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According to the article "Fatigue Testing of Condoms" (Polymer Testing, 2009: 567-571), "tests currently used for condoms are surrogates for the challenges they face in use," including a test for holes, an inflation test, a package seal test, and tests of dimensions and lubricant quality (all fertile territory for the use of statistical methodology!). The investigators developed a new test that adds cyclic strain to a level well below breakage and determines the number of cycles to break. A sample of 20 condoms of one particular type resulted in a sample mean number of 1584 and a sample standard deviation of 607 . Calculate and interpret a confidence interval at the \(99 \%\) confidence level for the true average number of cycles to break. [Note: The article presented the results of hypothesis tests based on the \(t\) distribution; the validity of these depends on assuming normal population distributions.]

Short Answer

Expert verified
The 99% confidence interval for the true average is (1195.55, 1972.45) cycles.

Step by step solution

01

Identify the Parameters

We are given a sample mean \(\bar{x} = 1584\), a sample standard deviation \(s = 607\), a sample size \(n = 20\), and need a 99% confidence interval for the true mean. This implies we are using a \(t\)-distribution due to the small sample size.
02

Determine the Critical Value

The critical value for a 99% confidence interval can be found using a \(t\)-distribution table with \(n - 1 = 19\) degrees of freedom. The critical value \(t^*\) for 99% confidence and 19 degrees of freedom is approximately 2.861 (use a table or statistical software for this).
03

Calculate the Standard Error

Calculate the standard error of the mean (SE) using the formula: \( SE = \frac{s}{\sqrt{n}} = \frac{607}{\sqrt{20}} \approx 135.7585 \).
04

Compute the Margin of Error

The margin of error (ME) is found by multiplying the critical value by the standard error: \( ME = t^* \times SE = 2.861 \times 135.7585 \approx 388.45 \).
05

Calculate the Confidence Interval

Calculate the confidence interval using the sample mean and the margin of error: \( \bar{x} \pm ME = 1584 \pm 388.45 \). This gives us a confidence interval of \((1195.55, 1972.45)\).
06

Interpret the Confidence Interval

We can say with 99% confidence that the true average number of cycles to break is between 1195.55 and 1972.45 cycles for this type of condom.

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Key Concepts

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

t-distribution
When working with small sample sizes, typically less than 30, or when the population standard deviation is unknown, we rely on the t-distribution. Unlike the normal distribution, the t-distribution has thicker tails. This accounts for the increased variability that tends to occur with smaller samples.
It ensures a better approximation of the population mean, especially when dealing with unknown variances.
Some key points about the t-distribution:
  • The shape of the distribution gets closer to a normal distribution as the sample size increases.
  • It is defined by the degrees of freedom, which in most cases is the sample size minus one ( - 1).
  • This distribution is crucial for calculating confidence intervals and hypothesis testing in small samples.
margin of error
The margin of error represents the amount we're allowing our estimate to vary. It's essentially the buffer zone around the sample mean where we believe the true population mean may lie.
In any statistical estimate, especially for confidence intervals, the margin of error helps us understand the potential error between the sample mean and the true mean.
To calculate the margin of error (ME):
  • Identify the critical value from the t-distribution table that corresponds to your confidence level and degrees of freedom.
  • Multiply this critical value by the standard error of the sample mean.
This value helps tell you how much your sample average is likely to differ from the actual population average.
standard error
The standard error measures how far the sample mean of the data is likely to be from the true population mean. It provides a way to quantify the uncertainty in our estimate of the mean.
The smaller the standard error, the closer the sample mean is likely to be to the true population mean.
Here's how to compute the standard error (SE):
  • Take the sample standard deviation, which tells us how much the sample values deviate from the mean.
  • Divide this value by the square root of the sample size.
This calculation provides an adjusted standard deviation that accounts for sample size, making our estimate more reliable.
sample mean
The sample mean is a central measure of the data set, representing the average value of a sample. It's a crucial element when estimating population parameters.
Unlike the population mean, which considers every member of the population, the sample mean uses only a part of the population, making it a more feasible statistic to work with.
The importance of the sample mean:
  • It serves as an unbiased estimator of the population mean if the sample is random.
  • The sample mean is utilized in constructing confidence intervals, which help infer the population parameters.
Therefore, accurately calculating and understanding the sample mean is foundational for performing further statistical analyses.
degrees of freedom
Degrees of freedom might sound complex, but they simply refer to the number of values in a calculation that are free to vary. In statistical tests that involve estimating parameters, the degrees of freedom play an integral role.
For example, when computing the variance or standard deviation with a sample, your degrees of freedom is the sample size minus one ( - 1). This adjustment accounts for the fact we're using sample data to estimate a population parameter.
Key elements regarding degrees of freedom:
  • They adjust the shape of the t-distribution used in your analysis.
  • They help to ensure that statistical estimates are unbiased.
Understanding this concept is vital because it influences how critical values are determined when calculating confidence intervals and performing hypothesis tests.

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Most popular questions from this chapter

Suppose that a random sample of 50 bottles of a particular brand of cough syrup is selected and the alcohol content of each bottle is determined. Let \(\mu\) denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the resulting \(95 \%\) confidence interval is \((7.8,9.4)\). a. Would a \(90 \%\) confidence interval calculated from this same sample have been narrower or wider than the given interval? Explain your reasoning. b. Consider the following statement: There is a \(95 \%\) chance that \(\mu\) is between \(7.8\) and \(9.4\). Is this statement correct? Why or why not? c. Consider the following statement: We can be highly confident that \(95 \%\) of all bottles of this type of cough syrup have an alcohol content that is between \(7.8\) and 9.4. Is this statement correct? Why or why not? d. Consider the following statement: If the process of selecting a sample of size 50 and then computing the corresponding \(95 \%\) interval is repeated 100 times, 95 of the resulting intervals will include \(\mu\). Is this statement correct? Why or why not?

The article "Measuring and Understanding the Aging of Kraft Insulating Paper in Power Transformers" (IEEE Electrical Insul. Mago, 1996: 28-34) contained the following observations on degree of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range: \(\begin{array}{lllllll}418 & 421 & 421 & 422 & 425 & 427 & 431 \\ 434 & 437 & 439 & 446 & 447 & 448 & 453 \\ 454 & 463 & 465 & & & & \end{array}\) a. Construct a boxplot of the data and comment on any interesting features. b. Is it plausible that the given sample observations were selected from a normal distribution? c. Calculate a two-sided \(95 \%\) confidence interval for true average degree of polymerization (as did the authors of the article). Does the interval suggest that 440 is a plausible value for true average degree of polymerization? What about 450 ?

Determine the \(t\) critical value(s) that will capture the desired \(t\)-curve area in each of the following cases: a. Central area \(=.95\), df \(=10\) b. Central area \(=.95\), df \(=20\) c. Central area \(=.99\), df \(=20\) d. Central area \(=.99\), df \(=50\) e. Upper-tail area \(=.01\), df \(=25\) f. Lower-tail area \(=.025\), df \(=5\)

The following observations are lifetimes (days) subsequent to diagnosis for individuals suffering from blood cancer ("A Goodness of Fit Approach to the Class of Life Distributions with Unknown Age," Quality and Reliability Engr. Intl., 2012: 761-766): a. Can a confidence interval for true average lifetime be calculated without assuming anything about the nature of the lifetime distribution? Explain your reasoning. [Note: A normal probability plot of the data exhibits a reasonably linear pattern.] b. Calculate and interpret a confidence interval with a \(99 \%\) confidence level for true average lifetime. [Hint: \(\bar{x}=1191.6\) and \(s=506.6\).]

A sample of 56 research cotton samples resulted in a sample average percentage elongation of \(8.17\) and a sample standard deviation of \(1.42(*\) An Apparent Relation Between the Spiral Angle \(\phi\), the Percent Elongation \(E_{1}\), and the Dimensions of the Cotton Fiber," Textile Research \(J_{.}, 1978: 407-410\) ). Calculate a \(95 \%\) large-sample CI for the true average percentage elongation \(\mu\). What assumptions are you making about the distribution of percentage elongation?

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