Question

5 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 6 793.9 793.1 793.8 790.4 792.8 791.7 B 794.4 788.6 803.3 786.7 797.3 790.3 (a) Why are these matched pairs data? A A. The same round was fired in every trial. B. The measurements (A and B) are taken by the same instrument C. Two measurements (A and B) are taken on the same round. D. All the measurements came from rounds fired from the same gun. (b) Is there a difference in the measurement of the muzzle velocity between device A and device B at the a= 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 d = A - B Identify the null and alternative hypotheses. Ho: Pa Hy Pa Determine the test statistic for this hypothesis test. 1- (Round to two decimal places as needed.) Find the P-value P-value=(Round to three decimal places as needed.) What is your conclusion regarding Ho? Ho. There sufficient evidence at the a = 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.) Interpret the confidence interval. Choose the correct answer below. A. One can be 1% confident that the mean difference in measurement lies in the interval found above. OB. One can be 99% confident that the mean difference in measurement is 0. C. One can be 99% confident that the mean difference in measurement is 0.01. D. One can be 99% confident that the mean difference in measurement lies in the interval found above. d) Draw a boxplot of the differenced data. Does this visual evidence support the results obtained in part (b)? ОО OA ов. -10-8-6 -10-8-84-2 Differences Differences OC D. o o os -10-8 -6 Differences Differences Does this visual evidence sunt the results obtained in part 2

Answer 1

Given that,
null, H0: Ud = 0
alternate, H1: Ud != 0
level of significance, α = 0.01
from standard normal table, two tailed t α/2 =4.032
since our test is two-tailed
reject Ho, if to < -4.032 OR if to > 4.032
we use Test Statistic
to= d/ (S/√n)
where
value of S^2 = [ ∑ di^2 – ( ∑ di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = -0.883
We have d = -0.883
pooled variance = calculate value of Sd= √S^2 = sqrt [ 150.41-(-5.3^2/6 ] / 5 = 5.399
to = d/ (S/√n) = -0.401
critical Value
the value of |t α| with n-1 = 5 d.f is 4.032
we got |t o| = 0.401 & |t α| =4.032
make Decision
hence Value of |to | < | t α | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -0.4006 ) = 0.7052
hence value of p0.01 < 0.7052,here we do not reject Ho
ANSWERS
---------------
a.
option:A
the same round was fired in every trail
b.
null, H0: Ud = 0
alternate, H1: Ud != 0
test statistic: -0.401 =-0.40
critical value: reject Ho, if to < -4.032 OR if to > 4.032
decision: Do not Reject Ho
p-value: 0.7052 =0.705
we do not have enough evidence to support the claim that difference in measurements of the muzzle velocity between device A and B.
c.
Confidence Interval
CI = d ± t a/2 * (Sd/ Sqrt(n))
Where,
d = ∑ di/n
Sd = Sqrt( ∑ di^2 – ( ∑ di )^2 / n ] / ( n-1 ) )
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
d = ( ∑ di/n ) =-5.29999999999995/6=-0.88
Pooled Sd( Sd )= Sqrt [ 150.41- (-5.3^2/6 ] / 5 = 5.4
Confidence Interval = [ -0.88 ± t a/2 ( 3.82/ Sqrt ( 6) ) ]
= [ -0.88 - 4.032 * (2.2) , -0.88 + 4.032 * (2.2) ]
= [ -9.77 , 8 ]
d.
option:B
diagram shows the box plot
yes,
because the zero contain in the box plot