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.

Obervation	1	2	3	4	5	6
A	791.2	790.7	793.9	793.7	794.3	793.7
B	795.1	786.0	802.5	789.8	797.5	793.1

Is there a difference in the measurement of the muzzle velocity between device A and device B at the α=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 di=Ai−Bi. The null and alternative hypotheses are identified.

Upper H0​: μd=0

Upper H1​: μd≠0

a)Determine the test statistic for this hypothesis test.

t0=

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

b) Find the P-value

P-value =

(Round to three decimal places as needed)

c) Construct a 99% confidence interval about the population mean difference. Compute the difference as A-B:

The lower bound is:

The upper bound is:

d) Construct a 95% confidence interval about the population mean difference. Compute the difference as A-B:

The lower bound is:

The upper bound is:

Answer 1

Solution:

a) Determine the test statistic for this hypothesis test.

Test statistic for paired t test is given as below:

t = (Dbar - µ_d)/[S_d/sqrt(n)]

From given data, we have

Dbar = -1.0833

S_d = 5.0996

n = 6

t = (-1.0833 – 0)/[ 5.0996/sqrt(6)]

t = -0.5204

Test statistic = t₀ = -0.52

b) Find the P-value

df = n – 1 = 6 – 1 = 5

P-value = 0.625

(by using t-table)

c) Construct a 99% confidence interval about the population mean difference.

Confidence interval for difference between two population means of paired samples is given as below:

Confidence interval = Dbar ± t*S_D/sqrt(n)

We are given

Dbar = -1.0833

S_d = 5.0996

n = 6

df = n – 1 = 5

Confidence level = 99%

Critical t value = 4.0321

(by using t-table)

Confidence interval = Dbar ± t*S_D/sqrt(n)

Confidence interval = -1.0833 ± 4.0321*5.0996/sqrt(6)

Confidence interval = -1.0833 ± 8.3945

Lower limit = -1.0833 - 8.3945 = -9.4778

Upper limit = -1.0833 + 8.3945 = 7.3112

The lower bound is: -9.4778

The upper bound is: 7.3112

d) Construct a 95% confidence interval about the population mean difference.

Confidence interval for difference between two population means of paired samples is given as below:

Confidence interval = Dbar ± t*S_D/sqrt(n)

We are given

Dbar = -1.0833

S_d = 5.0996

n = 6

df = n – 1 = 5

Confidence level = 95%

Critical t value = 2.5706

(by using t-table)

Confidence interval = Dbar ± t*S_D/sqrt(n)

Confidence interval = -1.0833 ± 2.5706*5.0996/sqrt(6)

Confidence interval = -1.0833 ± 5.3517

Lower limit = -1.0833 - 5.3517 =-6.4350

Upper limit = -1.0833 + 5.3517 =4.2684

The lower bound is: -6.4350

The upper bound is: 4.2684