Question 1
A statistics professor recently graded final exams for students in her introductory
statistics course. In a review of her grading, she found the mean score out of 100 points
was a x¯=77, with a margin of er
...
Question 1
A statistics professor recently graded final exams for students in her introductory
statistics course. In a review of her grading, she found the mean score out of 100 points
was a x¯=77, with a margin of error of 10.
Construct a confidence interval for the mean score (out of 100 points) on the final
exam.
That is correct!
$$(67, 87)
Answer Explanation
Correct answers:
$\left(67,\ 87\right)$(67, 87)
A confidence interval is an interval of values, centered on a point estimate, of the form
(pointestimate−marginof error,pointestimate+marginof error)
Using the given point estimate for the mean, x¯=77 and margin of error 10, the
confidence interval is:
(77−10,77+10)(67,87)
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Question 2
A random sample of adults were asked whether they prefer reading an e-book over a
printed book. The survey resulted in a sample proportion of p′=0.14, with a sampling
standard deviation of σp′=0.02, who preferred reading an e-book.
Use the empirical rule to construct a 95% confidence interval for the true proportion of
adults who prefer e-books.
That is correct!
$$(0.1, 0.18)
Answer Explanation
Correct answers:
$\left(0.10,\ 0.18\right)$(0.10, 0.18)
By the Empirical Rule, a 95% confidence interval corresponds to a z-score
of z=2. Substituting the given values p′=0.14and σp′=0.02, a confidence interval is
(p′−z⋅ σp′,p′+z⋅ σp′)(0.14−2⋅ 0.02,0.14+2⋅ 0.02)(0.14−0.04,0.14+0.04)(0.10,
0.18)
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Question 3
The pages per book in a library are normally distributed with an unknown population
mean. A random sample of books is taken and results in a 95% confidence interval
of (237,293) pages.
What is the correct interpretation of the 95% confidence interval?
That is correct!
We estimate with 95% confidence that the sample mean is between 237 and 293 pages.
We estimate that 95% of the time a book is selected, there will be
between 237 and 293 pages.
We estimate with 95% confidence that the true population mean is
between 237 and 293 pages.
Answer Explanation
Correct answer:
We estimate with 95% confidence that the true population mean is
between 237 and 293 pages.
Once a confidence interval is calculated, the interpretation should clearly state the
confidence level (CL), explain what population parameter is being estimated, and state
the confidence interval.
We estimate with 95% confidence that the true population mean is
between 237 and 293 pages.
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Question 4
The population standard deviation for the heights of dogs, in inches, in a city
is 3.7 inches. If we want to be 95% confident that the sample mean is within 2 inches
of the true population mean, what is the minimum sample size that can be taken?
z0.101.282z0.051.645z0.0251.960z0.012.326z0.0052.576
Use the table above for the z-score, and be sure to round up to the nearest integer.
That is correct!
$$14 dog heights
Answer Explanation
Correct answers:
$14\text{ dog heights}$14 dog heights
The formula for sample size is n=z2σ2EBM2. In this formula, z=zα2=z0.025=1.96, because
the confidence level is 95%. From the problem, we know that σ=3.7 and EBM=2.
Therefore, n=z2σ2EBM2=(1.96)2(3.7)222≈13.15. Use n=14 to ensure that the sample size
is large enough.
Also, the sample size formula shown above is sometimes written using an alternate
format of n=(zσE)2. In this formula, E is used to denote margin of error and the entire
parentheses is raised to the exponent 2.
Therefore, the margin of error for the mean can be denoted by "EBM" or by "E". Either
formula for the sample size can be used and these formulas are considered as
equivalent.
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Question 5
Clarence wants to estimate the percentage of students who live more than three miles
from the school. He wants to create a 98% confidence interval which has an error
bound of at most 4%. How many students should be polled to create the confidence
interval?
z0.10 z0.05 z0.02
5
z0.01 z0.005
1.2
82
1.6
45
1.9
60
2.3
26
2.576Use the table of values above.
That's not right.
$$847 students
Answer Explanation
Correct answers:
$846\ students$846 students
Given the information in the
question, EBP=0.04 since 4%=0.04 and zα2=z0.01=2.326 beca
use the confidence level is 98%. The values of p′ and q′ are
unknown, but using a value of 0.5 for p′ will result in the largest
possible product of p′q′, and thus the largest possible n.
If p′=0.5, then q′=1−0.5=0.5. Therefore,
n=z2p′q′EBP2=2.3262(0.5)(0.5)0.042=845.4
Round the answer up to the next integer to be sure the sample
size is large enough. The sample should include 846students.
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Question 6
The average score of a random sample of 87 senior business
majors at a university who took a certain standardized test follows
a normal distribution with a standard deviation of 28. Use Excel to
determine a 90% confidence interval for the mean of the
population. Round your answers to two decimal places and use
ascending order.
Score
516
536
462
461
519
496
517
488
521
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That is correct!
$$(509.30, 519.18)
Answer Explanation
Correct answers:
$\left(509.30,\ 519.18\right)$(509.30, 519.18)
A 90% confidence interval for μ is (x¯−zα/2σn‾√,x¯+zα/2σn‾√).
Here, α=0.1, σ=28, and n=87. Use Excel to calculate
the 90% confidence interval.
1. Open Excel, enter the given data in column A, and find the
sample mean, x¯, using the AVERAGE function. Thus, the
sample mean, rounded to two decimal places, is x¯=514.24.
2. Click on any empty cell,
enter =CONFIDENCE.NORM(0.1,28,87), and press ENTER.
3. The margin of error, rounded to two decimal places,
is zα/2σn‾√≈4.94. The confidence interval for the population
mean has a lower limit of 514.24−4.94=509.30 and an upper
limit of 514.24+4.94=519.18.
Thus, the 90% confidence interval for μ is (509.30, 519.18).
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