6. Resistance measurements were made on a sample of 81 wires of a certain type.

Question

6. Resistance measurements were made on a sample of 81 wires of a certain type. The sample mean resis- tance was 17.3 mS2, and the standard deviation was 1.2 mS2 a. Find a 95% confidence interval for the mean resis- tance of this type of wire. tance of this type of wire. (17.1. 17.5)? b. Find a 98% confidence interval for the mean resis- c. What is the level of the confidence interval d. How many wires must be sampled so that a 98% confidence interval will specify the mean to within ±0.1 ms2? How many wires must be sampled so that a 95% confidence interval will specify the mean to withirn ±0.1 m02 7. In a sample of 100 boxes of a certain type, the average compressive strength was 6230 N, and the standard deviation was 221 N a. Find a 95% confidence interval for the mean com pressive strength of boxes of this type. b. Find a 99% confidence interval for the mean com- pressive strength of boxes of this type c. An engineer claims that the mean strength is be- tween 6205 and 6255 N. With what level of confi- dence can this statement be made? d. Approximately how many boxes must be sampled so that a 95% confidence interval will specify the mean to within ±25 N?

Explanation / Answer

. n = 81, Mu = 17.3 , stdev = 1.2

a. 95% CI is given by :
Mu +/- Z*Sigma/sqrt(n)
=17.3 +/- 1.96*1.2/sqrt(81)
= 17.083 to 17.5613

b.98% CI is given by:
Mu +/- Z*Sigma/sqrt(n)
=17.3 +/- 2.33*1.2/sqrt(81)
= 16.9893 to 17.6107

c. MOE = 17.5 - (17.1+17.5)/2 = .2
So, .2 = Z*1.2/sqrt(81)
Z = 1.8/1.2 = 1.5, the cumulative probability is for 1.5 is .9332
Therefore confidence level is 2*(.9332-.5) = 86.64%

d. 1 <= 2.33*1.2/sqrt(n)
n >= (2.33*1.2)^2 = 7.82.
n >= 7.82.
Answer is round upp interger fof 7.82 ie. 8
Answer is 8

e. 1 <= 1.96*1.2/sqrt(n)
n = (1.96*1.2)^2 = 5.5319 or 6 people to round off to the higher interger
Answer is 6

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