Two Means - Unknown, Unequal Variance: Videos & Practice Problems

In hypothesis testing involving two samples, the primary focus shifts from a single mean to the difference between two sample means. The initial step involves formulating the null hypothesis, which posits that the two means are equal, expressed as \( H_0: \mu_1 = \mu_2 \). Alternatively, this can be represented as \( H_0: \mu_1 - \mu_2 = 0 \). The alternative hypothesis, \( H_a \), typically suggests that the means are not equal, leading to a two-tailed test.

Before proceeding with calculations, it is essential to verify certain conditions: the samples must be random and independent, the population standard deviations (\( \sigma_1 \) and \( \sigma_2 \)) are assumed unknown and unequal, and the samples should either be normally distributed or sufficiently large. In cases where the sample sizes are small, normality is assumed.

The test statistic for a two-sample t-test is calculated using the formula:

\( t = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \)

Here, \( \bar{x}_1 \) and \( \bar{x}_2 \) are the sample means, \( s_1 \) and \( s_2 \) are the sample standard deviations, and \( n_1 \) and \( n_2 \) are the sample sizes. For the null hypothesis, the difference in population means (\( \mu_1 - \mu_2 \)) is zero, simplifying the equation to:

\( t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \)

After calculating the t-value, the next step is to determine the p-value, which indicates the probability of observing the test statistic under the null hypothesis. For two samples, the degrees of freedom can be approximated by taking the smaller of the two sample sizes minus one, \( df = \min(n_1, n_2) - 1 \).

In a two-tailed test, the p-value is calculated as:

\( p = 2 \cdot P(T \leq -|t|) \)

Finally, the p-value is compared to the significance level (\( \alpha \)). If the p-value is less than \( \alpha \), the null hypothesis is rejected, indicating sufficient evidence to support the alternative hypothesis. For instance, if \( \alpha = 0.05 \) and the calculated p-value is \( 0.0005 \), the conclusion would be to reject the null hypothesis, suggesting a significant difference in means, such as the resting heart rates between males and females.

In hypothesis testing, particularly when comparing two population means, it's essential to follow a structured approach even when the context of the numbers is unclear. The first step involves formulating the null hypothesis (H₀) and the alternative hypothesis (H_a). In this scenario, the claim is that the mean of the first population (μ₁) is greater than the mean of the second population (μ₂). Therefore, the null hypothesis is that the means are equal (H₀: μ₁ = μ₂), while the alternative hypothesis is that μ₁ > μ₂.

Next, we calculate the test statistic using the t-score formula, which is appropriate when the population standard deviations are unknown and unequal. The t-score is calculated as follows:

t = \frac{\bar{x}_1 - \bar{x}_2 - 0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}

Here, \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means, \(s_1\) and \(s_2\) are the sample standard deviations, and \(n_1\) and \(n_2\) are the sample sizes. For this example, substituting the values gives:

t = \frac{462 - 431 - 0}{\sqrt{\frac{67^2}{32} + \frac{85^2}{19}}} = 1.359

After calculating the t-score, the next step is to determine the p-value, which represents the probability of observing a t-score as extreme as 1.359 under the null hypothesis. Since this is a right-tailed test, we look for the area to the right of the t-score in the t-distribution. With 18 degrees of freedom (the smaller of the two sample sizes minus one), the p-value is found to be approximately 0.0955.

Finally, we compare the p-value to the significance level (α = 0.1). Since the p-value (0.0955) is less than α, we reject the null hypothesis. This indicates that there is sufficient evidence to support the claim that μ₁ is greater than μ₂. In conclusion, despite the lack of context for the numbers, the statistical analysis suggests a significant difference between the two population means.

In statistical analysis, when comparing two samples, constructing a confidence interval for the difference in means is a crucial step. This process is similar to creating a confidence interval for a single mean, with some modifications to the point estimator and margin of error. The point estimator for the difference in means is calculated as the difference between the sample means, denoted as \( \bar{x}_1 - \bar{x}_2 \).

To begin, ensure that the samples are random and independent. For instance, if you are studying the mean resting heart rates of males and females, you can assume independence since the samples do not interact. Next, check that the populations are normally distributed or that the sample sizes are sufficiently large. If the sample sizes are small, as in the case of 10 males and 11 females, you can still assume normality if the population distributions are normal.

The next step involves finding the critical value, \( t_{\alpha/2} \), based on the desired confidence level. For a 90% confidence interval, you would look for the t-value that corresponds to 5% in each tail of the distribution. The degrees of freedom for this calculation is determined by the smaller sample size minus one. In this example, with 10 males, the degrees of freedom is 9, leading to a critical t-value of approximately 1.833.

Now, calculate the margin of error using the formula:

\[E = t_{\alpha/2} \times \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\]

Here, \( s_1 \) and \( s_2 \) are the sample standard deviations, and \( n_1 \) and \( n_2 \) are the sample sizes. For example, if the standard deviations are 5.8 for males and 6.4 for females, the margin of error can be computed accordingly. After calculating, you might find a margin of error of 4.88.

With the point estimator and margin of error determined, you can establish the confidence interval by subtracting and adding the margin of error to the point estimator. For instance, if the point estimator is -11.2, the lower bound would be -16.08 and the upper bound would be -6.32, resulting in a 90% confidence interval of (-16.08, -6.32).

This interval suggests that we are 90% confident that the true difference in mean resting heart rates between males and females lies within these bounds. To evaluate a claim regarding the equality of means, consider the null hypothesis \( H_0: \mu_1 - \mu_2 = 0 \). If the confidence interval does not include zero, it indicates a significant difference between the means, leading to the rejection of the null hypothesis. In this case, since the interval (-16.08, -6.32) does not include zero, we reject the null hypothesis, suggesting that there is a significant difference in mean resting heart rates between the two groups.