Question

he 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	Open in StatCrunch + Copy to Clipboard + Open in Excel +
A	793.2793.2	790.2790.2	793.7793.7	792.2792.2	793.8793.8	794.9794.9
B	802.9802.9	788.8788.8	802.3802.3	788.0788.0	801.0801.0	793.3793.3

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

A.

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

Your answer is correct.

B.

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

C.

The same round was fired in every trial.

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

▼

nothing

Upper H 1H1​:

mu Subscript dμd

▼

equals=

less than<

greater than>

not equals≠

nothing

Answer 1

Solution:-

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

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_d = 0

Alternative hypothesis: u_d ≠ 0

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.01. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).

s = 6.22404

SE = s / sqrt(n)

S.E = 2.541

DF = n - 1 = 6 -1

D.F = 5

t = [ (x₁ - x₂) - D ] / SE

t = - 1.168

where d_i is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 5 degrees of freedom is more extreme than -1.168; that is, less than - 1.168 or greater than 1.168.

Thus, the P-value = 0.295

Interpret results. Since the P-value (0.295) is greater than the significance level (0.01), we have to accept the null hypothesis.

Do not reject H0. The mean difference appears to be zero.