Insurance companies are interested in knowing the population proportion of drivers who always buckle up before riding in a car. They randomly survey 380 drivers and find that 288 claim to always buckle up. Construct a 95% confidence interval for the population proportion that claim to always buckle up. Do not round between steps. Round answers to at least 4 decimal places.
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The sample proportion is the number of drivers who claim to always buckle up divided by the total number of drivers surveyed. Sample proportion = 288/380 = 0.758 Show more…
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Use the sample data and confidence level given below to complete parts (a) through (d): A research institute poll asked respondents if they acted to annoy a bad driver. In the poll, n = 2460, and X = 1145 who said that they honked. Use a 95% confidence level. Click the icon to view a table of z scores. a) Find the best point estimate of the population proportion p. (Round to three decimal places as needed)
Jeremy G.
thoughtful Preserving the question text as it is already clean.
Ariana N.
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. Compute the margin of error for a 95% confidence interval and construct this confidence interval for the population proportion who claim they always buckle up. (i) State the margin of error. (Round your answers to four decimal places.) m=? (ii) State the confidence interval. (Round your answers to four decimal places.) ( ? , ? )
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