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	791.2791.2	791.5791.5	794.5794.5	791.5791.5	793.6793.6	793.7793.7
B	798.3798.3	787.9787.9	799.2799.2	789.6789.6	798.8798.8	793.6793.6

​(a) Why are these​ matched-pairs data?

A.

Two measurements​ (A and​ B) are taken on the same round.

Your answer is correct.

B.

The same round was fired in every trial.

C.

The measurements​ (A and​ B) are taken by the same instrument.

D.

All the measurements came from rounds fired from the same gun.

​(b) Is there a difference in the measurement of the muzzle velocity between device A and device B at the

alpha equals 0.01α=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

diequals=Aiminus−Bi.

Identify the null and alternative hypotheses.

Upper H 0H0​:

mu Subscript dμd

equals=

00

Upper H 1H1​:

mu Subscript dμd

not equals≠

00

Determine the test statistic for this hypothesis test.

t0=?

​(Round to two decimal places as​ needed.)

Answer 1

Answer:_ a) A.Two measurements (A and B) are taken on the same round. Hence here Date:--11/02/2019 we use matched pair data b) A-B 791.2 798.37.1 791.5 787.93.6 794.5 799.24.7 791.5 789.61.9 793.6 798.85.2 793.7 793.6 0.1 Hence from data we get n-6, d1.9, Sd-4.346 The hypothesis are The test statistic is 1.9/4.346/sqrt(6) Here df-n 6-1 5 The p value-p(t<-1.07)0.3335 so p value > α(0.01) Hence we tailed to reject null hypothesis Conclusion: There is no difference in the measurements of velocity between device A and device B