Problem 8 - An Introduction to Psychological Statistics

Question

You want to know if an employee's opinion about an organization is the same as the opinion of that employee's boss. You collect data from 18 employee- supervisor pairs and code the difference scores so that positive scores indicate that the employee has a higher opinion and negative scores indicate that the boss has a higher opinion (meaning that difference scores of 0 indicate no difference and complete agreement). You find that the mean difference score is \(\overline{X_{D}}=-3.15\) with a standard deviation of \(s_{D}=1.97\). Test this hypothesis at the \(\alpha=0.01\) level. Construct confidence intervals from a mean of \(\overline{X_{D}}=1.25\), standard error of \(s_{\overline{X_{D}}}=0.45\), and \(d f=10\) at the \(90 \%, 95 \%\), and \(99 \%\) confidence level.

Step-by-Step Solution

Answer

The mean difference is significant at \(\alpha=0.01\); 90% CI: [0.44, 2.06], 95% CI: [0.24, 2.26], 99% CI: [-0.18, 2.68].

Step 1: State the Hypotheses

We need to determine whether there is a significant difference between the employee's and supervisor's opinions. The null hypothesis is that there is no difference, which means the mean difference score is zero: \(H_0: \mu_D = 0\). The alternative hypothesis is that there is a difference: \(H_a: \mu_D eq 0\).

Step 2: Calculate the Test Statistic

The test statistic for a mean difference score uses \(t\)-distribution. The formula is \( t = \frac{\overline{X_D} - \mu_D}{s_D / \sqrt{n}} \), where \(\overline{X_D} = -3.15\), \(s_D = 1.97\), and \(n = 18\). Substituting, we calculate \( t = \frac{-3.15 - 0}{1.97 / \sqrt{18}} \approx -7.213\).

Step 3: Determine the Critical t-Value

The degrees of freedom (\(df\)) is \(n - 1 = 17\). For \(\alpha=0.01\) in a two-tailed test, we look up the critical \(t\)-value in the \(t\)-distribution table for \(df=17\). The critical values are approximately \(\pm 2.921\).

Step 4: Compare Test Statistic with Critical t-Value

The calculated \(t\)-value of \(-7.213\) is less than \(-2.921\) (falls in the rejection region), hence we reject the null hypothesis \(H_0\).

Step 5: Calculate Confidence Intervals

Confidence intervals are computed using \(\text{CI} = \overline{X} \pm (t^* \cdot s_{\overline{X_D}})\). We need the critical \(t^*\) values for each confidence level with \(df = 10\). The critical \(t^*\) values are approximately: 90% (1.812), 95% (2.228), 99% (3.169). The confidence intervals are: - 90% CI: \(1.25 \pm (1.812 \cdot 0.45) = [0.44, 2.06]\)- 95% CI: \(1.25 \pm (2.228 \cdot 0.45) = [0.24, 2.26]\)- 99% CI: \(1.25 \pm (3.169 \cdot 0.45) = [-0.18, 2.68]\).

Key Concepts

Mean Difference

The mean difference is a critical component in hypothesis testing, particularly when examining paired data. It represents the average difference between two paired observations. In this exercise, these observations are the opinions of employees and their supervisors. The mean difference, denoted as \(\overline{X_{D}}\), is calculated by finding the difference for each pair and averaging those values. Here, \(\overline{X_{D}} = -3.15\), indicating that, on average, supervisors rate the organization more favorably by 3.15 units compared to employees.
Understanding this concept is significant as it informs us whether there is a consistent bias in opinions. If the mean difference were zero, it would suggest that employees and supervisors generally agree in their ratings. However, a non-zero mean difference implies some level of disagreement between them. In our scenario, since \(\overline{X_{D}}\) is negative, it shows a trend where supervisors have higher ratings than employees.

Confidence Intervals

Confidence intervals provide a range of values within which we believe the true population parameter lies, with a certain level of certainty. In this problem, the mean difference confidence intervals help us assess the reliability of our sample statistics. By constructing them at various confidence levels—here at 90%, 95%, and 99%—we can evaluate how stable our findings are.
The formula for calculating a confidence interval is \(\text{CI} = \overline{X} \pm (t^* \cdot s_{\overline{X_{D}}})\). This formula uses the sample mean, \(\overline{X}\), the critical \(t\)-value \(t^*\), and the standard error \(s_{\overline{X_{D}}}\). For instance, the 90% confidence interval falls between 0.44 and 2.06, while the 99% interval is wider, ranging from -0.18 to 2.68. This illustrates that the higher the confidence level, the wider the interval.

• 90% CI: provides a more narrow, precise estimate, but with less certainty than higher intervals.
• 99% CI: offers a broader, less precise estimate but greater certainty that it contains the true parameter.

Thus, confidence intervals help in understanding the precision and reliability of our sample mean estimate in reflecting the true population mean difference.

T-Distribution

The \(t\)-distribution is fundamental to the process of hypothesis testing when dealing with sample data. It is especially useful for small sample sizes and unknown population variances. In our exercise, we rely on the \(t\)-distribution because the sample of employee-supervisor pairs is relatively small (\(n=18\)).
The \(t\)-distribution resembles the normal distribution, yet it has thicker tails, meaning there's greater likelihood of observing values far from the mean. This characteristic is crucial in providing more reliable estimates when we can’t meet the normality assumption due to our small sample size.

• The shape of the \(t\)-distribution depends on degrees of freedom (\(df\)), calculated as sample size minus one (\(n-1\)). In our case, \(df=17\).
• Lower \(df\) leads to heavier tails, accounting for more variability in small samples.
• As \(df\) increases, the \(t\)-distribution approximates a normal distribution.

Using the \(t\)-distribution, we can derive the test statistic and critical values needed to evaluate our hypothesis: determining if the observed mean difference is statistically significant.