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   A   B
1   791.9   801.8
2   794.1   791.2
3   793.5   797.9
4   793.9   789.4
5   791.1   799.3
6   794.1   789.3

(B) Determine the test statistic for this hypothesis test.

t0 = (BLANK)

Find the p value.

p0 = (BLANK)

What is your conclusion regarding a=.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.

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

(d) 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 0.01

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.

​

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

A.​Yes, because 0 is contained in the boxplot.

B.No, because the boxplot is too large.

C.No, because 0 is not containednot containedin the boxplot.

D.Yes, because the boxplot shows no outliers.

Answer 1

S. No	A	B	diff:(d)=x1-x2	d2
1	791.9	801.8	-9.9	98.01
2	794.1	791.2	2.9	8.41
3	793.5	797.9	-4.4	19.36
4	793.9	789.4	4.5	20.25
5	791.1	799.3	-8.2	67.24
6	794.1	789.3	4.8	23.04
	total	=	Σd=-10.3	Σd2=236.31
	mean dbar=	d̅     =	-1.7167
	degree of freedom =n-1                            =		5
	Std deviaiton SD=√(Σd2-(Σd)2/n)/(n-1) =		6.612539
	std error=Se=SD/√n=		2.6996
	test statistic            =	(d̅-μd)/Se         =	-0.6359
	p value	=	0.5528	from excel: tdist(0.636,5,2)

since p value >0.01 ; we cannot conclude that there is mean difference.

c)

for 99% CI; and 5 degree of freedom, value of t=			4.032
therefore confidence interval=sample mean -/+ t*std error
margin of errror          =t*std error=		10.884616
lower confidence limit                     =		-12.6013
upper confidence limit                    =		9.1679
from above 99% confidence interval for population mean =(-12.60 , 9.17)

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

d)

A.​Yes, because 0 is contained in the boxplot.