91Ó°ÊÓ

The manufacturer of hardness testing equipment uses steel-ball indenters to penetrate a metal that is being tested. However, the manufacturer thinks it would be better to use a diamond indenter so that all types of metal can be tested. Because of differences between the two types of indenters, it is suspected that the two methods will produce different hardness readings. The metal specimens to be tested are large enough so that two indentions can be made. Therefore, the manufacturer uses both indenters on each specimen and compares the hardness readings. Construct a \(95 \%\) confidence interval to judge whether the two indenters result in different measurements. $$ \begin{array}{lccccccccc} \text { Specimen } & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} & \mathbf{7} & \mathbf{8} & \mathbf{9} \\ \hline \text { Steel ball } & 50 & 57 & 61 & 71 & 68 & 54 & 65 & 51 & 53 \\ \hline \text { Diamond } & 52 & 56 & 61 & 74 & 69 & 55 & 68 & 51 & 56 \end{array} $$ Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers.

Step by step solution

01

Define the Problem

Determine if there is a significant difference between the hardness readings obtained using the steel-ball indenter and the diamond indenter by constructing a 95% confidence interval for the difference in means.

02

Calculate the Differences

First, calculate the differences between the hardness readings for each specimen. \( D = \text{Diamond} - \text{Steel} \)Specimen 1: \( 52 - 50 = 2 \)Specimen 2: \( 56 - 57 = -1 \)Specimen 3: \( 61 - 61 = 0 \)Specimen 4: \( 74 - 71 = 3 \)Specimen 5: \( 69 - 68 = 1 \)Specimen 6: \( 55 - 54 = 1 \)Specimen 7: \( 68 - 65 = 3 \)Specimen 8: \( 51 - 51 = 0 \)Specimen 9: \( 56 - 53 = 3 \)

03

Calculate the Mean and Standard Deviation of Differences

Next, calculate the mean and standard deviation of the differences.Mean of differences, \(\bar{D}\):\[\bar{D} = \frac{2 + (-1) + 0 + 3 + 1 + 1 + 3 + 0 + 3}{9} = \frac{12}{9} = 1.33 \]Standard deviation of differences, \( s_D \):\[s_D = \sqrt{\frac{\frac{1}{n - 1} ∑_{i=1}^{n}(D_i - \bar{D})^2}{}} \ = \sqrt{\frac{1}{9-1} ((2-1.33)^2 + (-1-1.33)^2 + (0-1.33)^2 + (3-1.33)^2 + (1-1.33)^2 + (1-1.33)^2 + (3-1.33)^2 + (0-1.33)^2 + (3-1.33)^2)} =\sqrt{\frac{1}{8}((0.67)^2 + (-2.33)^2 + (-1.33)^2 + (1.67)^2 + (-0.33)^2 + (-0.33)^2 + (1.67)^2 + (-1.33)^2 + (1.67)^2) }= 1.581 \]

04

Calculate the t-Value

For a 95% confidence interval with 8 degrees of freedom (n-1), using a t-table or calculator, the t-value is approximately 2.306.

05

Construct the Confidence Interval

Now, construct the 95% confidence interval for the mean difference:\[1.33 \pm t_{\alpha/2, =8} * \frac {s_D}\sqrt{n} \]\[1.33 \pm 2.306 * \frac {1.581}{\sqrt{9}} \ = \ 1.33 \pm 1.217 \ \]So, the confidence interval is:0.113 ≤ μD ≤ 2.547

06

Conclusion

Since the confidence interval (0.113, 2.547) does not include 0, there is sufficient evidence to suggest that there is a significant difference between the hardness readings obtained using the steel-ball indenter and the diamond indenter.

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Key Concepts

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

confidence interval

A confidence interval gives an estimated range of values which is likely to include an unknown population parameter. For instance, in our exercise, we aim to determine if there's a significant difference between the hardness readings of two indenters.
Using the samples from both indenters, we calculate a 95% confidence interval for the difference in their mean hardness readings. This interval helps quantify the precision of our estimate and indicates whether this difference is statistically meaningful.
In our case, the confidence interval was calculated as 0.113 to 2.547 for the mean difference in hardness readings.

mean difference

The mean difference is the average of the differences between paired observations. In our exercise, the differences are calculated by subtracting the steel-ball hardness readings from the diamond hardness readings for each specimen.
These differences are used to determine if there's a consistent difference in performance between the two indenters.
The mean difference in this exercise is calculated as 1.33. This shows, on average, how much the diamond indenter's reading differs from the steel-ball indenter's reading.

normal distribution

The normal distribution, also known as the Gaussian distribution, is a probability distribution that is symmetric about the mean. It's often used in statistics because many phenomena are approximately normally distributed.
In our exercise, the normality of the differences between the hardness readings was checked using a normal probability plot and boxplot. This step ensures that it's appropriate to apply the t-distribution for constructing the confidence interval.
Knowing that the data is approximately normally distributed allows for more accurate statistical inferences.

standard deviation

The standard deviation measures the amount of variation or dispersion in a set of values. In our exercise, we calculate the standard deviation of the differences in hardness readings to understand the variability in their measurements.
It is denoted by the symbol \(s_D\) and is calculated as 1.581 in our example.
A lower standard deviation would mean the hardness readings are closer to the mean difference, while a higher standard deviation indicates a wider spread of values.

t-distribution

The t-distribution is a type of probability distribution that is symmetric and bell-shaped, like the normal distribution, but has heavier tails. It's used when estimating population parameters when the sample size is small and the population standard deviation is unknown.
In our exercise, we use the t-distribution to calculate the critical value for constructing the confidence interval.
For a 95% confidence interval and 8 degrees of freedom, the critical t-value is approximately 2.306. This value is then used to find the range within which we expect the true mean difference to lie.