91Ó°ÊÓ

Have you ever been frustrated because you could not get a container of some sort to release the last bit of its contents? The article "Shake, Rattle, and Squeeze: How Much Is Left in That Container?" (Consumer Reports, May 2009: 8) reported on an investigation of this issue for various consumer products. Suppose five \(6.0 \mathrm{oz}\) tubes of toothpaste of a particular brand are randomly selected and squeezed until no more toothpaste will come out. Then each tube is cut open and the amount remaining is weighed, resulting in the following data (consistent with what the cited article reported): \(.53, .65, .46, .50, .37\). Does it appear that the true average amount left is less than \(10 \%\) of the advertised net contents? a. Check the validity of any assumptions necessary for testing the appropriate hypotheses. b. Carry out a test of the appropriate hypotheses using a significance level of \(.05\). Would your conclusion change if a significance level of .01 had been used? c. Describe in context type I and II errors, and say which error might have been made in reaching a conclusion.

Step by step solution

01

Calculate the Sample Mean and Standard Deviation

First, we need to calculate the sample mean \( \bar{x} \) and the sample standard deviation \( s \) for the data set. The data is: \( 0.53, 0.65, 0.46, 0.50, 0.37 \). The sample mean \( \bar{x} \) is given by:\[ \bar{x} = \frac{0.53 + 0.65 + 0.46 + 0.50 + 0.37}{5} = 0.502 \]To find the sample standard deviation \( s \), use the formula:\[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \]Calculating this, we get:\[ s \approx 0.1042 \].

02

Formulate Hypotheses

We need to formulate the null hypothesis \( H_0 \) and the alternative hypothesis \( H_a \). The claim is that the true average amount left is less than 10% of the tube's contents (0.6 oz). Thus,- \( H_0: \mu = 0.6 \) (the average amount left is 10% or more).- \( H_a: \mu < 0.6 \) (the average amount left is less than 10%).

03

Perform Hypothesis Test

We will use a one-sample t-test for the mean. The test statistic \( t \) is calculated using\[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \]where \( \mu_0 = 0.6 \), \( \bar{x} = 0.502 \), \( s = 0.1042 \), and \( n = 5 \).Substitute the values:\[ t = \frac{0.502 - 0.6}{0.1042/\sqrt{5}} \approx -2.096 \].Now, compare this with the critical value for \( df = 4 \) at \( \alpha = 0.05 \), which is approximately -2.132.

04

Conclusion from Hypothesis Test

Since the calculated t-value (-2.096) is greater than the critical t-value (-2.132), we do not reject the null hypothesis at \( \alpha = 0.05 \). For \( \alpha = 0.01 \), the critical value for \( df = 4 \) is about -3.747, and our test statistic (-2.096) is still greater than this, so again we would not reject \( H_0 \). Thus, using either significance level, there is not enough evidence to conclude that the true average amount left is less than 10%.

05

Describe Type I and Type II Errors and Conclusion Error Possibility

A Type I error would mean rejecting the null hypothesis when it is true, implying we believe less than 10% is left when in fact 10% or more might be left. A Type II error would mean failing to reject the null hypothesis when it is false, implying we believe 10% or more is left when less than 10% actually is. In our conclusion, given we did not reject the null, a Type II error might have been made if indeed the true average is less than 10%.

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Key Concepts

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

Type I and Type II Errors

Type I and Type II errors are essential concepts in hypothesis testing, relating to the accuracy of the test decisions. They help us understand the risks involved when we are making conclusions about our hypotheses.

A **Type I error** happens when we incorrectly reject a true null hypothesis. In simpler terms, it's when we think there is an effect when none actually exists. For example, in this toothpaste case, it is when we believe that the average leftover toothpaste is less than 10% when, in fact, it is not.

Conversely, a **Type II error** occurs when we fail to reject a false null hypothesis, meaning we miss detecting an effect that is there. In our earlier example, this would be concluding that no less than 10% remains when, in fact, the leftover amount is lesser.

• Type I error is often denoted by the symbol \( \alpha \), and is closely linked to the significance level.
• Type II error is represented by \( \beta \), and its complement (1-\( \beta \)) is called "power of the test".

In our scenario, given the decision was to not reject the null hypothesis, a Type II error might have been made, assuming the true average is indeed less than 10%.

One-Sample t-test

The one-sample t-test is a statistical test used when we want to determine if a sample mean is significantly different from a known or hypothesized population mean. It is most suitable when we have a small sample size and do not know the population standard deviation.

In the toothpaste problem, the **one-sample t-test** helps compare the sample mean of leftover toothpaste to the hypothesized mean of 10% or 0.6 oz. Here’s a simple look at the process:

• First, establish the null hypothesis \( H_0: \mu = 0.6 \) and the alternative hypothesis \( H_a: \mu < 0.6 \).
• Calculate the test statistic using the formula:
  \[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \]

The t-test gives us a t-value which we then compare against a critical value from the t-distribution table, considering our sample size and significance level.

This comparison helps us decide whether we can reject the null hypothesis, leading us to the conclusion about the true average left.

Sample Mean and Standard Deviation

The concepts of sample mean and standard deviation form the foundation of many statistical methods, including our hypothesis test. Understanding these helps in accurately assessing the data we have collected.

The **sample mean** is the average of the sample data. It provides a central tendency, representing a number that summarizes all data points. For the toothpaste residual problem, the sample mean was calculated as
\[ \bar{x} = \frac{0.53 + 0.65 + 0.46 + 0.50 + 0.37}{5} = 0.502 \]

This value tells us what the average leftover toothpaste is for the sampled tubes.

**Sample standard deviation**, on the other hand, measures variability around the sample mean. It indicates how much the individual data points fluctuate from the mean. The formula used is:
\[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \]
For our sample, this was calculated to be approximately 0.1042.

• Sample mean provides a summary of data.
• Standard deviation measures spread around the mean.

Both statistics are critical for conducting a meaningful t-test.

Significance Level

The significance level, often denoted as \( \alpha \), is a threshold set by the researcher which determines when to reject the null hypothesis. It represents the probability of making a Type I error.

For example, a significance level of \( 0.05 \) means there is a 5% risk of concluding that an effect exists when it actually does not. The toothpaste study used a significance level of 0.05 to decide the confidence we have in rejecting or not rejecting the null hypothesis.

• If \( \alpha = 0.05 \), we reject the null hypothesis if the p-value is less than 0.05, indicating strong evidence against \( H_0 \).
• If \( \alpha = 0.01 \), we require even stronger evidence (p-value < 0.01) to reject \( H_0 \).

Changing \( \alpha \) affects the critical value used in tests like the t-test, impacting whether conclusions are drawn about hypotheses. A higher significance level (e.g., 0.10) increases Type I error risk but more easily detects an effect, while a lower one (e.g., 0.01) decreases this risk but needs more substantial evidence to reject the null hypothesis.