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	794.4	790.8	794.1	794.1	791.8	790.4
B	799.0	787.2	801.0	791.3	795.6	785.7

​(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.

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 0.01α=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

▼

not equals≠

greater than>

equals=

less than<

nothing

Upper H 1H1​:

mu Subscript dμd

▼

nothing

Determine the test statistic for this hypothesis test.

t0equals=nothing

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

Find the​ P-value.

​P-valueequals=nothing

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

What is your conclusion regarding

Upper H 0H0​?

▼

Upper H 0H0.

There

▼

is not

is

sufficient evidence at the

alphaαequals=0.010.01

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

nothing.

The upper bound is

nothing.

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

Interpret the confidence interval. Choose the correct answer below.

A.

One can be​ 1% confident that the mean difference in measurement lies in the interval found above.

B.One can be​ 99% confident that the mean difference in measurement is

0.010.01.

C.

One can be​ 99% confident that the mean difference in measurement lies in the interval found above.

D.

One can be​ 99% confident that the mean difference in measurement is 0.

​d) Draw a boxplot of the differenced data. Does this visual evidence support the results obtained in part​ (b)?

A.

-8-6-4-20246Differences

x y graph

B.

-8-6-4-20246Differences

x y graph

C.

-8-6-4-20246Differences

x y graph

D.

-8-6-4-2024Differences

x y graph

Does this visual evidence support the results obtained in part​ (b)?

A.​Yes, because 0 is

containedcontained

in the boxplot.

B.

​Yes, because the boxplot shows no outliers.

C.

​No, because the boxplot is too large.

D.​No, because 0 is

not containednot contained

in the boxplot.

Answer 1

Given that,
null, H0: Ud = 0
alternate, H1: Ud != 0
level of significance, α = 0.01
from standard normal table, two tailed t α/2 =4.032
since our test is two-tailed
reject Ho, if to < -4.032 OR if to > 4.032
we use Test Statistic
to= d/ (S/√n)
where
value of S^2 = [ ∑ di^2 – ( ∑ di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = -0.7
We have d = -0.7
pooled variance = calculate value of Sd= √S^2 = sqrt [ 126.1-(-4.2^2/6 ] / 5 = 4.963
to = d/ (S/√n) = -0.345
critical Value
the value of |t α| with n-1 = 5 d.f is 4.032
we got |t o| = 0.345 & |t α| =4.032
make Decision
hence Value of |to | < | t α | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -0.3455 ) = 0.7438
hence value of p0.01 < 0.7438,here we do not reject Ho
ANSWERS
---------------
a.
matched-pairs data because both dependent variables
b.
null, H0: Ud = 0
alternate, H1: Ud != 0
test statistic: -0.345
critical value: reject Ho, if to < -4.032 OR if to > 4.032
decision: Do not Reject Ho
p-value: 0.7438
we do not have enough evidence to support the claim that difference in the measurement of the muzzle velocity between device A and device B
c.
Confidence Interval
CI = d ± t a/2 * (Sd/ Sqrt(n))
Where,
d = ∑ di/n
Sd = Sqrt( ∑ di^2 – ( ∑ di )^2 / n ] / ( n-1 ) )
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
d = ( ∑ di/n ) =-4.20000000000016/6=-0.7
Pooled Sd( Sd )= Sqrt [ 126.1- (-4.2^2/6 ] / 5 = 4.963
Confidence Interval = [ -0.7 ± t a/2 ( 2.865/ Sqrt ( 6) ) ]
= [ -0.7 - 4.032 * (2.026) , -0.7 + 4.032 * (2.026) ]
= [ -8.869 , 7.469 ]
d.
-8-6-4-20246 Differences
x y graph
A.
​Yes, because 0 iscontained in the boxplot.