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Suppose that insurance companies did a survey. They randomly surveyed 450 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. 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 an exact number as an integer, fraction, or decimal. n = (iii) Round your answer to four decimal places. 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 claim they buckle up, and P'is the number of people who 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 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. Part (c) Which distribution should you use for this problem? (Round your answer to four decimal places.) P'- Explain your choice. ○ The binomial distribution should be used because there are two outcomes, buckle up or do not buckle up. ○ The normal distribution should be used because we are interested in proportions and the sample size is large. ○ The Student's t-distribution should be used because √npq ≤ 10, which implies a small sample. ○ The Student's t-distribution should be used because we do not know the standard deviation. 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.) (ii) Sketch the graph. C.L.= 82 W. P' (iii) Calculate the error bound. (Round your answer 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. Individuals may choose to participate or not participate in the phone survey. Individuals may not tell the truth about buckling up. Children who do not answer the phone will not be included. ☐ Only people over the age of 24 would be included in the survey.

Suppose that insurance companies did a survey. They randomly surveyed 450 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. 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 an exact number as an integer, fraction, or decimal. n = (iii) Round your answer to four decimal places. 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 claim they buckle up, and P'is the number of people who 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 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. Part (c) Which distribution should you use for this problem? (Round your answer to four decimal places.) P'- Explain your choice. ○ The binomial distribution should be used because there are two outcomes, buckle up or do not buckle up. ○ The normal distribution should be used because we are interested in proportions and the sample size is large. ○ The Student's t-distribution should be used because √npq ≤ 10, which implies a small sample. ○ The Student's t-distribution should be used because we do not know the standard deviation. 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.) (ii) Sketch the graph. C.L.= 82 W. P' (iii) Calculate the error bound. (Round your answer 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. Individuals may choose to participate or not participate in the phone survey. Individuals may not tell the truth about buckling up. Children who do not answer the phone will not be included. ☐ Only people over the age of 24 would be included in the survey.

Holt Mcdougal Larson Pre-algebra: Student Edition 2012
Holt Mcdougal Larson Pre-algebra: Student Edition 2012
1st Edition
ISBN:9780547587776
Author:HOLT MCDOUGAL
Publisher:HOLT MCDOUGAL
Chapter11: Data Analysis And Probability
Section: Chapter Questions
Problem 8CR
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Suppose that insurance companies did a survey. They randomly surveyed 450 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. 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 an exact number as an integer, fraction, or decimal. n = (iii) Round your answer to four decimal places. 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 claim they buckle up, and P'is the number of people who 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 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. Part (c) Which distribution should you use for this problem? (Round your answer to four decimal places.) P'- Explain your choice. ○ The binomial distribution should be used because there are two outcomes, buckle up or do not buckle up. ○ The normal distribution should be used because we are interested in proportions and the sample size is large. ○ The Student's t-distribution should be used because √npq ≤ 10, which implies a small sample. ○ The Student's t-distribution should be used because we do not know the standard deviation. 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.) (ii) Sketch the graph. C.L.= 82 W. P' (iii) Calculate the error bound. (Round your answer 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. Individuals may choose to participate or not participate in the phone survey. Individuals may not tell the truth about buckling up. Children who do not answer the phone will not be included. ☐ Only people over the age of 24 would be included in the survey.
Transcribed Image Text:Suppose that insurance companies did a survey. They randomly surveyed 450 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. 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 an exact number as an integer, fraction, or decimal. n = (iii) Round your answer to four decimal places. 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 claim they buckle up, and P'is the number of people who 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 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. Part (c) Which distribution should you use for this problem? (Round your answer to four decimal places.) P'- Explain your choice. ○ The binomial distribution should be used because there are two outcomes, buckle up or do not buckle up. ○ The normal distribution should be used because we are interested in proportions and the sample size is large. ○ The Student's t-distribution should be used because √npq ≤ 10, which implies a small sample. ○ The Student's t-distribution should be used because we do not know the standard deviation. 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.) (ii) Sketch the graph. C.L.= 82 W. P' (iii) Calculate the error bound. (Round your answer 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. Individuals may choose to participate or not participate in the phone survey. Individuals may not tell the truth about buckling up. Children who do not answer the phone will not be included. ☐ Only people over the age of 24 would be included in the survey.
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