AP StatisticsConfidence Intervals with Proportions ReviewName:______

  1. The manager of an orchard expects about 70% of his apples to exceed the weight requirement for "Grade A" designation. At least how many apples must he sample to be 90% confident of estimating the true proportion within ± 4%?
  1. A statistics professor wants to see if more than 80% of her students enjoyed taking her class. At the end of the term, she takes a random sample of students from her large class and asks, in an anonymous survey, if the students enjoyed taking her class. Write the hypothesis statements for this situation.
  1. A recent Gallup poll found that 28% of U.S. teens aged 13-17 have a computer with Internet access in their rooms. The poll was based on a random sample of 1028 teens and reported a margin of error of ±3%. What level of confidence did Gallup use for this poll?
  1. One month the actual unemployment rate in France was 13.4%. If during that month you took a survey of 100 Frenchmen and constructed a confidence interval estimate of the unemployment rate, which of the following would be true?
  2. The center of the interval was .134
  3. The interval contained .134
  4. A 99% confidence interval contained .134
  1. In general, how does doubling the sample size change the confidence interval size?
  1. Under what conditions would it be meaningful to construct a confidence interval estimate when the data consists of the entire population?
  1. Two 95% confidence interval estimates are obtained: I (.128,.272) and II (.138, .262). If the sample proportions () are the same, which has the larger sample size?

a) Interval I has a larger sample size

b) Interval II has a larger sample size

c) The sample sizes are the same

d) More information is needed to answer this question

  1. A telephone survey of 400 registered voters showed that 256 had not yet made up their minds 1 month before the election. At what confidence level can we say that between 60% and 68% of the electorate were still undecided at that time?
  1. When an online news magazine asked viewers to click their agreement or disagreement, 300 out of 1200 respondents agreed with a statement that the most practical way of becoming a millionaire is winning a lottery. Immediate feedback stated that 25% of the viewers, with a margin of error of 2.5%, agreed with the statement. Fine print claimed 95% confidence. What is the proper conclusion?

a) We are 95% confident that the proportion of viewers who believe the most practical way of becoming a millionaire is winning a lottery is between .225 and .275.

b) Without knowing whether both np and n(1 – p) are > 10, the calculation is inappropriate.

c) Without knowing whether or not the 1200 respondents are 10% of all viewers, the calculation is inappropriate.

d) The z-distribution was inappropriately used.

e) The data was not a SRS, so the calculation is inappropriate

  1. There are 50,000 high school students in an extended metropolitan region. As each of their students came in to register for classes, guidance counselors were instructed to use a calculator to pick a random number between 1 and 100. If the number 50 was picked, the student was included in a survey. For one of the many survey questions, 30% of the students said they couldn’t live without instant messaging. Are the conditions met for constructing a confidence interval of the proportion of this region’s teenagers who believe they couldn’t live without instant messaging?

a) No, there is no guarantee that a representative random sample is chosen.

b) No, the sample size is not less than 10% of the population.

c) No, np and n(1 – p) are not both greater than 10.

d) No, there is not reason to assume that the population has a normal distribution.

e) Yes, all conditions are met, and a confidence interval can be constructed

  1. For a given large sample size, which of the following gives the smallest margin of error in calculating a confidence interval for a population proportion?

a) 90% confidence with =.15

b) 95% confidence with =.15

c) 99% confidence with =.15

d) 90% confidence with =.23

e) 95% confidence with =.23

  1. If 64% of a sample of 550 shoppers leaving a shopping mall claim to have spent over $25,
  2. determine a 99% confidence interval for the proportion of all shoppers who spend over $25.
  3. Shopping mall management claims that 75% of all shoppers spend over $25 at their mall per trip. What does your confidence interval say about this claim?
  1. A research study gives a 95% confidence interval for the proportion of subjects helped by a new anti-inflammatory drug as (0.56, 0.65).
  2. Interpret this interval in the context of the problem.
  3. What is the meaning of "95%" confidence interval as stated in the problem?