Suppose that insurance companies did a survey. They randomly surveyed 430 drivers and found that 330 claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up.
a) x = ? b) n = ? c) p' = ? d)
Which distribution should you use for this problem? (Round your answer to four decimal places.)
P- (_) (_,_)
e) Construct a 95% confidence interval for the population proportion who claim they always buckle up.
i. state the confidence interval
ii. sketch graph
iii. calculate error bound
Solution:
Given:
Sample size =n = 430
x = Number of drivers always buckle up = 330
Part a) x = ?
x = 330
Part b) n = ?
n = 430
Part c) p' = ?
Part d) Which distribution should you use for this problem?
Sampling distribution of sample proportions is approximate Normal distribution with mean of sample proportion:
and standard deviation of sample proportions is:
Thus
P ~ ( approximate Normal ) ( , )
Part e) Construct a 95% confidence interval for the population proportion who claim they always buckle up.
where
We need to find zc value for c=95% confidence level.
Find Area = ( 1 + c ) / 2 = ( 1 + 0.95) /2 = 1.95 / 2 = 0.9750
Look in z table for Area = 0.9750 or its closest area and find z value.
Area = 0.9750 corresponds to 1.9 and 0.06 , thus z critical value = 1.96
That is : Zc = 1.96
Thus
Thus
i) a 95% confidence interval for the population proportion who claim they always buckle up is between
ii) sketch graph
iii) calculate error bound
Suppose that insurance companies did a survey. They randomly surveyed 430 drivers and found that 330...
Suppose that insurance companies did a survey. They randomly surveyed 430 drivers and found that 330 claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up. a) x = ? b) n = ? c) p' = ? d) Which distribution should you use for this problem? (Round your answer to four decimal places.) e) Construct a 95% confidence interval for the population proportion who claim they always buckle up....
Suppose that insurance companies did a survey. They randomly surveyed 400 drivers and found that 320 claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up. Construct a 95% confidence interval for the population proportion who claim they always buckle up. (iii) Calculate the error bound. (Round your answer to four decimal places.)? I don't understand how to calculate the error bound
Suppose that several insurance companies conduct a survey. They randomly surveyed 400 drivers and found that 240 claimed to always buckle up. We are interested in the population proportion of drivers who claim to always buckle up a. (.20) n= b. (.20) p, = c. (.20) The standard deviation for p d. (.20) The z value for a 95% confidence interval is e. (.20) Construct a 95% confidence interval for the population proportion that claim to always buckle Fill in...
Suppose that insurance companies did a survey. They randomly surveyed 400 drivers and found that 320 claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up. which distribution should you use for this problem? (Round your answer to four decimal places.) P' ~( P', ?) ? I understand its normal distribution but i'm not sure what goes after the probability
Suppose that insurance companies did a survey. They randomly surveyed 410 drivers and found that 300 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.) Part (a) (i) Enter an exact number as an integer, fraction, or decimal. x = (ii) Enter...
Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 410 drivers and find that 304 claim to always buckle up. Construct a 91% confidence interval for the population proportion that claim to always buckle up show how to solve with and without calculator Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey...
Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 383 drivers and find that 296 claim to always buckle up. Construct a 87% 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 383 drivers and find that 296 claim to always buckle up. Construct a 87% 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 393 drivers and find that 305 claim to always buckle up. Construct a 87% 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 393 drivers and find that 305 claim to always buckle up. Construct a 87% confidence interval for the population proportion that claim to always buckle up. Use interval notation, for example, [1,5]