Mathematics, 16.05.2020 06:57, quay84
Suppose that insurance companies did a survey. They randomly surveyed 440 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.)
Which distribution should you use for this problem? (Round your answer to four decimal places.)
P' ~
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) Calculate the error bound. (Round your answer to four decimal places.)
(iiI) Sketch the graph.
Answers: 2
Mathematics, 21.06.2019 16:30, victoria8281
Answer the following for 896.31 cm= km 100cm = 1m 1000m = 1km a) 0.0089631 b) 0.0089631 c) 8.9631 d) 89.631
Answers: 1
Mathematics, 21.06.2019 21:30, kyandrewilliams1
Alcoa was $10.02 a share yesterday. today it is at $9.75 a share. if you own 50 shares, did ou have capital gain or loss ? how much of a gain or loss did you have ? express the capital gain/loss as a percent of the original price
Answers: 2
Suppose that insurance companies did a survey. They randomly surveyed 440 drivers and found that 300...
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