Question

In an experiment with oat, the mean yield was 52.8 bushels per acre and the experimental

error variance was 20.31 with 36 df. A similar experiment under somewhat different
circumstances might have ȳ = 48.4 and s
2
= 45.06with 24 df.

Assuming the error variances are not homogeneous, test the null hypothesis that the
two population means are equal using a t-test at the 95% level of confidence.

Answer 1

To perform a t-test for the equality of two population means with unequal variances, we can use the Welch's t-test. In this test, the formula for the t-statistic is:

t = (x1 - x2) / sqrt(s1^2/n1 + s2^2/n2)

Where x1 and x2 are the sample means, s1^2 and s2^2 are the sample error variances, and n1 and n2 are the degrees of freedom for the two samples.

Plugging in the given values, we have:

t = (52.8 - 48.4) / sqrt(20.31/36 + 45.06/24) = 4.25

Next, we need to find the critical t-value from the t-distribution table with degrees of freedom equal to the smaller of n1 and n2, which is 24. With a 95% confidence level, the critical t-value is approximately 2.492.

Since our calculated t-statistic (4.25) is greater than the critical t-value (2.492), we reject the null hypothesis that the two population means are equal. This suggests that there is strong evidence that the mean yield of oats is different between the two experiments.

Explanation: