Problem 10 - Basic Statistics for Business and Economics with Student

Question

Given the following hypothesis: $$\begin{aligned}H_{0}: \mu &=400 \\\H_{1}: \mu & \neq 400\end{aligned}$$ For a random sample of 12 observations, the sample mean was 407 and the sample standard deviation \(6 .\) Using the .01 significance level: a. State the decision rule. b. Compute the value of the test statistic. c. What is your decision regarding the null hypothesis?

Step-by-Step Solution

Answer

Reject the null hypothesis; the test statistic (4.04) exceeds the critical value (3.106).

Step 1: State the Decision Rule

The decision rule involves comparing the test statistic to the critical values from the t-distribution. Since the significance level \( \alpha \) is 0.01 and the test is two-tailed, we need to find the critical t-value for \( n - 1 = 12 - 1 = 11 \) degrees of freedom. Consult a t-table or use statistical software to find that the critical t-value for a two-tailed test at \( \alpha = 0.01 \) and 11 degrees of freedom is approximately 3.106. Thus, the decision rule is: if the calculated test statistic is greater than 3.106 or less than -3.106, reject the null hypothesis \( H_0 \).

Step 2: Compute the Test Statistic

We use the formula for the t-test statistic for a sample mean: \[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \]where:- \( \bar{x} \) is the sample mean (407),- \( \mu \) is the population mean under the null hypothesis (400),- \( s \) is the sample standard deviation (6),- \( n \) is the sample size (12).Substitute the values:\[ t = \frac{407 - 400}{6/\sqrt{12}} = \frac{7}{6/\sqrt{12}} = \frac{7}{1.732} \approx 4.04 \]

Step 3: Make the Decision Regarding Null Hypothesis

Compare the calculated test statistic \( t \approx 4.04 \) with the critical t-values from Step 1. Since 4.04 is greater than 3.106, we reject the null hypothesis \( H_0 \).

Key Concepts

T-distribution

In hypothesis testing, the t-distribution is a fundamental tool when dealing with small sample sizes. It's often used instead of the normal distribution because it does a better job of accounting for the additional uncertainty that comes with estimating the population parameters from a small sample. The t-distribution is similar to the normal distribution but has thicker tails. This means it is more prone to producing values that fall far from the mean. This characteristic becomes less pronounced as the sample size increases, and the t-distribution gradually approaches a normal distribution. The key parameter in a t-distribution is called degrees of freedom, which generally equals the sample size minus one (n-1). In our example, the degrees of freedom were 11 for a sample size of 12.

Significance Level

The significance level, denoted by \( \alpha \), is a threshold that determines whether we reject the null hypothesis in hypothesis testing. It's like a cutoff point for decision-making. If the results from your sample data produce a test statistic more extreme than what the significance level indicates, you reject the null hypothesis. A common choice for \( \alpha \) is 0.05, but in our exercise, a stricter threshold of 0.01 was used. A significance level of 0.01 means there's only a 1% risk of incorrectly rejecting a true null hypothesis. This low threshold is sometimes chosen in situations where the cost of making a mistake is high.

Null Hypothesis

The null hypothesis, represented by \( H_0 \), is the statement in hypothesis testing that assumes no effect or no difference. It's the default position that there is nothing new happening, and any observed effect is due to chance. In the given exercise, the null hypothesis is that the population mean \( \mu \) equals 400. Testing the null hypothesis involves determining whether observed sample data contradicts this assumption. If it's found that the sample data is very unlikely under the null hypothesis, we have grounds to reject \( H_0 \). The null hypothesis is vital because it provides a clear and unambiguous statement that can be tested.

Test Statistic

The test statistic is a value calculated from sample data that is used to decide whether to reject the null hypothesis. It's effectively a measure of how far your sample statistic is from what the null hypothesis predicts. In our example, the test statistic is calculated using the formula for the t-statistic: \[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \]This formula considers:

• The sample mean \( \bar{x} = 407 \)
• The assumed population mean \( \mu = 400 \)
• The sample standard deviation \( s = 6 \)
• And the sample size \( n = 12 \)

By plugging these values into the formula, we obtained a test statistic of approximately 4.04. This value is compared against critical values from the t-distribution to make a decision about the null hypothesis. Since our test statistic of 4.04 is greater than the critical t-value, we reject the null hypothesis. This indicates that the sample mean of 407 provides significant evidence against the assumption that the population mean is 400 at the 0.01 significance level.