for the Car type 1, use Number of
variables=1, Number of random numbers=1000, Distribution=Normal,
Mean=520 and...
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