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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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