Problem 28 - Basic Statistics for Business and Economics with Student

Question

Rutter Nursery Company packages its pine bark mulch n 50-pound bags. From a long history, the production department reports that the distribution of the bag weights follows the normal distribution and the standard deviation of this process is 3 pounds per bag. At the end of each day, Jeff Rutter, the production manager, weighs 10 bags and computes the mean weight of the sample. Below are the weights of 10 bags from today's production. $$\begin{array}{|lllllllll|}\hline 45.6 & 47.7 & 47.6 & 46.3 & 46.2 & 47.4 & 49.2 & 55.8 & 47.5 & 48.5 \\\\\hline \end{array}$$ a. Can Mr. Rutter conclude that the mean weight of the bags is less than 50 pounds? Use the .01 significance level. b. In a brief report, tell why Mr. Rutter can use the \(z\) distribution as the test statistic. c. Compute the \(p\) -value.

Step-by-Step Solution

Answer

Fail to reject \( H_0 \); mean is not significantly less than 50. Use \( z \)-distribution due to normality and known \( \sigma \). \( p \)-value is 0.027.

Step 1: Calculate Sample Mean

First, calculate the sample mean (\( \bar{x} \)) of the 10 bag weights. The data is: 45.6, 47.7, 47.6, 46.3, 46.2, 47.4, 49.2, 55.8, 47.5, and 48.5. Sum these values and then divide by the number of bags (10). ewline \( \bar{x} = \frac{45.6 + 47.7 + 47.6 + 46.3 + 46.2 + 47.4 + 49.2 + 55.8 + 47.5 + 48.5}{10} = 48.18 \).

Step 2: Formulate Hypotheses

State the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_1\)). ewline \(H_0: \mu = 50\) (the mean weight is 50 pounds) ewline \(H_1: \mu < 50\) (the mean weight is less than 50 pounds).

Step 3: Determine Test Statistic

Use the formula for the \( z \)-score: \( z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\), where \( \bar{x} = 48.18 \), \( \mu = 50 \), \( \sigma = 3 \), and \( n = 10 \). ewline \( z = \frac{48.18 - 50}{\frac{3}{\sqrt{10}}} = \frac{-1.82}{0.9487} = -1.92 \).

Step 4: Decision Rule & Critical Value

For a significance level of \( \alpha = 0.01 \), find the critical value from the \( z \)-table for a one-tailed test. The critical \( z \)-value is approximately -2.33. Compare the test statistic to the critical value.

Step 5: Conclusion on Null Hypothesis

Since \( z = -1.92 \) is greater than \(-2.33\), we fail to reject the null hypothesis \( H_0 \). There is not enough evidence at the 0.01 level of significance to conclude that the mean weight is less than 50 pounds.

Step 6: Reason for Using \( z \) Distribution

Mr. Rutter can use the \( z \)-distribution because the sample size is 10, which is reasonably large for approximation if the population is normally distributed. Additionally, the population standard deviation is known.

Step 7: Compute the p-Value

Find the \( p \)-value corresponding to \( z = -1.92 \). The \( p \)-value from the \( z \)-table is approximately 0.027. Compare this with the significance level.

Key Concepts

Normal Distribution

In statistics, the normal distribution is a continuous probability distribution that is symmetric around its mean, representing how data naturally clusters around a central point. It's often referred to as the "bell curve" due to its characteristic shape. When data follows a normal distribution, it allows for powerful statistical analysis methods, such as using the Z-score to test hypotheses.
The key properties of a normal distribution are:

• Symmetrical shape: Data is evenly distributed about the mean.
• Mean, median, and mode are all equal and located at the center.
• Characterized by its mean (\( \mu \)) and standard deviation (\( \sigma \)).

In Rutter's exercise, we assume the weights of the mulch bags follow a normal distribution based on historical data. This assumption justifies using the Z-score to determine if the sample mean significantly deviates from the population mean of 50 pounds.

Sample Mean

The sample mean, denoted as \( \bar{x} \), is the average value of a sample that estimates the population mean. It is calculated by summing all values in a sample set and dividing by the number of observations. In our exercise, Mr. Rutter calculated the sample mean of his 10 mulch bags to be 48.18 pounds.

• Formally, the formula for the sample mean is \( \bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} \), where \( x_i \) represents each observation and \( n \) is the sample size.
• The sample mean provides an unbiased estimation of the population mean if the sample is randomly selected.

The importance of the sample mean lies in its role as a foundational statistic for conducting hypothesis tests, as it gives us a measure against which we can compare our hypothesized population mean.

Z-Score

The Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It is expressed as the number of standard deviations a datapoint is from the mean. In hypothesis testing, Z-scores can be used to determine the likelihood of a sample statistic.

• To calculate a Z-score: \( z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \)
  Where: \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, \( \sigma \) is the standard deviation, and \( n \) is the sample size.
• A Z-score tells us how many standard deviations an element is from the mean. In this exercise, a Z-score of -1.92 indicates that our sample mean is 1.92 standard deviations below the hypothesized mean of 50 pounds.

This insight helps us understand whether a deviation observed in a sample is due to random chance or if it is statistically significant.

Significance Level

The significance level, denoted as \( \alpha \), defines the threshold for rejecting the null hypothesis. It represents the probability of rejecting a true null hypothesis, often known as a Type I error. Common significance levels include 0.05, 0.01, and 0.10.
In Mr. Rutter's analysis, the chosen significance level is 0.01, which indicates a 1% risk of concluding that the mean weight is less than 50 pounds when it is not.

• A lower \( \alpha \) value denotes more stringent criteria for rejecting the null hypothesis, thus reducing the chance of a Type I error.
• In practice, choosing a significance level depends on the context of the test and the acceptable risk of making an incorrect decision.

When comparing the calculated p-value (0.027) to the significance level (0.01), we find that the p-value is greater, meaning we do not have enough statistical evidence to reject the null hypothesis at this strict level.