A standardized math test was administered to two groups of 5th graders, one group sampled from classes whose teachers had followed the existing curriculum and one group sampled from classes whose teachers had followed a new curriculum, and the scores were compared using a t-test with the following results: Old Curriculum New Curriculum Mean 75.31914894 65.65957447 Variance 118.613321 801.1424607 Observations 47 47 Pooled Variance 459.8778908 Hypothesized Mean Difference 0 df 92 t Stat 2.1835887807 P(T<=)one tail 0.015768226 t Critical one-tail 1.661585397 P(T<=t) two-tail 0.031536452 t Critical two-tail 1.986086317 You decide to use the conventional p=.05 as your cutoff for statistical significance. What do you conclude from this analysis? a. The wrong t-test was conducted because of the relative sizes of the two groups' variances. b. We cannot be sufficiently confident that there is a relationship between the math curriculum followed and students' test scores.
A standardized math test was administered to two groups of 5th graders, one group sampled from classes whose teachers had followed the existing curriculum and one group sampled from classes whose teachers had followed a new curriculum, and the scores were compared using a t-test with the following results: Old Curriculum New Curriculum
Variance 118.613321 801.1424607
Observations 47 47
Pooled Variance 459.8778908
Hypothesized Mean Difference 0
df 92
t Stat 2.1835887807
P(T<=)one tail 0.015768226
t Critical one-tail 1.661585397
P(T<=t) two-tail 0.031536452
t Critical two-tail 1.986086317
You decide to use the conventional p=.05 as your cutoff for statistical significance. What do you conclude from this analysis?
a. The wrong t-test was conducted because of the relative sizes of the two groups' variances.
b. We cannot be sufficiently confident that there is a relationship between the math curriculum followed and students' test scores.
The t-test is based on the t-value, which is nothing but a ratio of the sample mean difference and the standard error of this mean difference. Therefore, the standard error is a function of the variance of the data that is the variance "within-group". And if the "within-group" variances cannot be assumed equal in both groups, it is not clear how the standard error should be calculated which is known as the Behrens-Fisher problem.
One may use the average variance as one would also do if the assumption of equal variances makes sense, to get a better estimate but that, one has to account for the additional uncertainty of the standard error as well. This is done by the Welch–Satterthwaite adjustment of the degrees of freedom known as Welch t-test.
Therefore, the correct t-test should be Welch t-test
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