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A newspaper reported that 4040​% of people say that some coffee shops are overpriced. The source of this information was a telephone survey of 5050 adults. a. Identify the population of interest in this study. b. Identify the sample for the study. c. Identify the parameter of interest in the study. d. Find and interpret a 9595​% confidence interval for the parameter of interest.

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Answer:

a. Identify the population of interest in this study.

  • all the population of the country

b. Identify the sample for the study.

  • 50 adults (the ones that were surveyed)

c. Identify the parameter of interest in the study.

  • the parameter of interest is the people that believe that coffee shops are overpriced

d. Find and interpret a 95​% confidence interval for the parameter of interest.

z score for a 95% confidence level = 1.96

margin of error (E) = z score x √{[0.4(1 - 0.4)] / 50} = 1.96 x 0.069282 = 0.13579

95% confidence level interval:

0.4 - 0.13579 = 0.26421

0.4 + 0.13579 = 0.53579

  • 95% confidence interval (0.26421 , 0.53579)

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