/*! 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 21 The article from which the data ... [FREE SOLUTION] | 91Ó°ÊÓ

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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?

Short Answer

Expert verified
Yes, use the one-sample t-test since the data is approximately normally distributed.

Step by step solution

01

Understand the Given Data

We have a set of data points representing the composite mass/outer fabric mass ratio. This set consists of 27 values.
02

Analyze the Ryan-Joiner Test Result

The Ryan-Joiner test is similar to the Shapiro-Wilk test for normality. A Ryan-Joiner statistic of 0.9852 close to 1 and a P-value greater than 0.10 suggests that the data does not significantly deviate from normality. This means the data is approximately normally distributed.
03

Assess Appropriateness for One-Sample t-Test

The one-sample t-test requires the sample to be approximately normally distributed when the sample size is relatively small. Since the P-value from the Ryan-Joiner test is greater than 0.10, indicating normality, and given the sample size, it is appropriate to use the one-sample t-test.

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Key Concepts

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

One-Sample T-Test
The one-sample t-test is a statistical method used for determining if the mean of a single sample is significantly different from a known or hypothesized population mean. This test is useful when you want to compare sample data to a theoretical expectation and is most effective when the sample size is small or moderate. To perform a one-sample t-test, you need:
  • A sample of continuous data
  • A known value or hypothesis for the population mean
The null hypothesis (\( H_0 \)) typically asserts that the sample mean is equal to the population mean, while the alternative hypothesis (\( H_a \)) posits that they are not the same. The test calculates a t-statistic and compares it to a critical value from the t-distribution. If the t-statistic falls into the rejection region (determined by your chosen significance level), you reject the null hypothesis. Make sure your data meet the assumptions of normality if your sample size is small.
Normality Test
A normality test is a statistical process used to determine if a dataset is well-modeled by a normal distribution. For many statistical methods, including the one-sample t-test, an assumption of normality underpins the validity of conclusions. Common normality tests include:
  • Shapiro-Wilk Test
  • Kolmogorov-Smirnov Test
  • Ryan-Joiner Test
In this context, the Ryan-Joiner test is employed. It operates similarly to other normality tests by comparing the sample data to a normal distribution structure. If the test statistic is close to 1 and the p-value is high (like greater than 0.05 or 0.10), it suggests that the dataset does not significantly deviate from a normal distribution—indicating that your dataset is approximately normal.
Statistical Hypothesis Testing
Statistical hypothesis testing is a method of making decisions or inferences about population parameters based on sample data. It commonly involves testing an assumption called the null hypothesis (\( H_0 \)), against an alternative hypothesis (\( H_a \)). Key steps in hypothesis testing:
  • Define the null and alternative hypotheses
  • Collect data and perform a statistically valid test
  • Calculate a test statistic and p-value
  • Compare the p-value with a significance level (commonly 0.05)
  • Decide to reject or fail to reject the null hypothesis
The objective is to evaluate if the observed data can occur under the null hypothesis. A low p-value indicates that such an event under the null is rare, leading to a rejection of the null hypothesis. In contrast, a high p-value suggests consistency with the null hypothesis.
Minitab
Minitab is a powerful statistical software widely used for data analysis, especially useful for students and professionals in statistical fields. It offers a wide range of statistical tools and graphical displays for interpreting data efficiently. With Minitab, you can:
  • Perform various forms of hypothesis tests, including the one-sample t-test
  • Conduct normality tests like the Ryan-Joiner test
  • Visualize data through comprehensive charts and graphs
Minitab simplifies the complex statistical calculations and provides a user-friendly platform for performing advanced analyses. When working with sample data, Minitab helps determine statistical significance by computing necessary tests and p-values, which aids in informed decision-making for hypothesis testing.

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Most popular questions from this chapter

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}

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 \(\alpha=.01\). $$ \begin{array}{lccccc} \hline & \multicolumn{5}{c}{\text { Years of Experience }} \\ \cline { 2 - 6 } \text { Gender } & \mathbf{1 - 3} & \mathbf{4 - 6} & \mathbf{7 - 9} & \mathbf{1 0 - 1 2} & \mathbf{1 3 +} \\ \hline \text { Male } & 202 & 369 & 482 & 361 & 811 \\ \text { Female } & 230 & 251 & 238 & 164 & 258 \\ \hline \end{array} $$

An article in Annals of Mathematical Statistics reports the following data on the number of borers in each of 120 groups of borers. Does the Poisson pmf provide a plausible model for the distribution of the number of borers in a group? [Hint: Add the frequencies for \(7,8, \ldots\), 12 to establish a single category " \(\geq 7 . "]\) $$ \begin{array}{l|llllllllllllll} \begin{array}{l} \text { Number } \\ \text { of Borers } \end{array} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline \text { Frequency } & 24 & 16 & 16 & 18 & 15 & 9 & 6 & 5 & 3 & 4 & 3 & 0 & 1 \end{array} $$

Say as much as you can about the \(P\)-value for an upper-tailed chi-squared test in each of the following situations: a. \(\chi^{2}=7.5, \mathrm{df}=2\) b. \(\chi^{2}=13.0, \mathrm{df}=6\) c. \(\chi^{2}=18.0, \mathrm{df}=9\) d. \(\chi^{2}=21.3, k=5\) e. \(\chi^{2}=5.0, k=4\)

The article "Susceptibility of Mice to Audiogenic Seizure Is Increased by Handling Their Dams During Gestation" (Science, 1976: 427-428) reports on research into the effect of different injection treatments on the frequencies of audiogenic seizures. \begin{tabular}{lc|c|c|c} & \multicolumn{1}{c}{ No } & \multicolumn{1}{c}{ Wild } & Clonic & Tonic \\ Treatment & Response & \multicolumn{1}{c}{ Running } & \multicolumn{1}{c}{ Seizure } & \multicolumn{1}{c}{ Seizure } \\ \cline { 2 - 5 } Thienylalanine & 21 & 7 & 24 & 44 \\ \cline { 2 - 5 } Solvent & 15 & 14 & 20 & 54 \\ \cline { 2 - 5 } Sham & 23 & 10 & 23 & 48 \\ \cline { 2 - 5 } Unhandled & 47 & 13 & 28 & 32 \\ \cline { 2 - 5 } & & & & \end{tabular} Does the data suggest that the true percentages in the different response categories depend on the nature of the injection treatment? State and test the appropriate hypotheses using \(\alpha=.005\).

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