As can be seen from the topic, population variance is unknown and belongs to an independent sample, so the homogeneity test of variance should be carried out first.

Step one:

From mode A, the variance is 1=5.96, the standard deviation is s 1=2.44, and the average value is 1=3.57, df 1=6, n 1=7.

From mode B, variance 2=7.8, standard deviation s2=2.793, mean 2=5.44, df2=8, n2=9.

F= variance (large)/variance (small) =7.8/5.96= 1.309. Looking up the F value table (bilateral test), F0.05/2 (6 6,8) = 5.6.

1.309 & lt； 5.6, so the variance difference between the two groups of data is not obvious and can be regarded as equal.

Step two:

Hypothetical topic:

H0:μA=μB

H 1:μA≠μB

1. Joint variance: s? p=n 1 S？ 1+n2 S？ 2/n 1+N2-2 =(7 * 5.96+9 * 7.8)/(7+9-2)= 7.99

Second, the standard error of two groups of mean difference distribution:

Standard deviation formula of two groups of mean difference distribution

Equal to 1.4248.

Third, substitute the t test formula:

T=- 1.3 1。

Since df=n 1+n2-2, that is, 7+9-2= 15.

Difference T table, bilateral test, T0.05/2 (15) = 2.131.

l- 1.3 1l & lt； 2. 13 1, so the H0 hypothesis is selected, and it is concluded that there is no obvious difference in the average number of errors between mode A and mode B.