AP Statistics – Chapter 12 Practice Free Response Test
AP Statistics – Chapter 12 Practice Free Response Test - ANSWERS
1. 1-Prop Z Test
Let p = the proportion of defective items produced after the modifications have been made.
p-hat = 16/300 = .053
Step 1: Hypotheses – Ho: p = .11 and Ha: p < .11
Step 2: Conditions – there must be at least 3000 items produced after the modifications
np = 300(.11) = 33 and n(1-p) = 300(.89) = 267. Both are greater than 10.
Step 3: Test Statistic – [pic] and the p-value = .0009
Step 4: Conclusion – With such a small p-value, we can reject the null hypothesis and conclude that the modifications have reduced the proportion of defects.
2. 2-Prop Z Interval
Let p1 = proportion of acetaminophen users who experience adverse symptoms
Let p2 = proportion of ibuprofen users who experience adverse symptoms
P1-hat = 44/650 = .068 and P2-hat = 49/347 = .141
a) Since this is a two-sample situation, counts of successes and failures must be at least 5. They all are. We also need an SRS and for the populations for each group to be 10 times the size of the sample. The number of users of these two drugs easily meets the minimum requirement.
b) The interval is [pic] = (-.108, -.039).
So we are 90% confident that between 3.9% and 10.8% more people suffer adverse symptoms from ibuprofen than from acetaminophen.
3. 2-Prop Z Test
Let p1 = proportion of Danish males with normal chromosomes who have criminal records
Let p2 = proportion of Danish males with abnormal chromosomes who have criminal records
P1-hat = 381/4096 = .093 and P2-hat = 8/28 = .286 and P-hat = (381+8) / (4096+28) = .094
Step 1: Hypotheses – Ho: p1 = p2 and Ha: p1 < p2
Step 2: Conditions – there must be at least 40,960 males in group 1 and 280 in group 2
All successes and failures are greater than 5 (the 8 is close though).
Step 3: Test Statistic – [pic] and the p-value = .0002
Step 4: Conclusion – With such a small p-value, we can reject the null hypothesis and conclude that chromosome abnormalities are associated with criminality.
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