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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, with the following data obtained: $$ \begin{array}{ccccccc} \text { Observation } & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} \\ \hline \mathbf{A} & 793.8 & 793.1 & 792.4 & 794.0 & 791.4 & 792.4 \\ \hline \mathbf{B} & 793.2 & 793.3 & 792.6 & 793.8 & 791.6 & 791.6 \\ \hline \end{array} $$ $$ \begin{array}{ccccccc} \text { Observation } & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline \text { A } & 791.7 & 792.3 & 789.6 & 794.4 & 790.9 & 793.5 \\ \hline \text { B } & 791.6 & 792.4 & 788.5 & 794.7 & 791.3 & 793.5 \\ \hline \end{array} $$ (a) Why are these matched-pairs data? (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. (c) Construct a \(99 \%\) confidence interval about the population mean difference. Interpret your results. (d) Draw a boxplot of the differenced data. Does this visual evidence support the results obtained in part (b)?

Step by step solution

01

Title - Identify Matched-Pairs Data

Explain why the data are considered matched pairs. Each observation from Device A has a corresponding measurement from Device B taken for the same round fired at the same instance, making the data paired.

02

Title - Calculate Differences

Calculate the difference for each paired observation: \( D_i = A_i - B_i \). For example, for the first observation, the difference is \( 793.8 - 793.2 = 0.6 \). Repeat for all observations.

03

Title - Calculate Mean and Standard Deviation of Differences

Determine the mean and standard deviation of the differences: \[ \bar{D} = \frac{\sum D_i}{n} \quad \text{and} \quad s_D = \sqrt{\frac{\sum (D_i - \bar{D})^2}{n-1}} \] where \( n = 12 \).

04

Title - Perform Hypothesis Test for Differences

Set up and perform the hypothesis test: \[ H_0: \mu_D = 0 \text{ versus } H_1: \mu_D eq 0 \] Calculate the test statistic: \[ t = \frac{\bar{D}}{s_D / \sqrt{n}} \]. Compare the calculated t-value with the critical value from the t-distribution table for 11 degrees of freedom at \(\alpha = 0.01\).

05

Title - Construct 99% Confidence Interval for Mean Difference

Find the 99% confidence interval for the population mean difference: \[ \bar{D} \pm t_{\alpha/2} \left( \frac{s_D}{\sqrt{n}} \right) \] where \(t_{\alpha/2}\) is the critical value from the t-distribution table.

06

Title - Draw Boxplot of Differences

Create a boxplot for the differences to visually inspect for any outliers or skewness. Based on this visual evidence, evaluate if it aligns with the hypothesis test results.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hypothesis Testing

Hypothesis testing is a powerful tool in statistics used to determine if there is enough evidence to reject a null hypothesis for a population. For our exercise, hypothesis testing helps us evaluate whether there's a significant difference in muzzle velocity measurements made by two devices, A and B. In this context, we frame our null hypothesis (H_0) and alternative hypothesis (H_1) as follows:

• H_0: 渭_D = 0 (No difference in measurements between the two devices.)
• H_1: 渭_D 鈮� 0 (There is a difference in measurements between the two devices.)

We need to calculate the test statistic (t) to test our hypotheses by finding the mean (\bar{D}) and standard deviation (s_D) of the differences. The formula is:\[ t = \frac{\bar{D}}{s_D / \sqrt{n}} \]Here, \bar{D} is the mean difference, s_D is the standard deviation of the differences, and n is the number of pairs.We compare the calculated t-value with the critical value from the t-distribution table for 11 degrees of freedom at \text伪 = 0.01. If the calculated t-value falls outside the range specified by the critical value, we reject the null hypothesis, suggesting a significant difference exists.

Confidence Interval

A confidence interval provides a range of values that is likely to contain a population parameter with a certain level of confidence. In our exercise, we aim to construct a 99% confidence interval for the mean difference (\bar{D}) in muzzle velocity measurements. It gives us an estimate of how different the measurements from Device A are compared to Device B.The formula for the 99% confidence interval is:\[\bar{D} \pm t_{\alpha/2} \left( \frac{s_D}{\sqrt{n}} \right)\]Here, \bar{D} is the mean of the differences, s_D is the standard deviation of the differences, and t_{\alpha/2} is the critical value for the 99% confidence level. This interval means that we are 99% confident the true mean difference falls within this range.Interpreting the confidence interval helps us understand whether the observed difference in our sample is significant enough to be generalized to the entire population.

Boxplot Analysis

A boxplot is a graphical representation that helps visualize the distribution, central value, and variability of a data set. For our exercise, creating a boxplot of the differences between measurements from Device A and Device B is an essential step. This visual analysis allows us to spot any outliers and assess the data's symmetry or skewness.A boxplot typically includes:

• A box spanning the interquartile range (IQR), which contains the middle 50% of the data.
• A line inside the box representing the median.
• Whiskers extending from the box to the smallest and largest values within 1.5 * IQR from the quartiles.
• Any points outside the whiskers are potential outliers.

In our boxplot of differences, we look for a symmetric distribution with no significant outliers, which supports the assumption that our differences are approximately normally distributed, as required by the paired sample t-test. If the visual evidence from the boxplot aligns with the results of our hypothesis test, it further strengthens our conclusions.