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In statistics, when comparing two sample means and assuming that both samples are drawn from populations with equal variances, we often use the concept of pooled variance. The pooled variance is a way of estimating the common variance between the samples by taking a weighted average of the variances of each sample. This method is crucial in a two-sample t-test as it provides a more accurate measure of variability.

To calculate the pooled variance, use the formula: \[s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}\] where:

• \(n_1\) and \(n_2\) are the sample sizes of the two groups.
• \(s_1^2\) and \(s_2^2\) are the sample variances of the two groups.
• The denominator \(n_1+n_2-2\)is the total degrees of freedom.

By pooling the variances, we assume that both populations have the same variance, allowing us to perform a two-sample t-test. This provides us with more statistical power, increasing the chances of correctly detecting a difference between the means.

A confidence interval provides a range of values within which the true difference between population means is likely to fall, based on sample data. Constructing a confidence interval allows us to estimate this difference with an associated level of confidence, usually expressed as a percentage like 95%.

For a 95% confidence interval using the pooled variance, the formula is: \[(\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2} \cdot s_p \cdot \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}\] Here:

• \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means.
• \(t_{\alpha/2}\) is the critical value from the t-distribution for the desired confidence level.
• \(s_p\) is the pooled standard deviation, the square root of the pooled variance.
• The expression \(\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}\) represents the standard error of the difference between means.

The resulting interval will include zero if there's no significant difference in means. For our example, the 95% confidence interval indicates that the true difference in mean rod diameters could be anywhere from -0.1 to 0.2, suggesting that there's no significant difference.

Hypothesis testing is a statistical method used to make decisions about the properties of a population, based on sample data. It's widely used to test assumptions and determine whether differences observed in the sample are statistically significant.

The process begins with defining two hypotheses:

• Null Hypothesis \(H_0\): This is the default assumption that there is no difference or effect. In our scenario, it suggests \(\mu_1 = \mu_2\) (the means are equal).
• Alternative Hypothesis \(H_a\): This suggests the presence of a difference or effect. Here, \(\mu_1 eq \mu_2\) (the means are not equal).

The next step is to calculate the test statistic, which measures how far the data deviates from the null hypothesis. We then compare this statistic to a critical value from the t-distribution, determined by the significance level \((\alpha)\).

In our example, with an \(\alpha\) of 0.05, the critical values were ±2.042. Our test statistic was 0.5202, which fell inside this range, leading us to "fail to reject" the null hypothesis. This outcome indicates that there is insufficient evidence to claim that the machines produce rods of different mean diameters. Consequently, hypothesis testing gives us a structured framework for making informed decisions based on sample data.