91Ó°ÊÓ

Use the data in the table below for sitting adult males and females (based on anthropometric survey data from Gordon, Churchill, et al.). These data are used often in the design of different seats, including aircraft seats, train seats, theater seats, and classroom seats. (Hint: Draw a graph in each case.) $$\begin{array}{|l|l|l|l|} \hline & \text { Mean } & \text { St. Dev. } & \text { Distribution } \\ \hline \text { Males } & 23.5 \mathrm{in} . & 1.1 \mathrm{in} . & \text { Normal } \\ \hline \text { Females } & 22.7 \mathrm{in} . & 1.0 \mathrm{in} . & \text { Normal } \\ \hline \end{array}$$ Significance Instead of using 0.05 for identifying significant values, use the criteria that a value \(x\) is significantly high if \(P(x \text { or greater) } \leq 0.01\) and a value is significantly low if \(P(x \text { or less }) \leq 0.01 .\) Find the back-to-knee lengths for males, separating significant values from those that are not significant. Using these criteria, is a male back-to- knee length of 26 in. significantly high?

Step by step solution

01

Understand the distribution and given data

The problem provides data for sitting adult males and describes the distribution as normal with a mean of 23.5 in. and a standard deviation of 1.1 in. We need to determine if a male back-to-knee length of 26 in. is significantly high using the criteria that a value is considered significantly high if the probability of observing such a value or greater is less than or equal to 0.01.

02

Standardize the value

Use the Z-score formula to standardize the value of 26 in. The Z-score formula is \[ Z = \frac{x - \mu}{\sigma} \] where \( x \) is the value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. Substituting the given values,\[ Z = \frac{26 - 23.5}{1.1} = \frac{2.5}{1.1} \approx 2.27 \]

03

Find the probability associated with the Z-score

Use the Z-score obtained in the previous step to find the probability from the standard normal distribution table. The Z-score of 2.27 corresponds to a cumulative probability of approximately 0.9884. This means that the probability of a back-to-knee length being less than 26 in. is 0.9884.

04

Calculate the probability for significant high

To find the probability of a back-to-knee length being 26 in. or greater, subtract the cumulative probability from 1. \[ P(X \geq 26) = 1 - P(X < 26) = 1 - 0.9884 = 0.0116 \] Compare this with the significance level of 0.01.

05

Determine significance

Since 0.0116 is greater than 0.01, the probability of having a back-to-knee length of 26 in. or greater is not less than or equal to 0.01. Therefore, 26 in. is not considered significantly high.

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 Distribution

The concept of a normal distribution is crucial in statistics. It is a type of continuous probability distribution that is symmetrical around its mean, indicating that data near the mean are more frequent in occurrence than data far from the mean. The shape of a normal distribution is often referred to as a 'bell curve'. For instance, in our exercise, both males and females' back-to-knee lengths follow a normal distribution. The mean (average) values are different for each group (23.5 inches for males and 22.7 inches for females), but they both have a similar bell-shaped curve. Understanding normal distribution helps us analyze probabilities and make inferences about data patterns.

Z-score Calculation

A Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It indicates how many standard deviations an element is from the mean. The formula for calculating the Z-score is: \[ Z = \frac{x - \mu}{\sigma} \] where \( x \) is the value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. For example, in our exercise, to determine if a back-to-knee length of 26 inches for males is significantly high, we calculate the Z-score. Substituting the given values: \[ Z = \frac{26 - 23.5}{1.1} = \frac{2.5}{1.1} \approx 2.27 \] This Z-score tells us that 26 inches is about 2.27 standard deviations away from the mean.

Significance Testing

Significance testing is a statistical technique used to identify whether a result is statistically significant. In our exercise, we use a significance level of 0.01 to determine if a back-to-knee length is significantly high or low. This means a value is significantly high if the probability of observing it or greater is less than or equal to 0.01. After calculating the Z-score, we check the cumulative probability from the standard normal distribution table. For example, a Z-score of 2.27 corresponds to a cumulative probability of approximately 0.9884. Thus, the probability of a back-to-knee length being 26 inches or greater is calculated as: \[ P(X \geq 26) = 1 - P(X < 26) = 1 - 0.9884 = 0.0116 \] Since 0.0116 is greater than 0.01, 26 inches is not considered significantly high.

Probability

Probability is the measure of the likelihood that an event will occur. In our exercise, determining the probability helps conclude whether a certain male back-to-knee length is significantly high. After finding the Z-score, we use the standard normal distribution table to find the cumulative probability associated with the Z-score. For example, a Z-score of 2.27 gives us a cumulative probability of 0.9884. This means there’s a 98.84% chance of a back-to-knee length being less than 26 inches. To find the probability of the length being 26 inches or more, we subtract this value from 1: \[ P(X \geq 26) = 1 - 0.9884 = 0.0116 \] This probability helps us in significance testing and making decisions based on statistical evidence.