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Chapter 17: Problem 38

Pulling Apart Wood. How heavy a load (in pounds) is needed to pull apart pieces of Douglas fir 4 inches long and \(1.5\) inches square? \begin{tabular}{|l|l|l|l|l|l|} \hline Here are data from students doing a laboratory exercise: & \\ \hline 33,190 & 31,860 & 32,590 & 26,520 & 33,290 \\ \hline 32,320 & 33,020 & 32,030 & 30,460 & 32,700 \\ \hline 23,040 & 30,930 & 32,720 & 33,650 & 32,340 \\ \hline 24,050 & 30,170 & 31,300 & 28,730 & 31,920 \\ \hline \end{tabular} We are willing to regard the wood pieces prepared for the lab session as an SRS of all similar pieces of Douglas fir. Engineers also commonly assume that characteristics of materials vary Normally. Suppose that the strength of pieces of wood like these follows a Normal distribution, with standard deviation 3000 pounds. a. Is there statistically significant evidence at the \(\alpha=0.10\) level against the hypothesis that the mean is 32,500 pounds for the two-sided alternative? b. Is there statistically significant evidence at the \(\alpha=0.10\) level against the hypothesis that the mean is 31,500 pounds for the two-sided alternative?

Short Answer

Expert verified
a. Reject \(H_0\); b. Fail to reject \(H_0\).

Step by step solution

01

Find the Sample Mean

Calculate the sample mean of the given data. First, sum all the values: \(33,190 + 31,860 + 32,590 + 26,520 + 33,290 + 32,320 + 33,020 + 32,030 + 30,460 + 32,700 + 23,040 + 30,930 + 32,720 + 33,650 + 32,340 + 24,050 + 30,170 + 31,300 + 28,730 + 31,920 = 619860\). There are 20 observations, so divide the sum by 20 to find the mean: \(\bar{x} = \frac{619860}{20} = 30993\).
02

Hypothesis Testing for Part (a)

The null hypothesis is \(H_0: \mu = 32500\) and the alternative hypothesis is \(H_a: \mu eq 32500\). Use the formula for the z-test: \( z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \), where \(\sigma = 3000\) and \(n = 20\). Calculate: \( z = \frac{30993 - 32500}{3000/\sqrt{20}} = \frac{-1507}{670.82} \approx -2.25 \).
03

Decision for Part (a)

For a two-sided test at \(\alpha = 0.10\), the critical z-values are approximately \(\pm 1.645\). Since \(-2.25\) is outside this range, we reject \(H_0\). There is statistically significant evidence against the mean being 32,500 pounds.
04

Hypothesis Testing for Part (b)

The null hypothesis is \(H_0: \mu = 31500\) and the alternative hypothesis is \(H_a: \mu eq 31500\). Calculate the z-test: \( z = \frac{30993 - 31500}{3000/\sqrt{20}} = \frac{-507}{670.82} \approx -0.76 \).
05

Decision for Part (b)

For a two-sided test at \(\alpha = 0.10\), the critical z-values are also \(\pm 1.645\). Since \(-0.76\) falls within this range, we fail to reject \(H_0\). There is no statistically significant evidence against the mean being 31,500 pounds.

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Key Concepts

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

Normal Distribution
In statistics, many natural phenomena, including the strength of Douglas fir wood, are assumed to be normally distributed. A normal distribution is a bell-shaped curve that is symmetrical around the mean. It's characterized by two main parameters: the mean, denoted by \( \mu \), which determines the center of the distribution, and the standard deviation, symbolized by \( \sigma \), which describes the spread or variability of the data around the mean.

This distribution is incredibly important in hypothesis testing because most inferential statistics rely on the assumption of normality. The central limit theorem also supports this assumption by stating that, for a sufficiently large sample size, the distribution of the sample mean will be normal, regardless of the original distribution of the data.

In our exercise, we assume that the strength of the wood samples follows a normal distribution with a standard deviation of 3000 pounds. This allows us to use the z-test in hypothesis testing, which is a statistical method that requires normally distributed data when the population standard deviation is known.
Sample Mean
The sample mean, denoted by \( \bar{x} \), is a crucial statistic used in hypothesis testing. It is calculated by summing all the individual data points and dividing by the number of observations. In context, it provides an estimate of the central tendency for our sample of wood strengths.

The sample mean serves as the best point estimate for the true population mean \( \mu \), which is typically unknown. It is foundational in drawing conclusions about the population from which the sample was taken.

For the given exercise, the sample mean was calculated to be 30993 pounds. By comparing this value with hypothesized population means (such as 32500 or 31500 pounds), we can conduct tests to infer whether our sample provides evidence against the assumed population mean.
Two-sided Test
A two-sided test, also known as a two-tailed test, is used in hypothesis testing when we want to determine whether a sample mean is significantly different from a population mean in either direction.

This type of test is appropriate when deviations in both directions are important. Both potential outcomes (being greater than or less than the hypothesized mean) are considered.

In the context of our exercise, two-sided tests were used to evaluate hypotheses about the mean strength of Douglas fir wood. When testing for two different population means (32500 and 31500 pounds), the two-sided test helped us decide whether the actual strength significantly differs from these values on both the higher and lower side.
Significance Level
The significance level, often denoted by \( \alpha \), is the probability of rejecting a true null hypothesis. It defines the threshold for determining statistical significance. A common significance level used in many hypothesis tests is 0.05 or 5%. However, the exercise used an \( \alpha \) level of 0.10 or 10%.

A lower \( \alpha \) level will mean a more stringent test, reducing the chances of a Type I error (incorrectly rejecting the null hypothesis). Conversely, a higher \( \alpha \) level increases the chance of identifying statistically significant effects easily, even if it slightly increases the risk of a Type I error.

In the exercise, the 0.10 significance level determined the critical values for the z-test. For a two-sided test, this level translates to critical z-values of approximately 卤1.645. We used these critical values to make decisions about rejecting or failing to reject the null hypotheses for the wood strength tests at 32500 and 31500 pounds.

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Most popular questions from this chapter

Experiments on learning in animals sometimes measure how long it takes mice to find their way through a maze. The mean time is 18 seconds for one particular maze. A researcher thinks that a loud noise will cause the mice to complete the maze faster. She measures how long each of 10 mice takes with a noise as stimulus. The sample mean is \(x=16.5\) seconds. The null hypothesis for the test of significance is a. \(H_{0}: \mu=18\). b. \(H_{0}: \mu=16.5\). c. \(H_{0}: \mu<18\).

Testing a Random Number Generator. A random number generator is supposed to produce random numbers that are uniformly distributed on the interval from 0 to 1 . If this is true, the numbers generated come from a population with \(\mu=0.5\) and \(\sigma=0.2887\). A command to generate 100 random numbers gives outcomes with mean \(x=0.5635\). Assume that the population \(\sigma\) remains fixed. We want to test $$ \begin{aligned} &H_{0}: \mu=0.5 \\ &H_{a}: \mu \neq 0.5 \end{aligned} $$ a. Calculate the value of the \(z\) test statistic. b. Use Table C: Is \(z\) statistically significant at the \(5 \%\) level \((\alpha=0.05) ?\) c. Use Table C: Is \(z\) statistically significant at the \(1 \%\) level \((\alpha=0.01)\) ? d. Between which two Normal critical values \(z^{*}\) in the bottom row of Table \(C\) does \(z\) lie? Between what two numbers does the \(P\)-value lie? Does the test give good evidence against the null hypothesis?

5% Versus 1\%. Sketch the standard Normal curve for the \(z\) test statistic and mark off areas under the curve to show why a value of \(z\) that is statistically significant at the \(1 \%\) level in a one-sided test is always statistically significant at the \(5 \%\) level. If \(z\) is statistically significant at the \(5 \%\) level, what can you say about its significance at the \(1 \%\) level?

Hotel Managers' Personalities. Successful hotel managers must have personality characteristics traditionally stereotyped as feminine (such as "compassionate") as well as those traditionally stereotyped as masculine (such as "forceful"). The Bem Sex-Role Inventory (BSRI) is a personality test that gives separate ratings for "female" and "male" stereotypes, both on a scale of 1 to 7 . Although the BSRI was developed at a time when these stereotypes were more pronounced, it is still widely used to assess personality types. Unfortunately, the ratings are often referred to as femininity and masculinity scores. A sample of 148 male general managers of three-star and four-star hotels had mean BSRI femininity score \(x=5.29 .2\) The mean score for the general male population is \(\mu=5.19\). Do hotel managers, on the average, differ statistically significantly in femininity score from men in general? Assume that the standard deviation of scores in the population of all male hotel managers is the same as the \(\sigma=0.78\) for the adult male population. a. State null and alternative hypotheses in terms of the mean femininity score \(\mu\) for male hotel managers. b. Find the \(z\) test statistic. c. What is the \(P\)-value for your \(z\) ? What do you conclude about male hotel managers?

Researchers investigated the effectiveness of oral zinc, as compared to a placebo, in reducing the duration of the common cold when taken within 24 hours of the onset of symptoms. The researchers found that those taking oral zinc had a statistically significantly shorter duration \((P<0.05)\) than those taking the placebo. \(-\) This means that a. the probability that the null hypothesis is true is less than \(0.05\). b. the value of the test statistic, the mean reduction in duration of the cold, is large. c. neither of the above is true.
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