Question

Suppose that insurance companies did a survey. They randomly surveyed 450 drivers and found that 340 claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up. NOTE: If you are using a Student's t-distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

A. (i) Enter an exact number as an integer, fraction, or decimal.

x =

(ii) Enter an exact number as an integer, fraction, or decimal.

n =

(iii) Round your answer to four decimal places.

p' =

(rounded to four decimal places)

B.

Which distribution should you use for this problem? (Round your answer to four decimal places.)

P' ~___ (________,_________)

C. Construct a 95% confidence interval for the population proportion who claim they always buckle up.

(i) State the confidence interval. (Round your answers to four decimal places.)

(ii) Sketch the graph.

a/2=

C.L.=

(iii) Calculate the error bound. (Round your answer to four decimal places.)

Answer #1

Insurance companies are interested in knowing the population
percent of drivers who always buckle up before riding in a car.
They randomly survey 388 drivers and find that 304 claim to always
buckle up. Construct a 97% confidence interval for the population
proportion that claim to always buckle up.

Insurance companies are interested in knowing the population
percent of drivers who always buckle up before riding in a car.
They randomly survey 402 drivers and find that 307 claim to always
buckle up. Construct a 84% confidence interval for the population
proportion that claim to always buckle up. Use interval notation,
for example, [1,5]

Insurance companies are interested in knowing the population
percent of drivers who always buckle up before riding in a car.
They randomly survey 420 drivers and find that 286 claim to always
buckle up. Construct a 84% confidence interval for the population
proportion that claim to always buckle up. Use interval notation,
for example, [1,5]

1. Insurance companies are interested in knowing the population
percent of drivers who always buckle up before riding in a car.
They randomly survey 380 drivers and find that 303 claim to always
buckle up. Construct a 87% confidence interval for the population
proportion that claim to always buckle up.
2. You recently sent out a survey to determine if the percentage
of adults who use social media has changed from 68%, which was the
percentage of adults who used...

1. Insurance companies are interested in knowing the population
percent of drivers who always buckle up before riding in a car.
When designing a study to determine this population proportion,
what is the minimum number of drivers you would need to survey to
be 95% confident that the population proportion is estimated to
within 0.04? (Round your answer up to the nearest whole
number.)
2. According to a poll, 75% of California adults (377 out of 506
surveyed) feel that...

Insurance companies are interested in knowing the population
percent of drivers who always buckle up before riding in a
car.
Part (a) When designing a study to determine this population
proportion, what is the minimum number of drivers you would need to
survey to be 95% confident that the population proportion is
estimated to within 0.04? (Round your answer up to the nearest
whole number.)

You wish to test the following claim (HaHa) at a significance
level of α=0.002α=0.002.
Ho:μ=58.2Ho:μ=58.2
Ha:μ<58.2Ha:μ<58.2
You believe the population is normally distributed, but you do not
know the standard deviation. You obtain a sample of size n=24n=24
with mean M=41M=41 and a standard deviation of
SD=20.6SD=20.6.
What is the p-value for this sample? (Report answer accurate to
four decimal places.)
p-value =
2. Insurance companies are interested in knowing the population
percent of drivers who always buckle...

1. Business Weekly conducted a survey of graduates from 30 top
MBA programs. On the basis of the survey, assume the mean annual
salary for graduates 10 years after graduation is 162000 dollars.
Assume the standard deviation is 31000 dollars. Suppose you take a
simple random sample of 70 graduates.
Find the probability that a single randomly selected policy has a
mean value between 153848.5 and 170892.5 dollars.
P(153848.5 < X < 170892.5)
= (Enter your answers as numbers accurate to...

1.
In the country of United States of Heightlandia, the height
measurements of ten-year-old children are approximately normally
distributed with a mean of 53.9 inches, and standard deviation of
8.5 inches.
A) What is the probability that a randomly chosen child has a
height of less than 59.05 inches?
Answer= (Round your answer to 3 decimal places.)
B) What is the probability that a randomly chosen child has a
height of more than 34.5 inches?
Answer= (Round your answer to...

You are a researcher studying the lifespan of a certain species
of bacteria. A preliminary sample of 35 bacteria reveals a sample
mean of ¯x=74x¯=74 hours with a standard deviation of s=6.8s=6.8
hours. You would like to estimate the mean lifespan for this
species of bacteria to within a margin of error of 0.55 hours at a
90% level of confidence.
What sample size should you gather to achieve a 0.55 hour margin
of error? Round your answer up to...

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