Question

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 =

  
  
  (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 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' ~

       ,  
  
  Explain your choice.The normal distribution should be used because we are interested in proportions and the sample size is large.The binomial distribution should be used because there are two outcomes, buckle up or do not buckle up.     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.

  

  (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.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.

i safari File Edit View History Bookmarks Window Help 令41%D Mon 7:15 AM OE webassign.net Cengage In six packages of A Magazine Publish.. www.cabril Suppose That An A... Quiz for Lesson 7 6. -/0.366 points illowskintrostati 8.НИ. 118. My Notes Ask Your T + 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.) D Part (a) () Enter an exact number as an integer, fraction, or decimal. (ii) Enter an exact number as an integer, fraction, or decimal. : AM (iii) Round your answer to four decimal places. (rounded to four decimal places) D 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. a...tin E Part (c) Which distribution should you use for this problem? (Round your answer to four decimal places.) Explain your choice. The normal distribution should be used because we are interested in proportions and the sample size is large. The binomial distribution should be used because there are two outcomes, buckle up or do not buckle up

Answer 1

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', $\sqrt{p'q'/n}$)

X ~ Normal ( 0.7317, 0.0219)

p' ~ ( 0.7317, 0.0219)

$\sqrt{npq} = \sqrt{410*0.7317*0.2683}$ = 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)

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