Questions: When testing the difference between means, is the data pooled or not pooled for the following data? Claim: μ₁=μ₂ ; α=0.05. Assume σ₁²=σ₂² Sample statistics: x̄₁=45.7, s₁=4.2, n₁=15 x̄₂=48.7, s₂=3.2, n₂=12 Pooled Not Pooled

Solution Steps

The standard error (SE) for the pooled variance t-test is calculated using the formula:

\[ SE = \sqrt{s_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} \]

Given that \(s_p^2\) (the pooled variance) is derived from the sample standard deviations, we find that:

\[ SE = 0.0 \]

The test statistic \(t\) is calculated using the formula:

\[ t = \frac{\bar{x}_1 - \bar{x}_2}{SE} \]

Substituting the values:

\[ t = \frac{45.7 - 48.7}{0.0} = -1048995303150328.2 \]

The degrees of freedom \(df\) for the t-test is calculated as:

\[ df = n_1 + n_2 - 2 = 15 + 12 - 2 = 25 \]

The P-value is calculated using the formula:

\[ P = 2(1 - T(|t|)) \]

Given that \(t = -1048995303150328.2\):

\[ P = 2(1 - T(1048995303150328.2)) = 0.0 \]

For a significance level of \(\alpha = 0.05\) and \(df = 25\), the critical value from the t-distribution table is:

\[ \text{Critical Value} = 2.0595 \]

Since the P-value \(0.0\) is less than \(\alpha = 0.05\), we reject the null hypothesis. This indicates that there is a statistically significant difference between the means of the two populations.

Final Answer