Problem 37 - Basic Statistics for Business and Economics with Student

Question

According to the Coffee Research Organization (http://www.coffeeresearch.org) the typical American coffee drinker consumes an average of 3.1 cups per day. A sample of 12 senior citizens revealed they consumed the following amounts of coffee, reported in cups, yesterday. $$\begin{array}{|llllllllllll|}\hline 3.1 & 3.3 & 3.5 & 2.6 & 2.6 & 4.3 & 4.4 & 3.8 & 3.1 & 4.1 & 3.1 & 3.2 \\\\\hline\end{array}$$ At the .05 significance level does this sample data suggest there is a difference between the national average and the sample mean from senior citizens?

Step-by-Step Solution

Answer

There is not enough evidence to suggest a significant difference from the national average.

Step 1: Define the Null and Alternative Hypotheses

We start by defining our hypotheses for the statistical test. The null hypothesis \(H_0\) states that there is no difference between the national average and the sample mean, so \( \mu = 3.1 \). The alternative hypothesis \(H_a\) states that there is a difference, so \( \mu eq 3.1 \).

Step 2: Calculate the Sample Mean

To find the sample mean, we sum up all the cups consumed and divide by the number of observations. The sample data is \(3.1, 3.3, 3.5, 2.6, 2.6, 4.3, 4.4, 3.8, 3.1, 4.1, 3.1, 3.2\). The sum is 40.1. The sample mean \(\bar{x}\) is:\[ \bar{x} = \frac{40.1}{12} = 3.3417 \text{ cups} \]

Step 3: Calculate the Sample Standard Deviation

First, calculate the variance. Use the formula for sample variance: \(s^2 = \frac{\sum{(x_i - \bar{x})^2}}{n-1}\). Calculate each difference \((x_i - \bar{x})\), square these differences, sum them up, and divide by \(n-1 = 11\) where \(n = 12\). Use these to find the sample standard deviation \(s\).

Step 4: Calculate the Test Statistic

We use a t-test for this sample. The formula is: \[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \]where \(\bar{x} = 3.3417, \mu = 3.1, s\) is the standard deviation calculated previously, and \(n = 12\). Substitute these values to find the \(t\)-value.

Step 5: Determine the Critical Value and Compare

Since our significance level is 0.05 and our test is two-tailed, we check the critical t-value for \(n-1 = 11\) degrees of freedom. Use a t-distribution table or calculator to find this value, typically around \( \pm 2.201 \). Compare the calculated \(t\)-value to the critical value.

Step 6: Draw a Conclusion

If the \(t\)-value is greater than 2.201 or less than -2.201, reject the null hypothesis \(H_0\). Otherwise, do not reject the null hypothesis. This helps us decide if the sample mean is significantly different from the population mean.

Key Concepts

Sample Mean

The sample mean is a fundamental concept in statistics used to determine the average value of a set of data points. It is often denoted by \( \bar{x} \). To calculate the sample mean, sum all observed values in the dataset and divide the total by the number of observations. In our exercise, we have a sample of coffee consumption from 12 senior citizens:

• 3.1, 3.3, 3.5, 2.6, 2.6, 4.3, 4.4, 3.8, 3.1, 4.1, 3.1, 3.2

By adding these values, we get a total of 40.1 cups. Dividing this sum by the number of observations (12), we find the sample mean to be:\[ \bar{x} = \frac{40.1}{12} = 3.3417 \text{ cups} \]This value represents the average coffee consumption for this particular group and is used to make inferences about the population from which the sample was drawn.

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. It tells us how spread out the numbers are in a dataset. A lower standard deviation indicates that the values are closer to the mean, while a higher standard deviation shows that they are more spread out. To calculate the sample standard deviation, follow these steps:

• Find the difference between each observation and the sample mean \((x_i - \bar{x})\).
• Square each of these differences.
• Sum all the squared differences.
• Divide by the number of observations minus one \((n-1)\) to find the variance.
• Take the square root of the variance to obtain the standard deviation \(s\).

This measure helps assess the reliability of the sample mean, as less variation makes the sample mean more representative of the population.

t-test

The t-test is a statistical test used to compare the sample mean to a known value or another sample mean. In this exercise, we use a t-test to determine if the average coffee consumption of senior citizens is significantly different from the national average of 3.1 cups. This test involves the following steps:

• Calculate the sample mean and standard deviation.
• Use these to determine the t-statistic with the formula:\[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \]where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size.
• Compare the calculated t-value against a critical value from the t-distribution table, which depends on the degrees of freedom \((n-1)\) and the chosen significance level.

If the t-value is beyond the critical range, we reject the null hypothesis, suggesting a significant difference.

Significance Level

The significance level, often denoted \(\alpha\), is a threshold used in hypothesis testing to determine whether to reject the null hypothesis. It represents the probability of making a Type I error, which is rejecting a true null hypothesis. Common significance levels are 0.05, 0.01, and 0.10. In our exercise, the significance level is set at 0.05, which indicates a 5% risk of concluding that a difference exists when there is none. The significance level is used together with the critical value from a statistical table to decide whether the observed data is statistically significant.To use the significance level in the t-test:

• Find the critical t-value for the given degrees of freedom and significance level.
• Compare the calculated t-statistic to this critical value.
• If the t-statistic exceeds this value, it indicates a statistically significant result, leading us to reject the null hypothesis.

By adhering to this process, we make informed conclusions based on data rather than chance.