How heavy a load (pounds) is needed to pull apart pieces of wood 4 inches long and 1.5 inches square? Here are data from students doing a laboratory exercise. 33,150 31,890 32,550 26,530 33,250 32,310 32,980 31,990 30,500 32,710 23,020 30,980 32,730 33,610 32,390 24,020 30,140 31,330 28,700 31,910 We are willing to regard the wood pieces prepared for the lab session as an SRS of all similar pieces of wood. Engineers also commonly assume that characteristics of materials vary Normally. Suppose that the strength of pieces of wood like these follows a Normal distribution with standard deviation 3000 pounds.

a. (29526.192 , 32155.808)

b. (29673.9301 , 32008.0699)

Step-by-step explanation:

on calculation, we find that sample mean = 30843 and sample standard deviation = 3018.504, and n=20

1)

Sample Mean = 30841

SD = 3000

Sample Size (n) = 20

Standard Error (SE) = SD/root(n) = 670.8204

alpha (a) = 1 - 0.95 = 0.05

z critical value for 95% confidence interval:

z(a/2) = z(0.025) = 1.96

Margin of Error (ME) = z(a/2) x SE = 1314.808

95% confidence interval is given by:

Sample Mean +/- (Margin of Error) = 30841 +/- 1314.808

= (29526.192 , 32155.808)

2)when std dev of population is not known, we use sample's, but we have to use t instead of z

Sample Mean = 30841

SD = 3018.504

Sample Size (n) = 20

Standard Error (SE) = SD/root(n) = 674.958

alpha (a) = 1-0.9 = 0.1

we use t-distribution as population standard deviation is unknown

t(a/2, n-1 ) = 1.7291

Margin of Error (ME) = t(a/2,n-1)x SE = 1167.0699

90% confidence interval is given by:

Sample Mean +/- (Margin of Error)

30841 +/- 1167.0699 = (29673.9301 , 32008.0699)