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.
\begin{tabular}{lcccccc}
Observation & 1 & 2 & 3 & 4 & 5 & 6 \\
A & 792.9 & 791.4 & 791.4 & 791.5 & 791.5 & 793.8 \\
B & 799.5 & 788.2 & 795.3 & 790.6 & 799.2 & 791.4
\end{tabular}
(a) Why are these matched -pairs data?
A. Two measurements ( $A$ and $B$ ) are taken on the same round
B. The same round was fired in every trial
C. All the measurements came from rounds fired from the same gun
D. The measurements ( $A$ and $B$ ) are taken by the same instrument
(b) Is there a difference in the measurement of the muzzle velocity between device A and device B at the $\alpha=0.01$ 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 $d_{i}=A_{i}-B_{i}$. Identify the null and alternative hypotheses
\[
\begin{array}{ll}
H_{0} \cdot \mu_{\mathrm{d}} & \square \square \\
H_{1} \cdot \mu_{\mathrm{d}} & \square \square
\end{array}
\]
Determine the test statistic for this hypothesis test
Answer
(a) The data are matched-pairs because two measurements (A and B) are taken on the same round. This means that each pair of measurements (one from A and one from B) corresponds to the same round. Therefore, option A is correct. (b) To determine if there is a difference in the measurement of the muzzle velocity between device A and device B at the α=0.01 level of significance, we need to perform a paired t-test. First, we calculate the differences (di = Ai - Bi) for each pair of observations: \begin{tabular}{lcccccc} Observation & 1 & 2 & 3 & 4 & 5 & 6 \\ d & -6.6 & 3.2 & -3.9 & 0.9 & -7.7 & 2.4 \end{tabular} The null hypothesis (H0) is that the mean difference in measurements between the two devices is zero (μd = 0), which means there is no difference in the measurements between the two devices. The alternative hypothesis (H1) is that the mean difference is not zero (μd ≠ 0), which means there is a difference in the measurements between the two devices. To determine the test statistic for this hypothesis test, we need to calculate the mean and standard deviation of the differences, and then use these to calculate the t-statistic. The mean (d̄) of the differences is (-6.6 + 3.2 - 3.9 + 0.9 - 7.7 + 2.4) / 6 = -1.95. The standard deviation (sd) of the differences can be calculated as sqrt[((-6.6 - (-1.95))^2 + (3.2 - (-1.95))^2 + (-3.9 - (-1.95))^2 + (0.9 - (-1.95))^2 + (-7.7 - (-1.95))^2 + (2.4 - (-1.95))^2) / (6 - 1)] = 4.68. The t-statistic is then calculated as (d̄ - μd) / (sd / sqrt(n)) = (-1.95 - 0) / (4.68 / sqrt(6)) = -1.26. Therefore, the test statistic for this hypothesis test is -1.26. This value would then be compared to the critical value for a t-distribution with 5 degrees of freedom (n - 1) at the α=0.01 level of significance to determine whether to reject the null hypothesis.
Step 1 ：(a) The data are matched-pairs because two measurements (A and B) are taken on the same round. This means that each pair of measurements (one from A and one from B) corresponds to the same round. Therefore, option A is correct. (b) To determine if there is a difference in the measurement of the muzzle velocity between device A and device B at the α=0.01 level of significance, we need to perform a paired t-test. First, we calculate the differences (di = Ai - Bi) for each pair of observations: \begin{tabular}{lcccccc} Observation & 1 & 2 & 3 & 4 & 5 & 6 \\ d & -6.6 & 3.2 & -3.9 & 0.9 & -7.7 & 2.4 \end{tabular} The null hypothesis (H0) is that the mean difference in measurements between the two devices is zero (μd = 0), which means there is no difference in the measurements between the two devices. The alternative hypothesis (H1) is that the mean difference is not zero (μd ≠ 0), which means there is a difference in the measurements between the two devices. To determine the test statistic for this hypothesis test, we need to calculate the mean and standard deviation of the differences, and then use these to calculate the t-statistic. The mean (d̄) of the differences is (-6.6 + 3.2 - 3.9 + 0.9 - 7.7 + 2.4) / 6 = -1.95. The standard deviation (sd) of the differences can be calculated as sqrt[((-6.6 - (-1.95))^2 + (3.2 - (-1.95))^2 + (-3.9 - (-1.95))^2 + (0.9 - (-1.95))^2 + (-7.7 - (-1.95))^2 + (2.4 - (-1.95))^2) / (6 - 1)] = 4.68. The t-statistic is then calculated as (d̄ - μd) / (sd / sqrt(n)) = (-1.95 - 0) / (4.68 / sqrt(6)) = -1.26. Therefore, the test statistic for this hypothesis test is -1.26. This value would then be compared to the critical value for a t-distribution with 5 degrees of freedom (n - 1) at the α=0.01 level of significance to determine whether to reject the null hypothesis.
