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The article "Compatibility of Outer and Fusible Interlining Fabrics in Tailored Garments (Textile Res. J., 1997: 137-142) gave the following observations on bending rigidity \((\mu \mathrm{N} \cdot \mathrm{m})\) for medium-quality fabric specimens, from which the accompanying MINITAB output was obtained: \(\begin{array}{rrrrrrrr}24.6 & 12.7 & 14.4 & 30.6 & 16.1 & 9.5 & 31.5 & 17.2 \\\ 46.9 & 68.3 & 30.8 & 116.7 & 39.5 & 73.8 & 80.6 & 20.3 \\ 25.8 & 30.9 & 39.2 & 36.8 & 46.6 & 15.6 & 32.3 & \end{array}\) Would you use a one-sample \(t\) confidence interval to estimate true average bending rigidity? Explain your reasoning.

Short Answer

Expert verified
Yes, a one-sample t confidence interval can be used, assuming the data is roughly normal or the sample size is close enough to 30.

Step by step solution

01

Check Data for Normality

First, verify if the distribution of bending rigidity is approximately normal, which is a prerequisite for using a t confidence interval. Given the sample size is moderately small (n = 24), we can check for normality by using graphical methods like a Q-Q plot or a normal probability plot, or run a statistical test like the Shapiro-Wilk test if available. Visual inspection can also help us gauge normality.
02

Evaluate Sample Size

Consider the sample size. The central limit theorem suggests that as the sample size increases, the sample mean will distributed approximately normally even if the original data is not. Generally, a sample size of 30 or more is considered sufficient for invoking the CLT, but here n = 24, which is slightly less than 30 but still reasonably close.
03

Consider Data Variability

Examine if there is significant variability in the data which may impact the normality assumption. Look at the range of data and standard deviation. A wide range and large standard deviation might indicate issues with normality and suggest a skewed distribution.
04

Conclusion

After observing the above parameters, determine that a one-sample t confidence interval can be used because the sample size is close to 30, and if data visualizations or tests do not show significant deviation from normality. The t distribution is robust to slight deviations from normality, especially with sample sizes around 25 or more.

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Key Concepts

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

t-distribution
The t-distribution is a probability distribution that is used to estimate population parameters when the sample size is small, and the population standard deviation is unknown. It's similar to the normal distribution but has heavier tails, meaning it is more spread out. This characteristic makes it a better fit for small sample sizes where the estimate of variability might be slightly off. This spread helps to account for uncertainty, giving wider confidence intervals when compared to the normal distribution. As the sample size increases, the t-distribution approaches a normal distribution.

In situations like estimating the true average bending rigidity of fabrics, where the sample size is not very large, a t-distribution offers a more accurate picture than a normal distribution. It helps in creating more accurate confidence intervals, thereby allowing for better statistical inference.
Confidence Interval
A confidence interval is a range of values, derived from a data sample, that is likely to contain the value of an unknown population parameter. When we say "a 95% confidence interval," we mean that if we were to take 100 different samples and compute a confidence interval for each sample, then approximately 95 of the 100 confidence intervals will contain the population mean.

The formula to calculate a confidence interval with a t-distribution is:\[\bar{x} \pm t_{(n-1, \alpha/2)} \times \left( \frac{s}{\sqrt{n}} \right)\]where \(\bar{x}\) is the sample mean, \(t_{(n-1, \alpha/2)}\) is the t-score from the t-distribution with \(n-1\) degrees of freedom, \(s\) is the sample standard deviation, and \(n\) is the sample size.

The confidence interval allows researchers to state the degree of certainty they have about their sample results being representative of the true population. For example, in estimating the fabric's bending rigidity, a confidence interval provides a range where the true average rigidity likely falls.
Normality Assumption
Normality assumption refers to the precondition that the data should follow a normal distribution to apply various statistical techniques accurately, especially when the sample size is small. This assumption is crucial for the validity of a t-confidence interval.

For small samples (like the sample size of 24 in the exercise), certain checks are vital to evaluate the normality assumption:
  • Visual methods: such as Q-Q plots or normal probability plots can be used. If the data points closely follow the reference line, the data is likely normal.
  • Statistical tests: Tests such as the Shapiro-Wilk test can statistically verify the normality assumption but should be used cautiously as they can be sensitive to large data sets.
A slight deviation from normality might be tolerable when using the t-distribution, given its robustness. Still, significant deviations may necessitate considering alternative approaches or transformations.
Sample Size
Sample size is a crucial consideration in statistical studies as it affects the accuracy and reliability of the results. In the context of statistical inference, a larger sample size offers:
  • More reliable estimates of population parameters.
  • Less variability in sample means, thus a tighter confidence interval.
  • Increased power to detect true effects or differences.
With a sample size of 24, as in the exercise, the sample is relatively small but close to the widely accepted threshold of 30. The Central Limit Theorem (CLT) tells us that as the sample size increases, the distribution of the sample mean approaches a normal distribution, even if the data itself is not perfectly normal.

For sample sizes around 25 or more, using the t-distribution for confidence intervals is generally considered acceptable, especially if the data doesn’t significantly deviate from normality. In practice, researchers often weigh these factors to decide the most suitable statistical method for their analysis.

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Most popular questions from this chapter

Each headlight on an automobile undergoing an annual vehicle inspection can be focused either too high \((H)\), too low \((L)\), or properly \((N)\). Checking the two headlights simultaneously (and not distinguishing between left and right) results in the six possible outcomes \(H H, L L, N N, H L, H N\), and \(L N\). If the probabilities (population proportions) for the single headlight focus direction are \(P(H)=\theta_{1}\), \(P(L)=\theta_{2}\), and \(P(N)=1-\theta_{1}-\theta_{2}\) and the two headlights are focused independently of each other, the probabilities of the six outcomes for a randomly selected car are the following: $$ \begin{aligned} &p_{1}=\theta_{1}^{2} \quad p_{2}=\theta_{2}^{2} \quad p_{3}=\left(1-\theta_{1}-\theta_{2}\right)^{2} \\ &p_{4}=2 \theta_{1} \theta_{2} \quad p_{5}=2 \theta_{1}\left(1-\theta_{1}-\theta_{2}\right) \\ &p_{6}=2 \theta_{2}\left(1-\theta_{1}-\theta_{2}\right) \end{aligned} $$ Use the accompanying data to test the null hypothesis $$ H_{0}: p_{1}=\pi_{1}\left(\theta_{1}, \theta_{2}\right), \ldots, p_{6}=\pi_{6}\left(\theta_{1}, \theta_{2}\right) $$ where the \(\pi_{i}\left(\theta_{1}, \theta_{2}\right)\) 's are given previously. \(\begin{array}{lllllll}\text { Outcome } & H H & L L & N N & H L & H N & L N \\\ \text { Frequency } & 49 & 26 & 14 & 20 & 53 & 38\end{array}\)

The article "Psychiatric and Alcoholic Admissions Do Not Occur Disproportionately Close to Patients' Birthdays"' (Psych. Rep., 1992: 944–946) focuses on the existence of any relationship between date of patient admission for treatment of alcoholism and patient's birthday. Assuming a 365day year (i.e., excluding leap year), in the absence of any relation, a patient's admission date is equally likely to be any one of the 365 possible days. The investigators established four different admission categories: (1) within 7 days of birthday, (2) between 8 and 30 days, inclusive, from the birthday, (3) between 31 and 90 days, inclusive, from the birthday, and (4) more than 90 days from the birthday. A sample of 200 patients gave observed frequencies of \(11,24,69\), and 96 for categories 1,2 , 3 , and 4, respectively. State and test the relevant hypotheses using a significance level of \(.01\).

Qualifications of male and female head and assistant college athletic coaches were compared in the article "Sex Bias and the Validity of Believed Differences Between Male and Female Interscholastic Athletic Coaches" (Res. Q. Exercise Sport, 1990: 259-267). Each person in random samples of 2225 male coaches and 1141 female coaches was classified according to number of years of coaching experience to obtain the accompanying two-way table. Is there enough evidence to conclude that the proportions falling into the experience categories are different for men and women? Use \(\alpha=.01\). $$ \begin{array}{lccccc} \hline & \multicolumn{5}{c}{\text { Years of Experience }} \\ \cline { 2 - 6 } \text { Gender } & \mathbf{1 - 3} & \mathbf{4 - 6} & \mathbf{7 - 9} & \mathbf{1 0 - 1 2} & \mathbf{1 3 +} \\ \hline \text { Male } & 202 & 369 & 482 & 361 & 811 \\ \text { Female } & 230 & 251 & 238 & 164 & 258 \\ \hline \end{array} $$

The article from which the data in Exercise 20 was obtained also gave the accompanying data on the composite mass/outer fabric mass ratio for highquality fabric specimens. \(\begin{array}{lllllll}1.15 & 1.40 & 1.34 & 1.29 & 1.36 & 1.26 & 1.22 \\ 1.40 & 1.29 & 1.41 & 1.32 & 1.34 & 1.26 & 1.36 \\ 1.36 & 1.30 & 1.28 & 1.45 & 1.29 & 1.28 & 1.38 \\ 1.55 & 1.46 & 1.32 & & & & \end{array}\) MINITAB gave \(r=.9852\) as the value of the Ryan- Joiner test statistic and reported that \(P\) value \(>.10\). Would you use the one-sample \(t\) test to test hypotheses about the value of the true average ratio? Why or why not?

Sorghum is an important cereal crop whose quality and appearance could be affected by the presence of pigments in the pericarp (the walls of the plant ovary). The article "A Genetic and Biochemical Study on Pericarp Pigments in a Cross Between Two Cultivars of Grain Sorghum, Sorghum Bicolor" (Heredity, 1976: 413-416) reports on an experiment that involved an initial cross between CK60 sorghum (an American variety with white seeds) and Abu Taima (an Ethiopian variety with yellow seeds) to produce plants with red seeds and then a self-cross of the red-seeded plants. According to genetic theory, this \(F_{2}\) cross should produce plants with red, yellow, or white seeds in the ratio \(9: 3: 4\). The data from the experiment follows; does the data confirm or contradict the genetic theory? Test at level \(.05\) using the \(P\)-value approach. \begin{tabular}{l|ccc} Seed Color & Red & Yellow & White \\ \hline Observed Frequency & 195 & 73 & 100 \end{tabular}

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