Can anybody help me to answer the question below?

Question

Tensile strength is the ability of a material to resist rupture when pressure is applied under specified conditions to one of its sides by an instrument. In the manufacture of a certain woven polypropylene, previous production runs indicate that the tensile strength is approximately normally distributed. The process is operating properly when the standard deviation of the tensile strength is 4 pounds per square inch. Measurements of the tensile strength on a random of 40 rolls of woven polypropylene produced a mean of 87.3 pounds.

(i) Find a 98% confidence interval for , the mean tensile strength of the material.

(ii)The precision of the 98% confidence interval obtained in (a) is quantified by its tolerable error. Suppose we wish to maintain the same validity-namely, 98%- but increase the precision so that the tolerable error is only 1.2 pounds per inch. How large should the sample size be?

Answer 1

(i) ANSWER:  98% Confidence Interval = [85.77, 88.83].

Why??

SMALL-SAMPLE CONFIDENCE INTERVAL FOR A POPLATION MEAN

98% Confidence Interval = x-bar +/- (t-critical value) * s/SQRT(n)

x-bar = SAMPLE MEAN [87.3]

s = STANDARD DEVIATION [4]

n = NUMBER OF SAMPLES [40]

df = DEGREES OF FREEDOM (n - 1 = 39) [2.423] from "look-up" Table (approximately)

98% Confidence Interval: 87.3 +/- 2.423 * 4 / SQRT(40) = [85.77, 88.83].  That is with a confidence interval of approximately 98% the "true mean" is within the interval of [85.77, 88.83]  and that the sample mean (which is an estimate of the "true mean") is 87.3.

(ii) ANSWER SAMPLE SIZE = 261

Why??

1/2 Tolerable Error = 0.6

n = [(2.423 * 4)/0.6]^2 = (approx) 261