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.

t0equals=negative 1.07−1.07

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

Find the​ P-value.

​P-valueequals=. 334.334

​(Round to three decimal places as​ needed.)What is your conclusion regarding

Upper H 0H0​?

Do not reject

Upper H 0H0.

There

is not

sufficient evidence at the

alphaαequals=0.01

level of significance to conclude that there is a difference in the measurements of velocity between device A and device B.

​(c) Construct a​ 99% confidence interval about the population mean difference. Compute the difference as device A minus device B. Interpret your results.

The lower bound is

?.

The upper bound is

?.

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

Answer 1

(a)

Correct option:

A.

Two measurements were taken on the same period.

(b)

Correct option:

H0:

H1:

From the given data, d values are calculated as follows:

- 7.1, 3.6, - 4.7, 1.9, - 5.2, 0.1

From the d values, the following statistics are calculated:

n = 6

= - 11.4/6 = - 1.9

s_d = 4.3465

SE = s_d/

= 4.3465/ = 1.7745

test statistic is:

t = - 1.9/1.7745 = - 1.07

So,

t₀ = - 1.07

ndf = 6 - 1= 5

Two Tail Test

By Technology, p - value = 0.334

Correct option:

Do not reject H0. There is not sufficient evidence at the = 0.01 level of significance to conclude there is a difference in measurements of velocity between device A and device B.

(c)

= 0.01

ndf = 5

From Table, critical values of t = 4.03

The lower bound is:

- 1.9 - (4.03 X 1.7745) = - 1.9 - 7.15 = - 9.05

The lower bound is:

- 1.9 + (4.03 X 1.7745) = - 1.9 + 7.15 = 5.25