A contractor involved in high

–

rise building construction

orders thousands of steel rods each year from

a supplier. The strength, X, of the rods is a normally distributed random variable. When a shipment

arrives, the contractor selects a random sample of 15 rods and determines the sample mean strength

and the s

ample standard deviation. These results are then used to compute a confidence interval for the

population mean strength of the rods.

1.

For a particular shipment, the sample mean strength of the selected rods

is =

990 pounds and the

sample standard deviation is

s

= 14.5 pounds. What is the margin of error for a 95% confidence

interval for the population mean strength? (Round your answer to 2 decimal places.)

A)

8.03

lbs

B)

7.98 lbs

C)

7.34 lbs

D)

8.31 lbs

E)

2.07 lbs

2.

The con

tractor would like to have an interval that has a

smaller

margin of error than the one above.

Which of the following would contribute to accomplishing this goal? (Choose the most

appropriate response.)

A)

Both decreasing the size of the sample mean

strength and increasing the size of the sample

to 30 rods would contribute to a reduction in the margin of error.

B)

Increase the population mean strength.

C)

Decrease the size of the sample mean strength.

D)

Increase the size of the sample to 30 rods.

E)

Increase the

level of confidence to 99%

1)
Given that:
The

Z Critical

n 15, x 990 and s 14.5
value at the 95% level of significance is: 1.96

The margin of error is,

n

Margin Error E Z Critical

14.5

15
1.96 3.74388…