A simple random sample of size n is drawn from a population that is known to be
ID: 3236794 • Letter: A
Question
A simple random sample of size n is drawn from a population that is known to be normally distributed. The sample variance, s^2, is determined to be 12.9. Complete pars (a) through (c).
(a) Construct a 90% confidence intercal for variance if the sample size, n, is 20.
lower bound [ ] (round to two decimal places as needed)
upper bound [ ] ( round to two decimal places as needed)
(b) Construct a 90% confindence interval for varience if the sample size, n, is 30.
The lower bound is [ ].
The upper bound is [ ].
How does increaing the sample size affect the width of the interval?
o The width decreases
o The width increases
o The width does not change
(c) Construct a 98% confidence interval for variance if the sample size, n, is 20.
lower bound [ ]
upper bound [ ]
Compare the results with those obtained in part (a). How does ncreasing the level of confidence affect the confidence interval?
o The width does not change
o The width decreases
o The width increases
Explanation / Answer
Answers to the question:
a. A 90% interval is given by :
xbar +/- 1.645*12.9/sqrt(20) = xbar +/- 5.745
Upper bound = xbar + 5.745
Lower bound = xbar - 5.745
b. If you increase the sample size the width of the interval
- width decreases as the Margin of error decreases
c. A 98% interval is given by :
xbar +/- 2.33*12.9/sqrt(20) = xbar +/- 6.72
d. Increasing the level of confidence :
the widht of the interval increases
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