Problem 4 - 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 64 observations is selected from a normal population. The sample mean is \(215,\) and the population standard deviation is \(15 .\) Conduct the following test of hypothesis using the .03 significance level.$$\begin{array}{l}H_{0} ; \mu \geq 220 \\\H_{1}: \mu<220\end{array}$$

Step-by-Step Solution

Answer

(a) One-tailed test. (b) Reject \( H_0 \) if z < -1.88. (c) Test statistic is -2.67. (d) Reject \( H_0 \). (e) p-value is 0.0037, indicating strong evidence against \( H_0 \).

Step 1: Determine the Nature of the Test

The null hypothesis is set as \( H_0: \mu \geq 220 \) and the alternative hypothesis as \( H_1: \mu < 220 \). This situation suggests that the test is examining if the sample mean is less than 220, which makes it a one-tailed test.

Step 2: State the Decision Rule

Since it is a one-tailed test, we will reject \( H_0 \) if the test statistic is less than the critical value for a one-tailed test at the 0.03 significance level. This critical value can be found using the standard normal distribution tables.

Step 3: Calculate the Test Statistic

The test statistic for a hypothesis test concerning a population mean with a known standard deviation is calculated using the formula: \[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]Substitute the given values: \( \bar{x} = 215 \), \( \mu_0 = 220 \), \( \sigma = 15 \), and \( n = 64 \):\[ z = \frac{215 - 220}{\frac{15}{\sqrt{64}}} = \frac{-5}{1.875} \approx -2.67 \]

Step 4: Make a Decision Based on the Critical Value

For a one-tailed test with a significance level of 0.03, the critical z-value is approximately -1.88 (from standard normal distribution tables). Since the calculated z-value (-2.67) is less than -1.88, we reject \( H_0 \).

Step 5: Calculate and Interpret the p-value

The p-value is the probability of obtaining a test statistic as extreme as the sample result, under the assumption that \( H_0 \) is true. From z-tables, a z-value of -2.67 corresponds to a p-value less than 0.0037. Since the p-value (0.0037) is less than the significance level (0.03), this provides further evidence to reject \( H_0 \).

Key Concepts

One-tailed test

In hypothesis testing, determining whether a test is one-tailed or two-tailed is crucial. For a one-tailed test, we're examining the data in only one direction. This means we check if a sample statistic is significantly greater than or less than a specified value, not both.

In the given exercise, the hypotheses are stated as follows:

• Null Hypothesis (\( H_0 \)): \( \mu \ge 220 \)
• Alternative Hypothesis (\( H_1 \)): \( \mu < 220 \)

The alternative hypothesis \( ( H_1: \mu < 220 ) \) suggests that we are interested in whether the mean is less than 220. Hence, it's a lower-tailed, or simply, a one-tailed test.

This focus on one direction makes the test more "powerful" for detecting an effect in that specific direction.

p-value interpretation

The p-value in hypothesis testing is a measure of the evidence against the null hypothesis. It reflects the probability of obtaining test results at least as extreme as the observed data, assuming the null hypothesis is true.

In this exercise, the null hypothesis is that \( \mu \geq 220 \) and the alternative hypothesis is \( \mu < 220 \). After calculating the z-value to be -2.67, we look at standard normal distribution tables to find the p-value. It comes out as approximately 0.0037.

A p-value below the significance level (0.03 in this case) indicates strong evidence against the null hypothesis, thus suggesting it should be rejected. Here, since 0.0037 < 0.03, there is strong evidence that the population mean is indeed less than 220.

Decision rule

A decision rule in hypothesis testing helps determine whether to reject or not reject the null hypothesis. It involves comparing the calculated test statistic with a critical value.

For a one-tailed test at a 0.03 significance level, the critical z-value is approximately -1.88. Our decision rule is:

• Reject \( H_0 \) if the calculated z-value is less than -1.88.
• Otherwise, do not reject \( H_0 \).

These rules provide a clear boundary. If our test statistic falls in the "rejection region" (below -1.88), it suggests that the data are unlikely under the null hypothesis.

Given that the calculated z-value is -2.67, we fall into the rejection zone. Therefore, we reject the null hypothesis based on our decision rule.

Test statistic

The test statistic is a standard numerical measurement that indicates how far the sample mean deviates from the null hypothesis mean. This calculation standardizes the sample mean difference so it can be compared against a probability distribution.

In this exercise, the test statistic is calculated using the formula:\[ z = \frac{\bar{x} - \mu_{0}}{\frac{\sigma}{\sqrt{n}}} \]where:

• \( \bar{x}\)
  is the sample mean (215)
• \( \mu_{0}\)
  is the population mean under the null hypothesis (220)
• \( \sigma\)
  is the population standard deviation (15)
• \( n\)
  is the sample size (64)

By plugging in these values, we get:\[ z = \frac{215 - 220}{\frac{15}{\sqrt{64}}} = -2.67 \]This result of -2.67 shows the number of standard deviations the sample mean is from the null hypothesis mean.