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 790.2 793.3 791.6 793.9 794.7 793.1 B 794.8 789.0 796.2 793.5 802.9 791.3 (a) Why are these matched-pairs data? A. All the measurements came from rounds fired from the same gun. B. Two measurements (A and B) are taken on the same round. C. The measurements (A and B) are taken by the same instrument. D. The same round was fired in every trial. (b) Is there a difference in the measurement of the muzzle velocity between device A and device B at the α=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. Let diequals=Aiminus−Bi. Identify the null and alternative hypotheses. H0:μd=0 H1:μd≠0 Determine the test statistic for this hypothesis test. t0= −0.94 (Round to two decimal places as needed.) Find the P-value. P-value=0.390 (Round to three decimal places as needed.) What is your conclusion regarding H0? Do not reject H0. There is not sufficient evidence at the α=0.01 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_. The upper bound is_. (Round to two decimal places as needed.)

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

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	790.2	793.3	791.6	793.9	794.7	793.1
B	794.8	789.0	796.2	793.5	802.9	791.3

(a) Why are these matched-pairs data?

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

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

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

The same round was fired in every trial.

(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

diequals=Aiminus−Bi.

Identify the null and alternative hypotheses.

H0:μd=0

H1:μd≠0

Determine the test statistic for this hypothesis test.

t0= −0.94

(Round to two decimal places as needed.)

Find the P-value.

P-value=0.390

(Round to three decimal places as needed.)

What is your conclusion regarding

H0?

Do not reject H0. There is not sufficient evidence at the α=0.01 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_.

The upper bound is_.

(Round to two decimal places as needed.)

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.

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 4 steps with 1 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

$Fundamental Theorem$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.

Similar questions