Question

A random sample of 26 local sociology graduates scored an average of 458 on the GRE advanced sociology test, with a standard deviation of 20 . Is this significantly different from the national average \((\mu=440) ?\)

Short answer

The sample mean is significantly different from the national average.

Step-by-step solution

- State the Hypotheses

First, we need to set up our null hypothesis \(H_0\) and the alternative hypothesis \(H_a\). The null hypothesis will state that the sample mean is equal to the population mean, and the alternative hypothesis will state that the sample mean is different from the population mean. \[H_0: \mu = 440 \] \[H_a: \mu e 440 \]

- Determine the Test Statistic

Next, we calculate the test statistic using the formula for the t-test: \[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \] Where: \(\bar{x} = 458\), \(\mu = 440\), \(s = 20\), and \(n = 26\). Plugging in the values: \[ t = \frac{458 - 440}{20/\sqrt{26}} \]

- Calculate the Test Statistic Value

Perform the calculations: \[ t = \frac{18}{20/\sqrt{26}} = \frac{18}{3.92} \approx 4.59 \]

- Determine the Degrees of Freedom

The degrees of freedom \text{(df)} are calculated as: \(\text{df} = n - 1 = 26 - 1 = 25\)

- Determine the Critical Value

Consult a t-distribution table at a chosen significance level (commonly \(\alpha = 0.05\) for a two-tailed test) with 25 degrees of freedom. The critical value \(t_c\) for \(\alpha = 0.05\) is approximately \(t_c = 2.060\).

- Compare the Test Statistic to the Critical Value

Compare the calculated t-value to the critical t-value: \[t = 4.59 > t_c = 2.060\]

- Conclusion

Since the calculated t-value (4.59) is greater than the critical t-value (2.060), we reject the null hypothesis \(H_0\). The sample mean is significantly different from the national average.

Key concepts

Null Hypothesis

The null hypothesis, often symbolized as \( H_0 \), is a fundamental concept in hypothesis testing. It reflects the idea that there is no significant difference or effect present in the data. In our exercise, the null hypothesis states that the sample mean is equal to the national average: \[ H_0: \mu = 440 \] This assumption is our starting point. Until we find substantial evidence against it, we assume it to be true. Rejection of the null hypothesis suggests that the observed data is unlikely to have occurred under this assumption.

Alternative Hypothesis

The alternative hypothesis (\( H_a \)) represents what we seek to prove. It contradicts the null hypothesis by suggesting that there is indeed a significant difference or effect. In the given exercise, the alternative hypothesis is: \[ H_a: \mu \e 440 \] This implies that the average score of 458 for local sociology graduates is significantly different from the national average of 440. When our statistical test supports the alternative hypothesis, we can infer that there is a true effect or difference.

Degrees of Freedom

Degrees of freedom (\text{df}) are crucial in determining the reliability of our test statistic. It essentially accounts for the number of values that are free to vary in our calculations. For a sample size (\text{n}) of 26, the degrees of freedom are calculated as: \[ \text{df} = n - 1 \] Hence, \text{df} = 26 - 1 = 25 \. These degrees of freedom help us reference the correct t-distribution table to find the critical value for hypothesis testing.

Critical Value

The critical value is the threshold against which our test statistic is compared. It depends on the chosen significance level (\text{\alpha}), such as 0.05 for a 5% significance level, and the degrees of freedom. In our case, the critical value from the t-distribution table with \text{df} = 25 \ and \alpha = 0.05\ for a two-tailed test is approximately \text{t_c} = 2.060\. If the calculated test statistic exceeds this critical value, we reject the null hypothesis. In our example: \[ t = 4.59 \gt \text{t_c} = 2.060 \] thus, leading to the rejection of the null hypothesis.

Standard Deviation

Standard deviation (\text{s}) measures the spread or dispersion of a set of data points. In the given exercise, the standard deviation is 20. It helps in understanding how much the individual data points deviate from the sample mean. Standard deviation is essential in the formula for calculating the t-test statistic: \[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \] where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, \text{s} is the standard deviation, and \text{n} is the sample size. By considering the standard deviation, we account for the variability in our data, making our test statistic more accurate and reliable.