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AP Statistics Practice Exam 1 (page 3)

based on 13 ratings
By — McGraw-Hill Professional
Updated on Apr 18, 2011

Solutions to Practice Exam 1, Section II, Part A

Solution to #1

  1. We note that r2 = 0.954 r = 0.98, so there is a strong linear correlation between the variables. We know that the correlation is positive since the slope of the regression line is positive. Also, the residual plot shows no obvious pattern, indicating that the line is a good model for the data.
  2. Words = 3.371 + 2.1143(Age).
  3. Words = 3.371 + 2.1143(7.5) = 19.2.
  4. For each year a child grows, the number of words he or she knows about transportation is predicted to grow by 2.1.
  5. No. This would require extrapolating beyond the range of the data.

Solution to #2

  1. P(a student takes calculus) = 90/675 = 0.133.
  2. P(a student takes analysis or calculus given that he/she is in the 12th grade)
      .
  3. P(a student is in the 10th grade given that he/she is taking algebra II)
      .
  4. Let A = "A student takes geometry" and B = "A student is a 10th grader." A and B are independent events if and only if P(A) = P(A | B).
  5. .

    Thus, the events are not independent.

Solution to #3

  1. Let p = the true proportion of students who will sign up for upper division mathematics classes during the coming year.
  2. .

    We want to use a one-proportion z-test, at the 0.05 level of significance. We note that we are given that the sample was a random sample, and that np = 55(0.8) = 44 and 55 (1 – 0.8) = 11 are both larger than 5 (or 10). Thus the conditions needed for this test are present.

    (from Table A; on the TI-83/84, the P-value = normalcdf(1.35,100)). Since P > α, we cannot reject the null hypothesis. The survey evidence is not strong enough to justify adding another section of upper-division mathematics.

  3. If all five of the other students returned their surveys, there are two worst-case scenarios: all five say they will sign up; all five say they will not sign up. If all five say they will not sign up, then an even lower percentage say they need the class (48/60 rather than 48/55) and our decision not to offer another class would not change. If all five say they will sign up, then
  4.  

    At the 5% level of significance, this is still not quite enough to reject the null. However, it's very close, and the assistant principal might want to generate some additional data before making a final decision.

Solution to #4

  1. Randomly divide your 150 volunteers into two groups. One way to do this would be to put each volunteer's name on a slip of paper, put the papers into a box, and begin drawing them out. The first 75 names selected would be in group A, and the other 75 would be in group B. Alternatively, you could flip a coin for each volunteer. If it came up heads, the volunteer goes into group A; if tails, the volunteer goes into group B. The second method would likely result in unequal size groups.
  2. Administer one group the new medication (treatment group), and administer the old medication to the other group (control) for a period of time. After enough time has passed, have each volunteer evaluated for reduction in the symptoms of depression. Compare the two groups.

  3. Because we know that being in therapy can affect the symptoms of depression, block by having the 60 people who have been in therapy be in one block and the 90 who have not been in therapy be in the other block. Then, within each block, conduct an experiment as described in part (a).

Solution to #5

H0: Month of birth and draft number are independent.

HA: Month of birth and draft number are not independent.

This is a two-way table with 12 rows and 2 columns. A chi-square test of independence is appropriate. The numbers of expected values are given in the table below:

 

Because all expected values are greater than 5, the conditions for the chi-square goodness-of-fit test are present.

.

 

df = (12–1)(2–1) = 11 0.001 < P-value < 0.0025 (from Table C; using the TI-83/84, we have X2cdf (31.14,1000,11)= 0.00105).

Because the P-value is so small, we have evidence to reject the null and conclude that birth month and draft number are not independent. That is, month of birth is related to draft numbers. Observation of the data indicates that the lottery was biased against people born later in the year. That is, people born later in the year were more likely to get drafted.

Solutions to Practice Exam 1, Section II, Part B

Solution to #6

  1. The situation described is pictured below.
  2.  

    Let X be the length of a fish. Then P(X > 11.5) = =1– 0.7967 = 0.2033 (from Table A; using the TI-83/84, we have P(X > 11.5) = normalcdf (0.83, 100) or normalcdf (11.5, 1000, 11, 0.6)). Rounding to the nearest tenth, P(X > 11.5) = 0.2.

  3. Because the probability that the fish is large enough to keep is 0.2, let the digits 0 and 1 represent a fish that is large enough to keep and the digits 2, 3, 4, 5, 6, 7, 8, and 9 represent fish that must be released. Begin at the first line of the table and count the number of digits required before five of the digits 0, 1 are found.
  4. On the random number table below, successes (a large enough fish) are in boldface. Backslashes separate the various trials. The number of catches it took to get two sufficiently large fish is indicated under each separate trial.
  5.  

  6. Based on the five trials, an estimate of the average number of trials required to get two fish of minimum size is (26 + 15 + 21 + 34 + 14)/5 = 22.
  7. The expected wait to catch a single fish with P = 0.2 is 1/0.2 = 5 fish. The expected wait to catch five fish is then 5(5) = 25.

 

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