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Ground beef is packaged in small trays, intended to hold 1 pound of meat. A random sample of 35 packages in the small tray produced weight measurements with an average of 1.01 pounds and a standard deviation of .18 pound. a. If you were the quality control manager and wanted to make sure that the average amount of ground beef was indeed 1 pound, what hypotheses would you test? b. Find the \(p\) -value for the test and use it to perform the test in part a. c. How would you, as the quality control manager, report the results of your study to a consumer interest group?

Short Answer

Expert verified
a. \(H_0\): The average weight of the ground beef packages is 1 pound (\(\mu = 1\)). b. \(H_a\): The average weight of the ground beef packages is not 1 pound (\(\mu \neq 1\)).

Step by step solution

01

Setting up the hypotheses

First, we need to set up the null hypothesis (\(H_0\)) and alternative hypothesis (\(H_a\)). \(H_0\): The average weight of the ground beef packages is 1 pound, i.e., \(\mu = 1\). \(H_a\): The average weight of the ground beef packages is not 1 pound, i.e., \(\mu \neq 1\).
02

Calculate the test statistic

Next, we need to calculate the test statistic. Since we have a small sample size and don't know the population standard deviation, we will use the t-statistic. The formula for the t-statistic is: \[t = \frac{(\bar{x} - \mu_0)}{s / \sqrt{n}}\] Here, \(\bar{x}\) is the sample mean, \(\mu_0\) is the hypothesized population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size. Plugging in the values, we get: \(t = \frac{(1.01 - 1)}{0.18 / \sqrt{35}}\) Calculate the t-statistic value.
03

Calculate the \(p\)-value

The \(p\)-value is the probability of obtaining a test statistic as extreme or more extreme than the one obtained if the null hypothesis is true. Because it is a two-tailed test (we are testing for \(\mu \neq 1\)), we need to find the area in both tails of the t-distribution. Using a t-distribution table or calculator, find the \(p\)-value corresponding to the calculated t-statistic value and degrees of freedom (df) which is equal to \(n - 1\) (in this case, 34).
04

Perform the hypothesis test

To perform the test in part a, decide on a level of significance, say, \(\alpha = 0.05\). Compare the \(p\)-value with this level of significance to make a decision. If \(p\)-value \(< \alpha\), reject the null hypothesis. This means there is enough evidence to support that the average weight of the ground beef packages is different from 1 pound. If \(p\)-value \(> \alpha\), fail to reject the null hypothesis. This means there isn't enough evidence to support the claim that the average weight is different from 1 pound.
05

Report the results

As the quality control manager, report the results of the study to a consumer interest group. Include the conclusions from the hypothesis test (reject or fail to reject the null hypothesis) and mention the \(p\)-value in your report. If the null hypothesis was rejected, discuss actions to be taken on the packaging process to ensure the target average weight of 1 pound is met.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis, symbolized as \begin{math}H_0\begin{math}, represents a statement of no effect or no difference—it is the presumption to be tested. For instance, if a quality control manager wishes to confirm whether the average weight of ground beef packages is consistent with the claimed amount, they would set this assertion as the null hypothesis. Given the scenario in our exercise, the null hypothesis (\(H_0\)) is that the average weight (\(\begin{\text}\text{\text{\textmu}}_0\begin{\text}\text{\) of the ground beef packages is 1 pound. This is not to be mistaken as claiming that the null hypothesis is true, but rather it is what we assume to be true for the sake of the test. If the evidence from the sample is strong enough to contradict this hypothesis, the null may be rejected, thereby suggesting an alternative explanation.
t-Statistic
In situations where the sample size is relatively small and the population standard deviation is unknown, the t-statistic comes into play for hypothesis testing. It is calculated by comparing the sample mean to the hypothesized population mean, adjusted for the sample size and variability (standard deviation). The formula is as follows:
\[t = \frac{(\bar{x} - \begin{\text}\text{\text{\textmu}}_0\begin{\text}\text{\)}{s / \begin{\text}n\begin{\text}n\text{sqrt}{\text}n{n}{\text}}\]
where (\bar{x}\) is the average from our sample, (\begin{\text}\text{\text{\textmu}}_0\begin{\text}\text{\) represents the supposed mean of the population, (s) is the standard deviation of the sample, and (n) is the sample size. In this exercise, our calculated t-statistic would tell us how the observed average weight from the sample compares to the hypothetical average stated in the null hypothesis.
p-Value
The p-value is a crucial concept in hypothesis testing, representing the probability of observing our statistic, or one more extreme, if the null hypothesis is true. A small p-value indicates that such an extreme observed outcome would be rare under the null hypothesis, and thus provides evidence against \(H_0\). When performing a two-tailed test, as we are in this case for (\(\begin{\text}\text{\textmu eq 1}\begin{\text}\text{\)), we look at the probability of getting a sample mean that is either greater or less than our hypothesized mean, hence the term 'two-tailed'. To determine our decision in the hypothesis test, we compare the p-value to a predetermined level of significance, typically (\(\begin{\text}\text{\textalpha = 0.05}\begin{\text}\text{\)). If the p-value falls below our significance level, we have reason to reject the null hypothesis in favor of the alternative.
Standard Deviation
Standard deviation is a measure that quantifies how spread out the values in a data set are. If the standard deviation is small, it suggests that the data points are clustered closely around the mean, implying minimum variability. In contrast, a large standard deviation indicates that the data points are more widely spread out from the mean, showing more variability in the data set. In the context of our exercise, the sample standard deviation (s) of the weights of the ground beef packages is 0.18 pounds. This statistic incorporates the variability of the sample weights into the calculation of the t-statistic, which would be essential for determining if the observed average weight is statistically different from the hypothesized value of 1 pound. Understanding how to use and interpret standard deviation is key in assessing the reliability and significance of statistical data.

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Most popular questions from this chapter

Define the level of significance.

List the five parts of a statistical test.

To determine whether there is a significant difference in the weights of boys and girls beginning kindergarten, random samples of 50 boys and 50 girls aged 5 years produced the following information: \({ }^{12}\) $$ \begin{array}{lccc} \hline & & \text { Standard } & \text { Sample } \\ & \text { Mean } & \text { Deviation } & \text { Size } \\ \hline \text { Boys } & 19.4 \mathrm{~kg} & 2.4 & 50 \\ \text { Girls } & 17.0 \mathrm{~kg} & 1.9 & 50 \end{array} $$ a. Do you have a preconceived idea of what to expect when examining the average weights of 5 -year-old boys and girls? Based on your answer, state the null and alternative hypotheses to be tested. b. Test the hypothesis in part a using \(\alpha=.05\).

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