Problem 1 - Basic Statistics for Business and Economics with Student

Question

A sample of 40 observations is selected from one population with a population standard deviation of \(5 .\) The sample mean is \(102 .\) A sample of 50 observations is selected from a second population with a population standard deviation of \(6 .\) The sample mean is \(99 .\) Conduct the following test of hypothesis using the .04 significance level. $$ \begin{array}{l} H_{0}: \mu_{1}=\mu_{2} \\ H_{1}: \mu_{1} \neq \mu_{2} \end{array} $$ a. Is this a one-tailed or a two-tailed test? b. State the decision rule. c. Compute the value of the test statistic. d. What is your decision regarding \(H_{0}\) e. What is the \(p\) -value?

Step-by-Step Solution

Answer

It's a two-tailed test, reject \(H_0\); \(p\)-value is \(0.006\).

Step 1: Determine the Type of Test

Since the alternative hypothesis \(H_1\) is \(\mu_1 eq \mu_2\), this is a two-tailed test. We are checking if the means are different in either direction.

Step 2: State the Decision Rule

For a two-tailed test at the \(0.04\) significance level, the critical value for \(Z\) based on a standard normal distribution table is approximately \(\pm 2.05\). We will reject \(H_0\) if the computed test statistic \(Z\) is less than \(-2.05\) or greater than \(2.05\).

Step 3: Compute the Test Statistic

The formula for the test statistic in a two-sample Z-test is:\[Z = \frac{(\bar{X}_1 - \bar{X}_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\]Substitute the given values:\(\bar{X}_1 = 102\), \(\bar{X}_2 = 99\), \(\sigma_1 = 5\), \(\sigma_2 = 6\), \(n_1 = 40\), \(n_2 = 50\).Thus,\[Z = \frac{(102 - 99)}{\sqrt{\frac{5^2}{40} + \frac{6^2}{50}}}\approx \frac{3}{\sqrt{0.625 + 0.72}} \approx \frac{3}{1.096} \approx 2.74\]

Step 4: Decision Regarding Hypothesis

The computed \(Z\)-value is approximately \(2.74\), which is greater than the critical value of \(2.05\). Therefore, we reject \(H_0\).

Step 5: Compute the p-value

Using a standard normal distribution table or calculator, the p-value corresponding to \(Z = 2.74\) is approximately \(0.006\). Since this is a two-tailed test, we compare \(0.006\) to our significance level of \(0.04\). The p-value is less than \(0.04\), which confirms our decision to reject \(H_0\).

Key Concepts

Two-Tailed Test

A two-tailed test is used when we want to determine if there is a difference between two groups, but we don't know which group will have a higher or lower mean. In other words, we're checking for any significant difference, regardless of direction.
For instance, in our exercise, we're testing if the means of two populations are equal or not. The alternative hypothesis, denoted as \(H_1\), indicates that the means are not equal:\(\mu_1 eq \mu_2\). This suggests a deviation in either direction, making it a two-tailed test.
The decision rule in a two-tailed test involves checking if the test statistic falls into either of the critical regions on both ends of the probability distribution curve. If you think of the curve as a bell shape, you're essentially looking at both the left and right ends of it.

• If the test statistic is much smaller than the left critical value or much larger than the right one, we reject the null hypothesis \(H_0\).
• In the exercise, the critical values for this two-tailed test at a 0.04 significance level were approximately \(\pm 2.05\).

Z-Test

The Z-test is a statistical test used to determine if there's a significant difference between sample and population means or between the means of two samples. It's appropriate when the sample size is large (usually over 30) and the population standard deviations are known.
For the given exercise, we're conducting a two-sample Z-test that compares the means of two independent groups. The formula for calculating the Z-statistic in this context is given by:
\[Z = \frac{(\bar{X}_1 - \bar{X}_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\]
Each symbol represents specific values:

• \(\bar{X}_1\) and \(\bar{X}_2\) are the sample means.
• \(\sigma_1\) and \(\sigma_2\) are the known population standard deviations.
• \(n_1\) and \(n_2\) are the sizes of each sample.

The Z-test helps determine if the difference between sample means is large enough to be statistically significant, beyond random chance. In our scenario, the calculated Z value was 2.74, indicating a more substantial deviation than what would be expected under the null hypothesis.

Significance Level

In hypothesis testing, the significance level, commonly denoted as \(\alpha\), represents the probability of mistakenly rejecting the null hypothesis when it is true. It's essentially the "risk" we're willing to take for making this error.
Common choices for \(\alpha\) are 0.05, 0.01, or 0.10, but in the given exercise, a significance level of 0.04 was used. This indicates a 4% risk of concluding there is a notable difference when there isn't one.

• A lower significance level means stricter criteria for rejecting the null hypothesis, reducing the risk of Type I errors (false positives).
• Conversely, a higher significance level may increase the likelihood of finding a difference, even if it doesn't truly exist.

Choosing the right significance level depends on the context and the consequences of errors in decision-making. In our example, a p-value (probability value) is compared to this significance level. The p-value for the computed Z statistic was approximately 0.006. Since this p-value is less than the significance level of 0.04, it provided further evidence to reject the null hypothesis \(H_0\).