Question

o 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 B A 790.5 794.1 790.5 792.5 790.2 792.7 B 791.6 791.9 799.4 792.1 799.9 791.1 Click the icon to view the table of critical t-values. a) Why are these matched pairs data? A. All the measurements came from rounds fired from the same gun. B. The same round was fired in every trial. C. The measurements (A and B) are taken by the same instrument. D. Two measurements (A and B) are taken on the same round. 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. Hoi Hd Hq : Ho Determine the test statistic for this hypothesis test. to-(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 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 confidence interval is CD (Round to two decimal places as needed.) Interpret the confidence interval. Choose the correct answer below. A. One can be 99% confident that the mean difference in measurement is 0.01. B. One can be 1% confident that the mean difference in measurement lies in the interval found above. OC. One can be 99% confident that the mean difference in measurement lies in the interval found above. D. One can be 99% confident that the mean difference in measurement is o. d) Draw a boxplot of the differenced data. Does this visual evidence support the results obtained in part (b)? a OA OB. -10-8 Differences Differences OC. D. 310-8 0 -6 4 Differences -10-3 -6 -2 Differences Does this visual evidence support the results obtained in part (b)?

Answer 1

a) D. Two measurements(A and B) are taken on the same round.

b) $\small H_0:\mu_d=0\ vs\ H_1:\mu_d\neq 0$

The sample mean of the difference is -2.5833 and the sample standard deviation is 5.33. Thus the test statistic is given as:

$\small t_0=\frac{\bar{d}}{\frac{s_d}{\sqrt{6}}}=\frac{-2.5833*\sqrt{6}}{5.33}=-1.19$. Thus the test statistic is -1.19

The critical value for the test is $\small \pm t_{0.005,5}$ =(-4.03, 4.03).

Conclusion regarding H₀: Fail to reject H₀. There is not sufficient evidence at the $\small \alpha$ =0.01.

c) The 99% confidence interval is given by

$\small \bar{d}\pm t_{0.005,5}\frac{s_d}{\sqrt{6}}=(-2.5833-4.032*\frac{5.33}{\sqrt{6}},-2.5833+4.032*\frac{5.33}{\sqrt{6}})=(-11.36,6.19)$

Interpretation of the confidence interval:

C. One can be 99% confident that the mean difference in the measurement lies in the interval found above.

d) B is the correct boxplot

Visual evidence:

D. Yes, because 0 is contained in the boxplot.