Question

1. two separate samples, each with ( n = 9 ) individuals, receive different treatments. after treatment, the first sample has ( ss = 546 ) and the second has ( ss = 606 ). a. find the pooled variance for the two samples. b. compute the estimated standard error for the sample mean difference.

Explanation:

Step1: Calculate degrees of freedom for each sample

For a sample, the degrees of freedom \(df=n - 1\). Given \(n_1=n_2 = 9\), then \(df_1=9 - 1=8\) and \(df_2=9 - 1=8\)

Step2: Calculate pooled variance (\(s_p^2\))

The formula for pooled variance is \(s_p^2=\frac{SS_1+SS_2}{df_1 + df_2}\). Substitute \(SS_1 = 546\), \(SS_2=606\), \(df_1 = 8\), \(df_2=8\) into the formula:

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Step3: Calculate the estimated standard error (\(s_{M_1 - M_2}\))

The formula for the estimated standard error for the sample mean difference is \(s_{M_1 - M_2}=\sqrt{\frac{s_p^2}{n_1}+\frac{s_p^2}{n_2}}\). Since \(n_1=n_2 = 9\) and \(s_p^2 = 72\)

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Answer:

a. The pooled variance \(s_p^2 = 72\)
b. The estimated standard error \(s_{M_1 - M_2}=4\)