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.)

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter2: Linear Equations

Section2.7: Litereal Equations And Dimensional Analysis

Problem 27PPS

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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	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.

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