/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 Fusible interlinings are being u... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Fusible interlinings are being used with increasing frequency to support outer fabrics and improve the shape and drape of various pieces of clothing. The article "Compatibility of Outer and Fusible Interlining Fabrics in Tailored Garments" (Textile Res. J., 1997: \(137-142\) ) gave the accompanying data on extensibility (\%) at \(100 \mathrm{gm} / \mathrm{cm}\) for both high-quality (H) fabric and poor-quality (P) fabric specimens. \(\begin{array}{rrrrrrrrrr}\mathrm{H} & 1.2 & .9 & .7 & 1.0 & 1.7 & 1.7 & 1.1 & .9 & 1.7 \\ & 1.9 & 1.3 & 2.1 & 1.6 & 1.8 & 1.4 & 1.3 & 1.9 & 1.6 \\ & .8 & 2.0 & 1.7 & 1.6 & 2.3 & 2.0 & & & \\ \mathrm{P} & 1.6 & 1.5 & 1.1 & 2.1 & 1.5 & 1.3 & 1.0 & 2.6 & \end{array}\) a. Construct normal probability plots to verify the plausibility of both samples having been selected from normal population distributions. b. Construct a comparative boxplot. Does it suggest that there is a difference between true average extensibility for high-quality fabric specimens and that for poor-quality specimens? c. The sample mean and standard deviation for the highquality sample are \(1.508\) and \(.444\), respectively, and those for the poor-quality sample are \(1.588\) and \(.530\). Use the two-sample \(t\) test to decide whether true average extensibility differs for the two types of fabric.

Short Answer

Expert verified
Construct plots and perform a t-test: normality is plausible; boxplot and t-test suggest no significant difference between fabrics.

Step by step solution

01

Calculate Probability Plot Data for H

To construct the normal probability plot for the high-quality fabric, arrange the data in increasing order: \(0.7, 0.8, 0.9, 0.9, 1.0, 1.1, 1.3, 1.3, 1.4, 1.6, 1.6, 1.6, 1.7, 1.7, 1.7, 1.9, 1.9, 2.0, 2.1, 2.3\). Determine the percentiles and corresponding z-scores for these ordered data points, which act as the theoretical quantiles. Plot these data points against the z-scores to form the plot.
02

Calculate Probability Plot Data for P

For the poor-quality fabric, order the data: \(1.0, 1.1, 1.3, 1.5, 1.5, 1.6, 2.1, 2.6\). Calculate the percentiles and match them with their z-scores. Plot these against the data to create the normal probability plot for the poor-quality fabric.
03

Interpret Probability Plots

Look for linearity in both probability plots. If the points roughly follow a straight line, it indicates that the sample data plausibly comes from a normally distributed population. Examine the residuals and outliers indicated in the plots to determine normality.
04

Construct Comparative Boxplot

Draw a boxplot for each fabric type using their data. Display the minimum, first quartile, median, third quartile, and maximum for both the high-quality and poor-quality fabrics. Investigate the boxplot for differences in median lines, spread (interquartile range), and outliers.
05

Interpret Boxplot Comparison

Compare the two boxplots to assess differences in median, spread, and overall distribution shape. Determine if there is a significant visual difference in these aspects between the high-quality and poor-quality samples.
06

Formulate Hypotheses for Two-sample t-test

Define the null hypothesis \(H_0: \mu_H = \mu_P\) (no difference in average extensibility) and the alternative hypothesis \(H_a: \mu_H eq \mu_P\) (there is a difference).
07

Calculate t-test Statistics

Use the formula for the t-statistic: \[ t = \frac{\bar{x}_H - \bar{x}_P}{\sqrt{\frac{s_H^2}{n_H} + \frac{s_P^2}{n_P}}} \]where \(\bar{x}_H = 1.508\), \(s_H = 0.444\), \(n_H = 20\), \(\bar{x}_P = 1.588\), \(s_P = 0.530\), and \(n_P = 8\) are means, standard deviations, and sample sizes for the fabric types. Calculate \(t\).
08

Determine Critical Value and Conclusion

With the degrees of freedom found using a formula or calculator, find the critical value for the two-tailed test at a typical significance level (such as 0.05). Compare the calculated t-statistic to this critical value. Reject the null hypothesis if the t-statistic exceeds the critical value, suggesting a significant difference in means.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Normal Probability Plot
A normal probability plot is a graphical tool used in statistics to assess whether a set of data is approximately normally distributed. To create a normal probability plot, the data must first be organized in ascending order. The data points are then plotted against a corresponding set of theoretical quantiles, usually derived from a standard normal distribution.

The key to interpreting a normal probability plot is checking the linearity. - If the points in the plot closely follow a straight line, this suggests that the data might come from a normal distribution. - Deviations from a straight line can indicate departures from normality, such as skewness, heavy tails, or outliers.

For the fabrics in this example, creating normal probability plots helps verify whether the high-quality and poor-quality fabric data come from normal populations. To achieve this, the z-scores (standard normal deviates) are paired with the ordered data values. Understanding and applying this method properly equips students to establish normalcy in other datasets beyond textile studies.
Two-Sample t Test
The two-sample t test is a statistical method used to compare the means of two independent groups and decide if they statistically differ from each other. This test is particularly useful for determining if differences in sample means reflect real effects in the population.

In this exercise, we aim to assess whether the average extensibility of high-quality fabric is different from that of poor-quality fabric. - **Null Hypothesis (H_0):** No difference in the means (ar{x}_H = ar{x}_P).- **Alternative Hypothesis (H_a):** There is a difference (ar{x}_H neq ar{x}_P).

The formula for the t-statistic involves comparing the difference in sample means to a weighted measure of variability across the samples. The t-statistic is calculated as\[t = \frac{\bar{x}_H - \bar{x}_P}{\sqrt{\frac{s_H^2}{n_H} + \frac{s_P^2}{n_P}}}\]where \(\bar{x}\) is the sample mean, \(s\) is the standard deviation, and \(n\) is the sample size.

After calculating the t-statistic, it is compared to a critical value derived from the t-distribution. If the absolute value of the t-statistic is greater than the critical value for the chosen significance level (like 0.05), the null hypothesis is rejected, indicating a significant difference between group means.
Comparative Boxplot
A comparative boxplot is a convenient graphical method to visualize and compare the distributions of two or more groups side by side. It displays summary statistics such as median, quartiles, and potential outliers, offering a snapshot of data distribution.

In the context of fabric extensibility: - **Box**: Represents the interquartile range (IQR) where the central 50% of the data lies. - **Median Line**: Depicted inside the box, indicating the central tendency. - **Whiskers**: Extend to the minimum and maximum data points within 1.5 times the IQR from the lower and upper quartiles. - **Outliers**: Points outside the whiskers are often marked as potential outliers.

When constructing a comparative boxplot for high-quality and poor-quality fabrics, looking at median alignment and variability is crucial. - Differences in the median position signify potential shifts in central tendency. - Varying box or whisker lengths suggest differences in data spread.

In this statistics exercise, understanding comparative boxplots helps deduce if there is a visible difference in extensibility between fabric types, even before statistical testing.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A study was carried out to compare two different methods, injection and nasal spray, for administering flu vaccine to children under the age of 5 . All 8000 children in the study were given both an injection and a spray. However, the vaccine given to 4000 of the children actually contained just saltwater, and the spray given to the other 4000 children also contained just saltwater. At the end of the flu season, it was determined that \(3.9 \%\) of the children who received the real vaccine via nasal spray contracted the flu, whereas \(8.6 \%\) of the 4000 children receiving the real vaccine via injection contracted the flu. a. Why do you think each child received both an injection and a spray? b. Does one method for delivering the vaccine appear to be superior to the other? Test the appropriate hypotheses. [Note: The study was described in the article "Spray Flu Vaccine May Work Better Than Injections for Tots," San Luis Obispo Tribune, May 2, 2006.]

Toxaphene is an insecticide that has been identified as a pollutant in the Great Lakes ecosystem. To investigate the effect of toxaphene exposure on animals, groups of rats were given toxaphene in their diet. The article "Reproduction Study of Toxaphene in the Rat"" \((J\). of Emiron. Sei. Health, 1988: 101-126) reports weight gains (in grams) for rats given a low dose (4 ppm) and for control rats whose diet did not include the insecticide. The sample standard deviation for 23 female control rats was \(32 \mathrm{~g}\) and for 20 female low-dose rats was \(54 \mathrm{~g}\). Does this data suggest that there is more variability in low-dose weight gains than in control weight gains? Assuming normality, carry out a test of hypotheses at significance level .05.

Sometimes experiments involving success or failure responses are run in a paired or before/after manner. Suppose that before a major policy speech by a political candidate, \(n\) individuals are selected and asked whether \((S)\) or not \((F)\) they favor the candidate. Then after the speech the same \(n\) people are asked the same question. The responses can be entered in a table as follows: where \(x_{1}+x_{2}+x_{3}+x_{4}=n\). Let \(p_{1}, p_{2}, p_{3}\), and \(p_{4}\) denote the four cell probabilities, so that \(p_{1}=P(S\) before and \(S\) after), and so on. We wish to test the hypothesis that the true proportion of supporters \((S)\) after the speech has not increased against the alternative that it has increased. a. State the two hypotheses of interest in terms of \(p_{1}, p_{2}\), \(p_{3}\), and \(p_{4}\). b. Construct an estimator for the after/before difference in success probabilities. c. When \(n\) is large, it can be shown that the rv \(\left(X_{i}-X_{j}\right) / n\) has approximately a normal distribution with variance given by \(\left[p_{i}+p_{j}-\left(p_{i}-p_{j}\right)^{2}\right] / n\). Use this to construct a test statistic with approximately a standard normal distribution when \(H_{0}\) is true (the result is called McNemar's test). d. If \(x_{1}=350, x_{2}=150, x_{3}=200\), and \(x_{4}=300\), what do you conclude?

The National Health Statistics Reports dated Oct. \(22 .\) 2008 , included the following information on the heights (in.) for non-Hispanic white females: \begin{tabular}{lccc} Age & Sample Size & Sample Mean & Std. Error Mean \\ \hline \(20-39\) & 866 & \(64.9\) & \(.09\) \\ 60 and older & 934 & \(63.1\) & \(.11\) \\ \hline \end{tabular} a. Calculate and interpret a confidence interval at confidence level approximately \(95 \%\) for the difference between population mean height for the younger women and that for the older women. b. Let \(\mu_{1}\) denote the population mean height for those aged 20-39 and \(\mu_{2}\) denote the population mean height for those aged 60 and older. Interpret the hypotheses \(H_{0}: \mu_{1}-\mu_{2}=1\) and \(H_{a}: \mu_{1}-\mu_{2}>1\), c. Based on the \(P\)-value calculated in (b) would you reject the null hypothesis at any reasonable significance level? Explain your reasoning. d. What hypotheses would be appropriate if \(\mu_{1}\) referred to the older age group, \(\mu_{2}\) to the younger age group, and you wanted to see if there was compelling evidence for concluding that the population mean height for younger women exceeded that for older women by more than 1 in.?

A mechanical engineer wishes to compare strength properties of steel beams with similar beams made with a particular alloy. The same number of beams, \(n\), of each type will be tested. Each beam will be set in a horizontal position with a support on each end, a force of \(2500 \mathrm{lb}\) will be applied at the center, and the deflection will be measured. From past experience with such beams, the engineer is willing to assume that the true standard deviation of deflection for both types of beam is .05 in. Because the alloy is more expensive, the engineer wishes to test at level .01 whether it has smaller average deflection than the steel beam. What value of \(n\) is appropriate if the desired type II error probability is \(.05\) when the difference in true average deflection favors the alloy by \(.04\) in.?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.