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 793.3, 791.7, 791.3, 791.4, 793.7, 790.1

B 803.1, 791.1, 800.9, 789.1, 795.2, 788.7

(b) Is there a difference in the measurements of the muzzle velocity between device A and device B at the a+0.01 level if significance? A normal probability plot and box plot of the data indicate that the differences are approximately normally distributed with no outliers. let d1=A1-B1. Identify the null and alternative hypotheses.

Answer 1

This example of two sample t-test

Because we are checking the mean difference of both sample is similar or not.

i.e. Muzzle Velocity passing form both gun is same or not.

Testing of Hypothesis:

H0: 1 - 2 = 0 i.e. Muzzle Velocity passing form both gun is same

against,

H1: 1 - 2 != 0 i.e. Muzzle Velocity passing form both gun is not same

Test Statistics:

T= (A¯ − B¯) / √(S1^2 * 1/N1+ s2^2 * 1/N2)) ...... If variance of sample is not equal

where, A¯ and B¯ are the mean of A and B samples

s1 and s2 are sample variance of A and B

N1 and N2 the no of samples from A and B

T= (A¯ − B¯) / (S * √(1/N1+ 1/N2)) ...... If variance of sample is equal

where, S = [(N1−1) * s1^2 + (N2−1) * s2^2 ] / (N1+N2−2)

We have given,

A = (793.3, 791.7, 791.3, 791.4, 793.7, 790.1)

B = (803.1, 791.1, 800.9, 789.1, 795.2, 788.7)

A¯ =  791.9167

B¯ = 794.6833

N1 = N2 = 6

s1^2 = 1.817667

s2^2 = 37.91367

variance of both sample is not equal so use the first formula of t test.

T = (A¯ − B¯) / √(S1^2 * 1/N1+ s2^2 * 1/N2))

= -1.0751

Decision Rule:

If p-value < level of significance then reject the null hypothesis.

P-value two sample t-test statistic with level of significance 0.01

p-value = 0.3274

i.e. p-value (0.3274) > 0.01 (level of significance)

i.e reject the null hypothesis

i.e. 1 - 2 != 0 i.e. Muzzle Velocity passing form both gun is not same

Consider d1 = A - B

d1 = (-9.8 0.6 -9.6 2.3 -1.5 1.4)

We can check the d1 is normally distributed or not using normal probability plot and boxplot,

Normal Probability Plot:

Box-Plot:

From the Normal probability we can see that line is not straight line.

And from boxplot graph is not looking symmetric.

therefore variable d not follows normal distribution.  

From the above box-plot we can see that point are not above above the max threshold and not below the min threshold so there is no outlires in data.