Question

The following data represent the muzzle velocity​ (in feet per​ second) of rounds fired from a​ 155-mm gun. For each​ round, two measurements of the velocity were recorded using two different measuring​ devices, resulting in the following data. Complete parts​ (a) through​ (d) below. Observation 1 2 3 4 5 6 A 793.3 792.5 793.2 793.1 793.5 793.9 B 794.1 790.9 800.5 788.7 796.2 791.5 ​(a) Why are these​ matched-pairs data? A. The same round was fired in every trial. B. Two measurements​ (A and​ B) are taken on the same round. Your answer is correct.C. All the measurements came from rounds fired from the same gun. D. The measurements​ (A and​ B) are taken by the same instrument. ​(b) Is there a difference in the measurement of the muzzle velocity between device A and device B at the alpha equals 0.01 level of​ significance? ​Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers. Let diequalsAiminusBi. Identify the null and alternative hypotheses. Upper H 0​: mu Subscript d equals 0 Upper H 1​: mu Subscript d not equals 0 Determine the test statistic for this hypothesis test. t0=(Round to two decimal places as​ needed.)

Answer 1

(a)

Correct option:(B) Two measurements (A and B) are taken on the same round.

(b)

Values of di = Ai - Bi are got as follows:

- 0.8, 1.6, - 7.3, 4.4, - 2.7, 2.4

From d values, the following statistics are calculated:

n = 6

= - 0.4

s_d = 4.1938

SE = s_d/

= 4.1938/

= 1.7121

Test statistic is:

t = - 0.4/1.7121 = - 0.2336

= 0.01

ndf = n - 1 = 6 - 1 = 5

From Table, critical values of t = 4.0321

Since calculated value of t = - 0.2336 is greater than critical value of t = - 4.0321, the difference is not significant. Fail to reject null hypothesis.

Conclusion:

The data do not support the claim that there is a difference in the measurement of the muzzle velocity between device A and device B.