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AP Statistics
7.2 Assignment
1. Suppose a large candy machine has 45% orange candies.
A. Would you be surprised if a sample of 25 candies from the machine contained 8 orange candies (that’s
32% orange)? How about 5 orange candies (20% orange)? Explain.
The probability that I would get 8 or less candies is about 0.0957. I would expect to get 8 or less
orange candies about 10% of the time and would not be surprised. The probability that I would get 5
or less candies is about 0.00599. I would expect to get 5 or less orange candies less than 1% of the
time and would be surprised
B. Which is more surprising: getting a sample of 25 candies in which 32% are orange or getting a sample
of 50 candies in which 32% are orange? Explain
It is more surprising to get 32% orange candies in a sample of 50 than an sample of 25. The variation of
sample proportions would be smaller with a larger sample, so it would be more unlikely to get a value
away from the population proportion (.45).
C. Imagine taking an SRS of 25 candies from the machine and observing the sample proportion p̂ of
orange candies.
i. What is the mean of the sampling distribution of p̂ . Why?
 pˆ  p  .45
ii. Find the standard deviation of the sampling distribution of p̂ . Check to see if the 10% condition is
met.
It is reasonable to assume the population of candies in the machine is greater than 25(10) = 250.
p 1  p 
Therefore,  pˆ 

.45 1  .45 
 0.0995
n
25
iii. Is the sampling distribution of p̂ approximately Normal? Check to see if the Normal condition is
met.
The sampling distribution is approximately normal because np  25 .45  11.25  10 and
n 1  p   25 1  .45  13.75  10
iv. If the sample size were 50 rather than 25, how would this change the sampling distribution of p̂ ?
If the sample size were 50 instead of 25, the sampling distribution would still be approximately
normal with a mean of 0.45, but the standard deviation would be
 pˆ 
p 1  p 
n

.45 1  .45 
50
 0.0704
2. In the game of Scrabble, each player begins by drawing 7 tiles from a bag containing 100 tiles. There are
42 vowels, 56 consonants, and 2 black tiles in the bag. Rhonda chooses an SRS of 7 tiles. Let p̂ be the
proportion of vowels in her sample.
A. Is the 10% condition met in this case? Justify your answer.
Yes, we are drawing 7 out of 100 tiles and 7(10) = 70 < 100
B. Is the Normal condition met in this case? Justify your answer.
No, the sample size is 7 so both np and n(1 – p) are less than 10.
3. A USA Today Poll asked a random sample of 1012 U.S. adults what they do with the milk in the bowl after
they have eaten the cereal. Of the respondents, 67% said that they drink it. Suppose that 70% of U.S.
adults actually drink the cereal milk. Let p̂ be the proportion of people in the sample who drink the cereal
milk.
A. What is the mean of the sampling distribution of p̂ ? Why?
 pˆ  p  .7
B. Find the standard deviation of the sampling distribution of p̂ . Check to see if the 10% condition is
met.
It is reasonable to assume the population of U.S. adults is greater than 1012(10) = 101200. Therefore,
p 1  p 
 pˆ 

.7 1  .7 
 0.0144
n
1012
C. Is the sampling distribution of p̂ approximately Normal? Check to see if the Normal condition is met.
The sampling distribution is approximately normal because np  1012 .7   708.4  10 and
n 1  p   1012 1  .7   303.6  10
D. Find the probability of obtaining a sample of 1012 adults in which 67% or fewer say they drink the
cereal milk. Do you have any doubts about the result of this poll?
.67  .70 

P  pˆ  0.67   P  z 
  0.0186
0.0144 

E. What sample size would be required to reduce the standard deviation of the sampling distribution to
one-half the value you found in part B? Justify your answer.
.7 1  .7 
n
.7 1  .7 
n

1  .7 1  .7  
 

4  1012 
1
1

n 4 1012 
n  4048
1 .7 1  .7 
2
1012