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91Ó°ÊÓ

A drug manufacturer forms tablets by compressing a granular material that contains the active ingredient and various fillers. The hardness of a sample from each batch of tablets produced is measured to control the compression process. The target value for the hardness is \(\mu=11.5 .\) The hardness data for a random sample of 20 tablets are \(\begin{array}{lllll}11.627 & 11.613 & 11.493 & 11.602 & 11.360 \\ 11.374 & 11.592 & 11.458 & 11.552 & 11.463 \\ 11.383 & 11.715 & 11.485 & 11.509 & 11.429 \\ 11.477 & 11.570 & 11.623 & 11.472 & 11.531\end{array}\) Is there convincing evidence at the \(5 \%\) level that the mean hardness of the tablets differs from the target value?

Short Answer

Expert verified
There is no convincing evidence at the 5% level that the mean hardness differs from 11.5.

Step by step solution

01

Define the Hypotheses

We need to determine if the mean tablet hardness significantly differs from the target value of 11.5. Here, the null hypothesis ("\(H_0\)") posits that the mean tablet hardness is equal to the target value, \(\mu = 11.5\). The alternative hypothesis ("\(H_a\)") suggests that the mean tablet hardness is different from the target value, \(\mu eq 11.5\).
02

Set the Significance Level

Determine the significance level for the test. The significance level (\(\alpha\)) is given as 0.05 or 5%. This will be used to evaluate the p-value from the hypothesis test.
03

Calculate Sample Mean and Standard Deviation

Calculate the sample mean (\(\bar{x}\)) and standard deviation (\(s\)) from the provided data. The sample mean is calculated as \(\bar{x} = \frac{\sum x_i}{n}\), where \(n = 20\), the number of tablets. Also calculate the standard deviation, \(s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}\).
04

Conduct the t-test

Perform a two-tailed t-test as the population standard deviation is unknown and the sample size is relatively small (n < 30). Use the formula for the t-statistic: \(t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}\), where \(\mu = 11.5\) is the target mean. This value will indicate how much the sample mean deviates from the target mean.
05

Determine Degrees of Freedom and Critical Value

Determine the degrees of freedom (\(df\)) which is \(n-1 = 19\). Use a t-distribution table to find the critical t-value for a two-tailed test at \(\alpha = 0.05\).
06

Compare t-Statistic with Critical Value

Compare the calculated t-statistic with the critical t-value from Step 5. If the absolute t-statistic is greater than the critical t-value, then reject the null hypothesis (\(H_0\)); otherwise, do not reject \(H_0\). Additionally, consider the p-value as an alternative means of determining statistical significance.

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Key Concepts

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

Understanding the t-test
A t-test is a statistical method used to determine if there is a significant difference between the means of two groups. In this exercise, the t-test helps us decide if the mean hardness of the tablets is significantly different from the target hardness of 11.5.

This test is particularly useful when you have a small sample size, like the 20 tablets in the problem. Because we're dealing with a sample mean and an unknown population standard deviation, the t-test is the appropriate choice here.
  • It compares the sample mean to a known value (the target, in this case).
  • It helps assess whether deviations are due to random chance or indicate a real difference.

In this exercise, we use a two-tailed t-test since we are interested in differences that could be either higher or lower than the target value of 11.5.
The Role of Significance Level
The significance level, denoted by \(\alpha\), is a crucial part of hypothesis testing. It represents the probability of rejecting the null hypothesis when it is actually true. For this exercise, the significance level is set at 5%, or \(\alpha = 0.05\).

This value sets the threshold for deciding whether the test results are statistically significant. Essentially, a 5% significance level indicates that we are willing to accept a 5% chance of incorrectly rejecting the null hypothesis.
  • A lower significance level means you require more evidence to reject the null hypothesis.
  • The level you choose can affect the results; common choices are 0.01, 0.05, or 0.10 depending on the field of study.

In this context, a p-value lower than 0.05 would mean there's convincing evidence that the mean hardness is not equal to 11.5.
Understanding Degrees of Freedom
Degrees of freedom (df) are an essential concept in statistics, especially in the context of a t-test. They refer to the number of values in the final calculation of a statistic that are free to vary.

In our exercise, the degrees of freedom for a one-sample t-test is calculated as \(\text{df} = n - 1\), where \(n\) is the sample size. So, with 20 tablets, the degrees of freedom would be 19.
  • Degrees of freedom helps in selecting the correct distribution curve from a statistical table.
  • It influences the shape of the t-distribution used to determine the critical value for the test.

Understanding degrees of freedom allows us to interpret the t-statistic correctly and helps in decision-making during hypothesis testing.
What is the Null Hypothesis?
The null hypothesis, often symbolized as \(H_0\), is a statement about a population parameter that serves as a hypothesis to be tested.

In the case of our drug manufacturer's tablet hardness test, the null hypothesis states that the mean hardness of the tablets is equal to the target hardness of 11.5. This provides a baseline or a claim that the test tries to challenge.
  • It is often a statement of no effect or no difference.
  • In hypothesis testing, the null hypothesis is assumed true until evidence suggests otherwise.

The goal is to determine whether the sample data provide enough evidence against the null hypothesis to support an alternative hypothesis (which in this case suggests that the mean tablet hardness is different from 11.5). If we reject the null hypothesis, it implies the sample provides sufficient evidence to support that there is a significant difference.

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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 appropriate hypotheses for the significance test are (a) \(H_{0}: \mu=18 ; H_{a}: \mu \neq 18\). (b) \(H_{0}: \mu=18 ; H_{a}: \mu>18\). (c) \(H_{0}: \mu<18 ; H_{a}: \mu=18\) (d) \(H_{0}: \mu=18 ; H_{a}: \mu<18\). (e) \(H_{0}: \bar{x}=18 ; H_{a}: \bar{x}<18\).

The French naturalist Count Buffon \((1707-1788)\) tossed a coin 4040 times. He got 2048 heads. That's a bit more than one-half. Is this evidence that Count Buffon's coin was not balanced? To find out, Luisa decides to perform a significance test. Unfortunately, she made a few errors along the way. Your job is to spot the mistakes and correct them. $$ \begin{array}{l} H_{0}: \mu>0.5 \\ H_{a}: \bar{x}=0.5 \end{array} $$ \(\bullet\quad\) \(10 \%: 4040(0.5)=2020\) and \(4040(1-0.5)=2020\) are both at least 10 . \(\bullet\quad\) Large Counts: There are at least 40,400 coins in the world. \(t=\frac{0.5-0.507}{\sqrt{\frac{0.5(0.5)}{4040}}}=-0.89 ; P\) -value \(=1-0.1867=0.8133\) Reject \(H_{0}\) because the \(P\) -value is so large and conclude that the coin is fair.

A manufacturer of compact discs (CDs) wants to be sure that their CDs will fit inside the plastic cases they have bought for packaging. Both the CDs and the cases are circular. According to the supplier, the plastic cases vary Normally with mean diameter \(\mu=4.2\) inches and standard deviation \(\sigma=0.05\) inches. The CD manufacturer decides to produce CDs with mean diameter \(\mu=4\) inches. Their diameters follow a Normal distribution with \(\sigma=0.1\) inches. (a) Let \(X=\) the diameter of a randomly selected \(\mathrm{CD}\) and \(Y=\) the diameter of a randomly selected case. Describe the shape, center, and spread of the distribution of the random variable \(X-Y\). What is the importance of this random variable to the CD manufacturer? (b) Compute the probability that a randomly selected CD will fit inside a randomly selected case. (c) The production process actually runs in batches of 100 CDs. If each of these CDs is paired with a randomly chosen plastic case, find the probability that all the CDs fit in their cases.

Vigorous exercise helps people live several years longer (on average). Whether mild activities like slow walking extend life is not clear. Suppose that the added life expectancy from regular slow walking is just 2 months. A statistical test is more likely to find a significant increase in mean life expectancy if (a) it is based on a very large random sample and a \(5 \%\) significance level is used. (b) it is based on a very large random sample and a \(1 \%\) significance level is used. (c) it is based on a very small random sample and a \(5 \%\) significance level is used. (d) it is based on a very small random sample and a \(1 \%\) significance level is used. (e) the size of the sample doesn't have any effect on the significance of the test.

Your company markets a computerized device for detecting high blood pressure. The device measures an individual's blood pressure once per hour at a randomly selected time throughout a 12 -hour period. Then it calculates the mean systolic (top number) pressure for the sample of measurements. Based on the sample results, the device determines whether there is convincing evidence that the individual's actual mean systolic pressure is greater than \(130 .\) If so, it recommends that the person seek medical attention. (a) State appropriate null and alternative hypotheses in this setting. Be sure to define your parameter. (b) Describe a Type I and a Type II error, and explain the consequences of each.

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