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Chapter 16: Problem 569

All boxes of a particular type of detergent indicate that they contain 21 ounces of detergent. A government agency receives many consumer complaints that the boxes contain less than 21 ounces. To check the consumers' complaints at the \(5 \%\) level of significance, the government agency buys 100 boxes of this detergent and finds that \(\underline{\mathrm{X}}=20.5\) ounces and \(\mathrm{S}=2\) ounces. Should the government agency order the seller to put more detergent into its boxes?

Short Answer

Expert verified
There is sufficient evidence at the \(5\%\) level of significance to conclude that the detergent boxes contain less than 21 ounces. The government agency should order the seller to put more detergent into its boxes.

Step by step solution

01

Null hypothesis (H0) and alternative hypothesis (H1)

The null hypothesis states that the population mean (\(\mu\)) is equal to 21, and the alternative hypothesis states that the population mean is less than 21: H0: \(\mu = 21\) H1: \(\mu < 21\) #Step 2: Identify the test statistic and critical value#
02

t-test

Since we don't know the population standard deviation, we will use a t-test to perform the hypothesis test. The level of significance is \(\alpha = 0.05\), and the degree of freedom is \(df=n-1=100-1=99\). Find the critical value (t-critical) using a t-table or calculator. t-critical = -1.660 #Step 3: Calculate the sample t-statistic#
03

Formula to calculate t-statistic

t = \(\frac{\underline{\mathrm{X}} - \mu}{\frac{\mathrm{S}}{\sqrt{n}}}\) Insert the values \(\underline{\mathrm{X}}=20.5\), \(\mu = 21\), \(\mathrm{S}=2\), and \(n = 100\): t = \(\frac{20.5 - 21}{\frac{2}{\sqrt{100}}}\) #Step 4: Compute the t-statistic#
04

Calculate t-statistic

t = -2.50 #Step 5: Compare test statistic with critical value and make a decision#
05

Decision rule

Compare the test statistic to the critical value: t-statistic < t-critical -2.50 < -1.660 Since the t-statistic is less than the t-critical value, we reject the null hypothesis (H0) in favor of the alternative hypothesis (H1). #Step 6: Conclusion#
06

Interpretation of the hypothesis test

Since we rejected the null hypothesis (H0), there is sufficient evidence at the 5% level of significance to conclude that the detergent boxes contain less than 21 ounces. The government agency should order the seller to put more detergent into its boxes.

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Key Concepts

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

T-Test
The t-test is a statistical method used to determine if there is a significant difference between the means of two groups or to compare a sample mean to a known value, which is typically the population mean. It is most appropriate when the sample size is small and the population standard deviation is unknown. In the given exercise, the t-test is used to ascertain whether the average content of the detergent boxes is less than the advertised 21 ounces. As the sample standard deviation (S) is provided instead of the population standard deviation, the t-test is the suitable choice for analyzing the data. The computed t-statistic will be compared against a critical value to reach a conclusion.
Significance Level
The significance level, denoted by \( \alpha \), is the probability of rejecting the null hypothesis when it is actually true. Think of it as the threshold for considering the results statistically significant. In this scenario, the significance level is set at 5%, or \( \alpha = 0.05 \). If the probability of observing the sample results, or something more extreme, is less than 5% assuming the null hypothesis is true, then the null hypothesis can be rejected. This indicates that there is less than a 5% chance that the sample results are due to random variation alone.
Null Hypothesis
The null hypothesis, represented as \( H_0 \), is a statement that there is no effect or no difference, and it serves as the default assumption that a test seeks to challenge. In the given exercise, the null hypothesis is that the true mean weight of the detergent boxes is equal to the claimed 21 ounces (\( \mu = 21 \)). The null hypothesis is what we test against the alternative hypothesis, and it remains assumed true unless evidence suggests otherwise.
Alternative Hypothesis
The alternative hypothesis, denoted as \( H_1 \) or \( H_a \), represents the opposite claim of the null hypothesis and is what you suspect might be true instead. For our exercise, the alternative hypothesis posits that the detergent boxes contain less than the stated 21 ounces (\( \mu < 21 \)). If evidence strongly supports the alternative hypothesis, the null hypothesis can be rejected.
Critical Value
The critical value is a point on the distribution that is compared to the test statistic to determine whether to reject the null hypothesis. It represents the threshold beyond which the observed data is considered statistically significant. In the solution, the critical value (t-critical) is -1.660. This number comes from the t-distribution table which correlates to our test's significance level and degrees of freedom. If our calculated t-statistic is beyond this critical value, it indicates the results are not likely to have occurred by chance, and we may reject the null hypothesis.
Degrees of Freedom
Degrees of freedom is a concept tied to the estimation of population parameters and it accounts for the number of values in the final calculation of a statistic that are free to vary. Mathematically, for a single-sample t-test, the degrees of freedom (df) are calculated as the sample size (n) minus one. In our example, with 100 boxes of detergent, the degrees of freedom are \( n - 1 = 100 - 1 = 99 \). The degrees of freedom are used to determine the shape of the t-distribution which, in turn, helps to evaluate the critical value and p-value for the test.

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Most popular questions from this chapter

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