Estimating Population Mean Sigma Known
Quiz
•
Mathematics
•
11th Grade
•
Hard
Anthony Clark
FREE Resource
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15 questions
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1.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
Overproduction of uric acid in the body can be an indication of cell breakdown. This may be an advance indication of illness such as gout, leukemia, or lymphoma. Over a period of months, an adult male patient has taken eight blood tests for uric acid, which had the mean concentration of 5.3 mg/dl. The distribution of uric acid in healthy adult males can be assumed to be normal with a standard deviation of 1.85 mg/dl. The level of confidence sis 95%. What is the margin of error E?
4.07 mg/dl
1.82 mg/dl
1.48 mg/dl
1.28 mg/dl
2.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
Suppose an archaeologist discovers seven fossil skeletons from a previously unknown species of miniature horse. Reconstructions of the skeletons of these seven miniature horses show the shoulder heights (in centimeters) to be 45.3, 47.1, 44.2, 46.8, 46.5, 45.5, 47.6 Assume that the population of shoulder heights is approximately normal. Find the 99% confidence interval for the mean of the shoulder heights of the entire population of such horses.
(45.4 cm, 48.7 cm)
(46 cm, 48 cm)
(45 cm, 48 cm)
(44.5 cm, 47.8 cm)
3.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
When all other quantities remain the same, how does increasing the sample size affect the width of a confidence interval?
nothing changes
CI width increases
CI width decreases
4.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
When all other quantities remain the same, how does increasing the population standard deviation affect the width of a confidence interval?
nothing changes
CI width increases
CI width decreases
5.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
Select the formula to calculate the minimum sample size n, to obtain some maximal margin of error.
a
b
c
d
e
6.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
0.9130
0.1788
0.1814
0.5596
7.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
A certain variety of table grapes has fruit diameters that are distributed normally with mean 13 mm and standard deviation 2 mm. If you take a sample of 16 grapes, approximately what value represents the 80th percentile for the sample mean?
12.60
12.58
13.42
13.40
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