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.
$\begin{array}{lcccccc}\text { Observation } & 1 & 2 & 3 & 4 & 5 & 6 \\ \text { A } & 791.8 & 794.5 & 791.3 & 791.8 & 793.5 & 793.8 \\ \text { B } & 798.2 & 794.1 & 797.2 & 790.7 & 796.1 & 791.4\end{array}$
Click the icon to view the table of critical t-values.
Determine the test statıstıc for thıs hypothesis test.
$t_{0}=-1.20$ (Round to two decimal places as needed.)
Find the critical value(s) for this hypothesis test.
(Use a comma to separate answers as needed. Round to two decimal places as needed.)
What is your conclusion regarding $\mathrm{H}_{0}$ ?
Do not reject $H_{0}$. There is not sufficient evidence at the $\alpha=0.01$ 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 confidence interval is (Round to two decimal places as needed.)
Interpret the confidence interval. Choose the correct answer below.
Answer
Final Answer: The test statistic for this hypothesis test is \(\boxed{-1.34}\).
Step 1 ：Given two sets of measurements A and B, we have A = [791.8, 794.5, 791.3, 791.8, 793.5, 793.8] and B = [798.2, 794.1, 797.2, 790.7, 796.1, 791.4].
Step 2 ：Calculate the mean of A and B, we get \(\bar{A} = 792.78\) and \(\bar{B} = 794.62\).
Step 3 ：Calculate the standard deviation of A and B, we get \(s_{A} = 1.31\) and \(s_{B} = 3.09\).
Step 4 ：Since the sample sizes of A and B are both 6, we have \(n_{A} = 6\) and \(n_{B} = 6\).
Step 5 ：Calculate the standard error of the difference, we get \(SE_{diff} = 1.37\).
Step 6 ：Calculate the test statistic using the formula \(t_{stat} = \frac{\bar{A} - \bar{B}}{SE_{diff}}\), we get \(t_{stat} = -1.34\).
Step 7 ：Since the calculated test statistic is close to the given test statistic of -1.20, the difference between the means of the two sets of measurements is not statistically significant at the 0.01 level of significance.
Step 8 ：Therefore, we do not reject the null hypothesis that there is no difference in the measurements of velocity between device A and device B.
Step 9 ：Final Answer: The test statistic for this hypothesis test is \(\boxed{-1.34}\).
