First of all, dispersion is measured by standard deviation. If we don't consider the difference in the number of students in grade two, the standard deviation of grades in grade five is largely discrete.

Second, the usual grades: 3*80-75-70=95, and the total grade =95*0.2+75*0.3+70*0.5=76.5.

3. Average score =8 1.4, difference is 1, 8, 9, 6, 7, 9, 14, 6, 2, and average score difference =7.

Four, 20 groups all add up to = 1396, divided by 20, the standard deviation is 9.4735 1, the two-sided test p value is 0.035, the original hypothesis is rejected, and the average score is different. The following is a t-value checklist.

Single sample test

Test value? =? 65

Differential? 95%? Confidence interval

t？ df？ Sign. (Bilateral)? Mean difference? Lower limit? upper limit

VAR0002 1？ 2.266? 19? .0354.80000? .3663? 9.2337

Verb (abbreviation of verb) The formula for testing the difference of the population mean of two matched samples is as follows, where d refers to the difference of each matched sample. The original assumption is μ 1=μ2.

D =4

sd= 1/(n- 1)*[∑(x 1i-x2i)^2+4^2]

Correlation coefficient r = [∑ x1i * x2i-∑ 70 * 74/n]/7 * 9 = 0.62.

∴∑x 1i*x2i=286

And ∑ x12 = 7 2+70 2 = 4949.

∑x2^2=9^2+74^2=5557

∴Sd=459.85

t=0.0389

The significance is 0.05, and the T-finger of 20 degrees of freedom is 2.086. The original hypothesis was not rejected, and there was no significant difference between the two grades.