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 =
n =
p' =
(rounded to four decimal places)Part (b)
In words, define the random variables X and P'.X is the proportion of people in the sample who do not buckle up, and P' is the number of people who do not buckle up.X is the number of people who do not buckle up, and P' is the proportion of people in the sample who do not buckle up. X is the number of people who claim they buckle up, and P' is the proportion of people in the sample who buckle up.X is the proportion of people in the sample who claim they buckle up, and P' is the number of people who buckle up.
Part (c)
Which distribution should you use for this problem? (Round your answer to four decimal places.)P' ~
npq |
Part (d)
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.)Part (e)
If this survey were done by telephone, list three difficulties the companies might have in obtaining random results.The individuals in the sample may not accurately reflect the population.Only people over the age of 24 would be included in the survey.Individuals may not tell the truth about buckling up.Children who do not answer the phone will not be included.Individuals may choose to participate or not participate in the phone survey.
Part (a)
(i) Enter an exact number as an integer, fraction, or decimal.
x = 300
(ii) Enter an exact number as an integer, fraction, or decimal.
n = 410
(iii) Round your answer to four decimal places.
p' = x/n = 300/410 = 0.7317
q' = 1 - p' = 1 - 0.7317 = 0.2683
Part (c)
Which distribution should you use for this problem? (Round your answer to four decimal places.)
X ~ Normal (p', )
X ~ Normal ( 0.7317, 0.0219)
p' ~ ( 0.7317, 0.0219)
= 8.9715
Which is ≤ 10, which implies a small sample.The Student's t-distribution should be used because we do not know the standard deviation.
Part (d)
Suppose that insurance companies did a survey. They randomly surveyed 410 drivers and found that ...
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...
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. 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
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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.
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