Question

A random sample of 423 Chinese Americans drawn from the population of a particular state has finished an average of \(12.7\) years of formal education, with a standard deviation of \(1.7\). Is this significantly different from the statewide average of \(12.2\) years?

Short answer

Yes, the mean difference is significant.

Step-by-step solution

Define the Hypotheses

Set up the null and alternative hypotheses. Null hypothesis (\text{H}_0): The sample mean is equal to the population mean, \ \text{H}_0: \ \text{H}_0: \ \ \text{H}_0\ Alternative hypothesis (\text{H}_a): The sample mean is not equal to the population mean, \ \text{H}_a: \text{}}.$$ H$$:

Determine the Test Statistic

Calculate the test statistic using the formula for the one-sample t-test: \( t = \frac{\overline{x} - \mu}{s / \sqrt{n}} \) Where: \( \overline{x} = 12.7 \text{ years} \mu = 12.2 \text{ years} s = 1.7 \text{ years} n = 423 \) \( t = \frac{12.7 - 12.2}{1.7 / \sqrt{423}} \)

Compute the Test Statistic

Calculate the value: \( t = \frac{12.7 - 12.2}{1.7 / \sqrt{423}} = \frac{0.5}{1.7 / 20.57} \approx \frac{0.5}{0.0827} \approx 6.05 \)

Determine the Degrees of Freedom

Find the degrees of freedom (df): \( df = n - 1 \) \( df = 423 - 1 = 422 \)

Find the Critical Value

Use a t-distribution table or calculator to find the critical value at a 0.05 significance level for a two-tailed test with 422 degrees of freedom. The critical t-value is approximately ±1.96.

Compare the Test Statistic to the Critical Value

Compare the calculated test statistic to the critical value: \( t = 6.05 \) Critical value: ±1.96 Since 6.05 > 1.96, reject the null hypothesis.

Conclusion

Since the test statistic exceeds the critical value, we reject the null hypothesis and conclude that the average years of formal education for Chinese Americans in this sample is significantly different from the statewide average of 12.2 years.

Key concepts

one-sample t-test

The one-sample t-test is a statistical method used to determine if the mean of a single sample is significantly different from a known or hypothesized population mean. It is particularly useful when dealing with small sample sizes or when the population standard deviation is unknown. In our example, we have a random sample of 423 Chinese Americans with an average of 12.7 years of formal education. We want to check if this is significantly different from the statewide average of 12.2 years. We begin by calculating the test statistic using the formula for the one-sample t-test:
\( t = \frac{\overline{x} - \mu}{s / \sqrt{n}} \)
Here, \( \overline{x} \) is the sample mean, \( \mu \) is the population mean, \( s \) is the sample standard deviation, and \( n \) is the sample size.

hypothesis testing

Hypothesis testing is a statistical method that allows us to make inferences or draw conclusions about a population based on sample data. It involves setting up two competing hypotheses:
\[ \text{Null hypothesis } (H_0): \overline{x} = \mu \ \]
\[ \text{Alternative hypothesis } (H_a): \overline{x} eq \mu \ \]
In our case, the null hypothesis (\(H_0\)) states that the average years of education for Chinese Americans is equal to the statewide average of 12.2 years. The alternative hypothesis (\(H_a\)) posits that the average years of education is different from 12.2 years. The goal is to use sample data to decide whether or not to reject the null hypothesis.

degrees of freedom

Degrees of freedom (df) are an important concept in hypothesis testing and refer to the number of values in a calculation that are free to vary. In the context of the one-sample t-test, the degrees of freedom are calculated as:
\( df = n - 1 \)
where \( n \) is the sample size. For our sample of 423 individuals, the degrees of freedom would be:
\[ df = 423 - 1 = 422 \]
Degrees of freedom impact the shape of the t-distribution, which in turn affects the critical values used in hypothesis testing.

significance level

The significance level (\( \alpha \)) is a threshold set by the researcher before conducting a hypothesis test. It represents the probability of rejecting the null hypothesis when it is actually true (Type I error). A common choice for the significance level is 0.05, which implies a 5% risk of making a Type I error. In our example, we are using a 0.05 significance level. This means that we are willing to accept a 5% chance of concluding that the average years of education for Chinese Americans is different from the statewide average when it actually isn't.

critical value

The critical value is a point on the test distribution compared to the test statistic to decide whether to accept or reject the null hypothesis. For a two-tailed test at a 0.05 significance level with 422 degrees of freedom, the critical value can be found using a t-distribution table or calculator. For simplicity, the approximate critical value is ±1.96. If the calculated test statistic is greater than the critical value in absolute terms, we reject the null hypothesis. In our example, the calculated test statistic is 6.05, which is much greater than 1.96. Therefore, we reject the null hypothesis and conclude that the difference in education levels is significant.