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.Observation123456 A792.6792.8791.1791.9790.8790.5 B796.7790.5793.9790.6799.7789.7 (a) Why are these matched-pairs data? A.The same round was fired in every trial. B.All the measurements came from rounds fired from the same gun. C.The measurements (A and B) are taken by the same instrument. D.Two measurements (A and B) are taken on the same round.Your answer is correct.(b) Is there a difference in the measurement of the muzzle velocity between device A and device B at theα=0.01level of significance? Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers. Letdi=Ai−Bi.Identify the null and alternative hypotheses. H0:μd▼ nothingH1:μd▼ nothing

Given: The muzzle velocity (in feet per second) of rounds fired from a 155-mm gun. x~N(μ,σ) α=0.01 a) The correct option is B. All the measurements came from rounds fired from the same gun. b) Hypothesis Ho:μd=0 H0: There no difference in the measurement of the muzzle velocity between device A and device B. Ho:μd≠0 H1: There is a difference in the measurement of the muzzle velocity between device A and device B. The difference table is given below. A B d=A-B 792.6 796.7 -4.1 792.8 790.5 2.3 791.1 793.9 -2.8 791.9 790.6 1.3 790.8 799.7 -8.9 790.5 789.7 0.8 using the excel formula "=AVERAGE(C2:C7)" the average of d is -1.9. using the excel formula "=STDEV.S(C2:C7)" the standard deviation of d is 4.238. Test statistic: t=d¯sdnt=-1.94.2386t=-1.098 Test statistic value =-1.098 using the excel formula "= 2*(1 - T.DIST(1.098, 5, TRUE))" p-value=0.322.Decision rule: If the p-value is greater than the significance level, there is no evidence to reject the null hypothesis. If the p-value is lesser than the significance level, the hypothesis is rejected. Conclusion: The p-value (0.322) is greater than the significance (0.01) Thus, there is no evidence to reject the null hypothesis. There no difference in the measurement of the muzzle velocity between device A and device B.

Answered: The following data represent the muzzle…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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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	792.6	792.8	791.1	791.9	790.8	790.5
B	796.7	790.5	793.9	790.6	799.7	789.7

(a) Why are these matched-pairs data?

The same round was fired in every trial.

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

The measurements (A and B) are taken by the same instrument.

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

Your answer is correct.

(b) 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.

Identify the null and alternative hypotheses.

H0:

μd

▼

nothing

H1:

μd

▼

nothing

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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