Problem 3 - Basic Statistics for Business and Economics with Student

Question

(a) Is this a one- or two-tailed test? (b) What is the decision rule? (c) What is the value of the test statistic? (d) What is your decision regarding \(H_{0} ?\) (e) What is the \(p\) -value? Interpret it. A sample of 36 observations is selected from a normal population. The sample mean is \(21,\) and the population standard deviation is \(5 .\) Conduct the following test of hypothesis using the .05 significance level. $$\begin{array}{l}H_{0}: \mu \leq 20 \\\H_{1}: \mu>20\end{array}$$

Step-by-Step Solution

Answer

One-tailed test; do not reject \(H_0\); \(p\)-value is 0.1151, not significant.

Step 1: Determine the Type of Test

The hypothesis stated includes a 'greater than' symbol for the alternative hypothesis: \(H_1: \mu > 20\). This indicates that the test is one-tailed, specifically focusing on whether the mean is greater than 20.

Step 2: Set the Decision Rule

Since this is a one-tailed test at the \(\alpha = 0.05\) significance level, we look up the critical value for a normal distribution. The z-value that corresponds to the 95th percentile is approximately 1.645. Our decision rule: Reject \(H_0\) if the test statistic is greater than 1.645.

Step 3: Calculate the Test Statistic

The test statistic is calculated using the formula: \[ z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} \] where \(\bar{x} = 21\), \(\mu_0 = 20\), \(\sigma = 5\), and \(n = 36\). Substituting these values, we have: \[ z = \frac{21 - 20}{\frac{5}{\sqrt{36}}} = \frac{1}{\frac{5}{6}} = \frac{1}{0.8333} \approx 1.20 \]

Step 4: Make a Decision Regarding \(H_0\)

The calculated z-value is 1.20, which is less than the critical value of 1.645. Thus, we do not reject the null hypothesis \(H_0: \mu \leq 20\).

Step 5: Calculate and Interpret the \(p\)-Value

The \(p\)-value for a z-value of 1.20 is approximately 0.1151 (from z-tables or a calculator). Since the \(p\)-value (0.1151) is greater than the significance level \(\alpha = 0.05\), we do not have enough evidence to reject the null hypothesis. This suggests there is a lack of statistical significance in supporting that the mean is greater than 20.

Key Concepts

one-tailed test

In hypothesis testing, a one-tailed test is used when we want to determine if there is a statistically significant increase or decrease in a parameter in one specific direction. We apply a one-tailed test when our alternative hypothesis (denoted as \( H_1 \)) suggests that the true parameter value is either greater than or less than a certain value, but not both.

For instance, if we hypothesize that a sample mean is greater than a specified value, as seen with \( H_1: \mu > 20 \), it indicates a one-tailed test. This is because we're only interested in knowing if the mean exceeds a particular threshold. This specific direction allows researchers to have more power to detect a significant effect in that one direction compared to a two-tailed test, where the interest would be in both directions – either an increase or a decrease.

test statistic

The test statistic is a standardized value that is calculated from sample data and used in hypothesis testing. It helps in determining whether to reject the null hypothesis \( H_0 \). The formula for calculating the test statistic in the context of a one-tailed test with a normal distribution is as follows:

\[ z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} \]

Here:

• \( \bar{x} \) is the sample mean
• \( \mu_0 \) is the population mean under the null hypothesis
• \( \sigma \) is the population standard deviation
• \( n \) is the sample size

In this exercise, substituting the given values, we calculate a test statistic of approximately 1.20. This value quantifies how far the sample mean deviates from the hypothesized population mean, standardized by the standard error of the mean.

p-value

The \( p \)-value is a crucial concept in hypothesis testing. It represents the probability of observing a test statistic as extreme as the one calculated (or more extreme) under the assumption that the null hypothesis \( H_0 \) is true. A smaller \( p \)-value indicates stronger evidence against the null hypothesis.

In this exercise, the calculated \( p \)-value is approximately 0.1151. This value is compared against the significance level \( \alpha = 0.05 \). Since the \( p \)-value (0.1151) is greater than \( \alpha \), there is not enough statistical evidence to reject the null hypothesis. This means we don't have sufficient evidence to claim that the mean is greater than 20 at the 5% significance level.

decision rule

The decision rule in hypothesis testing provides a clear guideline for when to reject the null hypothesis \( H_0 \). For a one-tailed test at a specified significance level \( \alpha \), we need the critical value from the standard normal distribution to set our decision rule.

In this scenario, the critical z-value for a one-tailed test with \( \alpha = 0.05 \) is approximately 1.645, which corresponds to the 95th percentile of the normal distribution. Thus, our decision rule is:

• Reject \( H_0 \) if the test statistic is greater than 1.645
• Do not reject \( H_0 \) if the test statistic is less than or equal to 1.645

Given the test statistic calculated as 1.20, which is less than 1.645, we do not reject the null hypothesis. This decision tells us that based on the sample data, there's insufficient evidence to conclude that the mean is greater than 20.