AP Statistics 2017

(Marvins-Underground-K-12) #1

  1. (B) The steeper part of the graph corresponds to the higher bars on a histogram. So the higher bars
    would be on the left with a tail stretching toward the right.

  2. (E) No causal relationship is established simply because of a correlation. But r 2 is 0.09.

  3. (C) Nonresponse bias is a problem because the kind of people who don’t respond might feel
    differently from those who do. Getting responses from more of the people that do respond won’t fix
    the problem.

  4. (B) Under the null hypothesis, the proportions are the same, so the standard error formula is based on


the pooled  proportion      .


  1. (E) Choice E implies a causal relationship, which is not justified because this is an observational
    study.

  2. (E) The margin of error is largest for = 0.5, so if ≠ 0.5, the margin of error will be even smaller.


So  use     .   Solve   for n and   round   up.


  1. (A) They have all the data for November, so there is no inference to do. 50% of the donations are
    above the median, not the mean. A is the only conclusion we can make.

  2. (A) n is not large enough to use a normal approximation. This is a binomial situation, and choice A
    correctly applies the binomial calculation.

  3. (E) Using s in the denominator makes Max’s an approximate t -distribution. That has heavier tails and
    more spread than Claire’s approximately normal distribution.

  4. (C) Multiplying affects both measures of location and spread. Adding affects only measures of
    location.

  5. (C) This is easiest to see in a two-way table.


The required    probability is  0.096/0.114 =   0.842.


  1. (D) The confidence interval is based on one sample of many possible samples. The process has a
    95% chance of capturing the true proportion of all Americans within the interval created. But there is
    nothing special about that one interval.

  2. (A) That is the multiplication rule for independent events.

  3. (C) Since these are four independent occurrences of W , the variances add. So the standard deviation
    is

  4. (A) The bottom path in the tree diagram shows that the probability that a patient has neither condition
    is 0.85 0.90 = 0.765. 1 – 0.765 = 0.235.

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