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Chapter 9: Problem 36

Two kinds of thread are being compared for strength. Fifty pieces of each type of thread are tested under similar conditions. Brand \(A\) had an average: tensile strength of 78.3 kilograms with a standard deviation of 5.6 kilograms, while brand \(B\) had an average tensile strength of 87.2 kilograms with a standard deviation of 6.3 kilograms. Construct a \(95 \%\) confidence interval for the difference of the population means.

Short Answer

Expert verified
The 95% confidence interval for the difference in average tensile strengths of the thread brands A and B is (-11.25 kg, -6.55 kg).

Step by step solution

01

Calculate the difference of sample means

Firstly, we need to calculate the difference of the sample means. In this case, our samples are the tensile strength of brands A and B. Hence, \(difference = \mu_{A} - \mu_{B} = 78.3 kg - 87.2 kg = -8.9 kg\)
02

Compute the standard errors

Next, we calculate the standard error for each of the samples. The formula for standard error SE is given by \(SE = \sqrt{\sigma_{a}^{2}/n_{a} + \sigma_{b}^{2}/n_{b}}\), where \(\sigma_{a}\) and \(\sigma_{b}\) are the standard deviations of samples a and b, and \(n_{a}\) and \(n_{b}\) are the sizes of the samples a and b respectively. Substituting the given values into the formula: \(SE = \sqrt{(5.6 kg)^{2}/50 + (6.3 kg)^{2}/50} = 1.2012 kg\)
03

Calculate the 95% Confidence Interval

The 95% confidence interval for the difference in population means is given by the formula: \(CI = difference \pm Z * SE\), where Z is the z-score that corresponds to the desired confidence level (which is 1.96 for 95% confidence level). Substituting the calculated values in the formula we get: \(CI = -8.9 kg \pm 1.96 * 1.2012 kg = -8.9 kg \pm 2.352 kg\). This gives us the confidence interval of (-11.25 kg, -6.55 kg).
04

Interpret the Confidence Interval

This confidence interval tells us that, with 95% confidence, the true difference in tensile strength between thread A and B lies between -11.25 kg and -6.55 kg. Note that the negative value indicates that thread B is stronger.

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Key Concepts

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

Tensile Strength
Tensile strength is the measurement of how much pulling force a material can withstand before breaking or snapping.
It is an important property for materials that are used in construction, manufacturing, and other applications where mechanical stress is common.
In this exercise, tensile strength is used to compare two brands of thread to determine which one can handle more stress.
  • Brand A has an average tensile strength of 78.3 kilograms, while
  • Brand B has a higher average tensile strength of 87.2 kilograms.
The higher the tensile strength, the stronger the thread, in this case.
By comparing these values, we understand the capability of each brand under similar testing conditions.
Standard Error
The standard error (SE) is a statistical measure that estimates the accuracy with which a sample mean represents the population mean.
The smaller the standard error, the more representative the sample mean is of the actual population mean.
In this exercise, the standard error helps in understanding the variability of the tensile strength between different samples of thread.
  • The formula to calculate the standard error for these threads involves the standard deviations and sample sizes of both brands.
  • For Brand A, the standard deviation is 5.6 kilograms, and for Brand B, it's 6.3 kilograms.
The calculated standard error of 1.2012 kilograms gives insight into the uncertainty associated with the mean difference between the two brands.
This helps in constructing a reliable confidence interval for the difference in tensile strength.
Z-score
The Z-score is a critical value that represents how many standard deviations an element is from the mean.
In the context of confidence intervals, the Z-score helps determine the margin of error.
For a 95% confidence level, the Z-score is usually 1.96.
  • A Z-score allows us to calculate how many standard errors to add or subtract from the sample mean to generate a confidence interval.
  • In this exercise, the Z-score helps to set the range within which we can be 95% confident the true difference in means lies.
By applying the Z-score to the standard error, we amplify the range of possible means, ensuring that the interval accounts for sampling variability.
Difference of Means
The difference of means is a simple calculation that involves subtracting one mean from another.
It quantifies how much two averages differ.
In this problem, we focus on the difference in tensile strength between two brands of thread.
  • The calculation shows that Brand A's average tensile strength is 8.9 kilograms less than Brand B's.
  • This difference ( |-8.9 kg| represents the more robust performance of Brand B when compared under similar test conditions.
Understanding this difference is crucial for statistical analyses that require comparing two or more samples.
It is also the basis for constructing the confidence interval, giving us a range of potential values for the true population difference.

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Most popular questions from this chapter

Consider two estimators of \(a^{2}\) in a sample \(x_{1}, x_{2}, \ldots,\) which is drawn from a normal distribution with mean \(\mu\) and variance \(a^{2} .\) The estimators are the unbiased estimator \(\left.s^{2}=\frac{1}{n-1} \sum_{1=1}^{n} x,-\vec{x}\right)^{2},\) and the maximum likelihood estimator \(\widehat{\mathrm{CT}}^{2}=\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}\) Discuss the variance properties of these two estimators.

A group of human factor researchers are concerned about reaction to a stimulus by airplane pilots with a certain cockpit arrangement. An experiment was conducted in a simulation laboratory and 15 pilots were used with average reaction time 3.2 seconds and sample standard deviation 0.6 seconds, It is of interest to characterize extremes (i.e., worst case scenario). To that end, answer the following: (a) Give a particular important one-sided \(99 \%\) confidence bound on the mean reaction time. What assumption, if any, must you make on the distribution of reaction time? (b) Give a \(99 \%\) one-sided prediction interval and give an interpretation of what it means. Must you make an assumption on the distribution of reaction time to compute this bound? (c) Compute a one-sided tolerance bound with \(99 \%\) confidence that involves \(95 \%\) of reaction times. Again, give interpretation and assumption on distribution if any. [Note: The one-sided tolerance limit values are also included in Table A.7].

According to USA Today (March 17. \(\lfloor 997\) ), women made up \(33.7 \%\) of the editorial staff at local TV stations in 1990 and \(36.2 \%\) in \(1994 .\) Assume 20 new employees were: hired as editorial staff. (a) Estimate the number that would have been women in each year, respectively, (b) Compute a \(95 \%\) confidence interval to see if there is evidence that the proportion of women hired as editorial staff in 1994 was higher than the proportion hired in 1990 .

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A random sample of 20 students obtained a mean of \(\bar{x}=72\) and a variance of \(s \sim^{2}=16\) on a college placement test in mathematics. Assuming the scores to be normally distributed, construct a \(98 \%\) confidence interval for \(\sigma^{2}\).
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