Stat 250.3 – Practice Quiz 3
Questions 1 to 4: Suppose that the weight of apples eaten by individual Americans can be described by a
normal probability distribution with mean  = 15 pounds and standard deviation  = 5 pounds per year.
1.
For a random sample of n = 16 people, what is the standard deviation of the sampling distribution of
the sample mean x ?
KEY: 1.25 pounds.
2.
For a random sample of n = 16 people, what is the mean of the sampling distribution of the sample
mean x ?
KEY: 15 pounds.
3. For a random sample of n = 16 people, what is the z-score for a sample mean x = 10 pounds?
KEY: 4.
4. For a random sample of n = 9 people, what is the z-score for a sample mean x = 18 pounds?
KEY: 1.80
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5. In a lottery, a real number from 0 to 100 is picked at random. If X is the number picked, then the mean
of X is 50 and the standard deviation of X is 28.9. For a random sample of 60 numbers, use the
Empirical Rule to find a range that will include the sample mean,
KEY: 42.5 to 57.5
6.
X , approximately 95% of the time.
In a creek, the proportion of turtles that are snake-necked turtles is 35%. In a random sample of 50
turtles, the sample proportion of snake-necked turtles, p̂ , is 50%. What is the z-score associated with
p̂ ?
KEY: 2.22
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Questions 7 to 9: In a pond with a large population of turtles, the percent of three different types are
Type of Turtle
Chicken Turtle
Spotted Turtle
Snapping Turtle
Percentage
30%
50%
20%
A random sample of 100 turtles is taken from the pond.
7. What is the standard deviation for the sample proportion, p̂ , of chicken turtles?
KEY: 0.0458
8. What is the standard deviation for the sample proportion, p̂ , of spotted turtles?
KEY: 0.050
9. What is the standard deviation for the sample proportion, p̂ , of snapping turtles?
KEY: 0.040
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Questions 10 to 12: Every student taking elementary statistics at a large university (about 1,100 students)
participated in a class project by rolling a 6-sided die 100 times. Each individual student determined the
proportion of his or her 100 rolls for which the result was a “1”. The instructor plans to draw a histogram
of the 1,100 sample proportions.
10. What will be the approximate shape of this histogram?
A. Skewed
B. Uniform
C. Normal (bell-shaped)
D. Chi-square
KEY: C
11. What will be the approximate mean for the 1,100 sample proportions?
A. 1/100
B. 1/6
C. 6/100
D. 6
KEY: B
12. What will be the approximate standard deviation for the 1,100 sample proportions?
A.
B.
C.
D.
(1 / 6)( 5 / 6)
1,100
(1 / 6)(5 / 6)
100
(1 / 100)( 99 / 100)
1,100
(1 / 100)( 99 / 100)
100
KEY: B
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13. Which statement is true about p and p̂ ?
A.
B.
C.
D.
KEY: C
They are both parameters.
They are both statistics.
p is a parameter and p̂ is a statistic.
p̂ is a parameter and p is a statistic.
14. When a random sample is to be taken from a population and a statistic is to be computed, the statistic
can also be thought of as:
A. A parameter
B. A random variable
C. A standard error
D. A sample
KEY: B
15. Imagine taking many different samples of n = 30 values from a population of resting pulse rate values.
Which one of these statistics is most likely to have a sampling distribution that is approximately a
normal distribution?
A. The maximum pulse rate in a sample.
B. The minimum pulse rate in a sample.
C. The mean pulse rate in a sample.
D. The mean pulse rate in the population.
KEY: C
16. Explain what a confidence level of 95% means in a 95% confidence interval.
KEY: The 95% confidence level means that, in the long run, the procedure used to determine the interval
will provide an interval that captures the population value about 95% of the time.
17. In a past General Social Survey, 37% of a random sample of n = 1035 respondents answered yes to the
question, “Have you ever been punched or beaten by another person?” Calculate an approximate 95%
confidence interval for the proportion of all Americans who have been punched: use the approximate
margin of error (or the “95% margin of error”).
KEY: (.340, .400)
18. For a randomly selected sample of 40 first-year students at a University, 20% received a grade of A in
their required course in English. Calculate a 90% confidence interval for the proportion of all firstyear students who received an A in English.
KEY: (.096, .304)
19. In a randomly selected sample of 100 students at a University, 85% had use of a laptop computer.
Calculate a 99% confidence interval for the proportion of all students who had use of a laptop
computer.
KEY: (.758, .942)
20. In a past General Social Survey, responders to the question “Have you ever been beaten or punched by
another person” gave the following 95% confidence intervals for the population proportions of women
and men who would say yes as .195 to .265 and .493 to 587, respectively. The sample sizes for men
and women were large (444 men and 591 women). Is it reasonable to conclude that there is a
difference in the population between men and women in the proportion that have ever been beaten or
punched by another person?
KEY: Yes, because the two intervals do not overlap.
21. Suppose that a standard error for a sample proportion is .02. Which of the following is the margin of
error for a 98% confidence interval for the corresponding population proportion?
A. 1.65 .02
B. 2 02
C. 2.33  .02
D. 2.58  .02
KEY: C
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Questions 35 to 37: A 95% confidence interval for the proportion of young adults who skip breakfast is .20
to .27.
22. Which of the following is a correct interpretation of the 95% confidence level?
A. In about 95% of all studies for which this procedure is used, the confidence interval will cover the
true population proportion, but there is no way to know if this interval covers the true proportion
or not.
B. There is a 95% probability that the proportion of young adults who skip breakfast is between .20
and .27.
C. If this study were to be repeated with a sample of the same size, there is a 95% probability that
the sample proportion would be between .20 and .27.
D. The proportion of young adults who skip breakfast 95% of the time is between .20 and .27.
KEY: A
23. Which of the following is the correct interpretation of the 95% confidence interval?
A. There is a 95% probability that the proportion of young adults who skip breakfast is between .20
and .27.
B. If this study were to be repeated with a sample of the same size, there is a 95% probability that
the sample proportion would be between .20 and .27.
C. We can be 95% confident that the proportion of young adults in the sample who skip breakfast is
between .20 and .27.
D. We can be 95% confident that the proportion of young adults in the population who skip
breakfast is between .20 and .27.
KEY: D
24. From the information provided, we can determine that:
p̂ = .235 and margin of error = .035
A.
p̂ = .235 and margin of error = .07
B.
C. p = .235 and margin of error = .035
D. p = .235 and margin of error = .07
KEY: A
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25. If the confidence level is increased, which of the following must also be increased?
A. sample estimate
B. multiplier
C. standard error
D. none of the elements would be increased
KEY: B
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Questions 26 to 27: In a past General Social Survey, a sample proportion p̂ = .22 (or 22%) from a random
sample of n = 1006 respondents answered yes to the question “Are you a member of any sports groups?” A
95% confidence interval for the population proportion of Americans who belonged to a sports group at that
time is 19.4% to 24.6%.
26. Based on these results, you can reasonably conclude that
A. Less than 50% of all Americans belong to sports clubs.
B. More than 50% of all Americans belong to sports clubs
C. 22% of all Americans belong to sports clubs.
D. None of the above
KEY: A
27. Based on these results, you can reasonably conclude that
A. 22% is an “acceptable” possibility for the proportion of all Americans belonging to sports clubs,
but 20% is not.
B. Neither 20% nor 22% are “acceptable” possibilities for the proportion of all Americans belonging
to sports clubs.
C. Both 20% and 22% are “acceptable” possibilities for the proportion of all Americans belonging to
sports clubs.
D. None of the above
KEY: C
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Equations that are going to be provided in the quiz:
Useful for sampling distribution:
s.d.( p̂ ) =
p(1  p)
n
s.d.( x ) =

n