A sample of 20 cans of tomato juice showed a standard deviation of 0.4 ounces. A
ID: 3154347 • Letter: A
Question
A sample of 20 cans of tomato juice showed a standard deviation of 0.4 ounces. A 95% confidence interval estimate for the variance of the population is: (a) 0.2313 to 0.8533. (b) 0.2224 to 0.7924. (c) 0.0889 to 0.3169. (d) 0.0925 to 0.3413. (e) 0.0936 to 0.1754. Use the following information for questions 17 - 18: A bottler of a ccrtain soft drink claims its equipment is accurate and that the variance of all filled bottles less than 0.05 ounces. In order to verify his claim, a random sample of 26 bottles was obtained. The standard deviation of the amount of drink in the 26 bottles was 0.2 ounces. 17. Rcspectively, which among the following are the numerical value of test statistic and the critical value when alpha = 0.05? (a) 20.00 and 14.61. (b) 20.80 and 14.61. (c) 37.65 and 38.88. (d) 20.00 and 37.65. (e) 20.80 and 37.65. 18. For alpha = 0.05, which among the following is the CORRECT decision? (a) Reject Ho and conclude that the variance of filled bottles is less than 0.05 ounces. (b) Reject Ho and conclude that the variance of filled bottles is greater than 0.05 ounces. (c) Fail to reject Ho and conelude that there is insufficient evidenec to conclude that the variance of filled bottles is less than 0.05 ounces. (d) Fail to rejcet Ho and conclude that there is insufficient evidcnce to conclude that the variance of filled bottles is greater than 0.05 ounces. (e) Reject Ho and conclude that the variance of filled bottles is 0.05 ounces. Final 2: 18. A machine manufactures bolts that are supposed to be 3 inches in length. Each day, a quality engineer selects a random sample of 36 bolts from the day's production, measures lengths, and performs a hypothesis test of Ho: mu = 3, where: mu notequalto 3, where mu is the mean length of all the bolts manufactured that day. Assume that the population standard deviation for bolt lengths is 0.1 inchcs. Ho is rejected, forcing the machine to be shut down and recalibrated, if the sample mean falls outside the range of 2.98Explanation / Answer
16) The X^20.975 at df=19 is 32.8523 and X^20.025 is 8.9065.
Compute lower bound and upper bound of 95% c.i
(n-1)s^2/X^20.975 and (n-1)s^2/X^20.025
(19*0.4^2)32.8523 and (19*0.4^2)/8.9065
0.0925 to 0.3413 (d)
Test statistic, T=(N-1)(s/sigma)^2, where, N is sample size, s is sample standard deviation, and sigm ai spopulation standard deviation.
=(26-1)(0.2/sqrt 0.05)^2
=20
Critical region, X^2alpha,N-1=37.652 (d)
Test statistic do not fall in critical region, fail to reject null hypothesis. (c)
Compute 1-sample Z test.
Z=(Xbar-mu)/(sigma/sqrt N), where, Xbar is sample mean, mu is population mean, N is sample size and sigma is population standard deviation.
=(2.98-3)/(3/sqrt 36)
=-0.04
Similarly, Z score corresponding to Xbar3.02 is 0.04.
P(2.98<X<3.02)=0.32