Question

Suppose that insurance companies did a survey. They randomly surveyed 440 drivers and found that 310 claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up.

Part A:

(i) The observed count of drivers who claimed they always buckle up in this sample is = ?

(ii) The sample size = ?
(iii) The observed sample proportion = ?

Part B:

In words, define the random variables X and p̂.

- X is the proportion of people in the sample who claim they buckle up, and p̂ is the number of people who buckle up.

- X is the number of people who claim they buckle up, and p̂ is the proportion of people in the sample who buckle up.    

- X is the proportion of people in the sample who do not buckle up, and p̂ is the number of people who do not buckle up.

- X is the number of people who do not buckle up, and p̂ is the proportion of people in the sample who do not buckle up.

Part C:

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.) (___,___)

Part D:

If this survey were done by telephone, list three difficulties the companies might have in obtaining random results. (Select all that apply.)

- Only people over the age of 24 would be included in the survey.

- The individuals in the sample may not accurately reflect the population.

- Individuals may choose to participate or not participate in the phone survey.

- Children who do not answer the phone will not be included.Individuals may not tell the truth about buckling up.

Answer 1

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