Question

A teacher asserts that students do not do as well in class using Method A in the same amount of time as they would using Method B. Seven students are randomly selected. Each student’s grades are recorded by the use of Method A and Method B.

Student	Method A	Method B
1	33	20
2	90	40
3	90	55
4	60	80
5	100	100
6	90	95
7	80	90

1. Specify the null and alternate hypotheses to test the teacher’s assertion.
2. At the 1% significance level, specify the critical value(s) and decision rule.
3. Assuming the completion difference of the methods is normally distributed, calculate the value of the test statistic.
4. Is the teacher’s assertion supported by the data?

Answer 1

we have that, from the probleme Student method A Method B 20 33 90 yo ss Go 80 100 loo 90 80 90 calculation of null hypotheses ; - Here, null hypotheses is to Ho: Mo=0. Null bypotheses is there is no significant mean difference between completing the job using method A & B. Alternative hypotheses :- Her , Alternative hypotheses is Ha. Ha: lo to

the two Hence to is the mean time difference at methods (A&B). Hypotheses are specified to test the meanin managing assestion. critical value - 1.1. at level of significa we know that, critical vale at 1% significance level Two tailed test degree of freedom is df = n-1=7-1 (n=7). de = 6 t-critical value for a two tailed dest, to- significance live at X=010) tc = -3.707 , 6c = 3.707. Mean St.Dev. SE mean. method O z 77-571 21.6059 8.166 10 724 method-B 28.374 7 - 6 769 -2.558 Difference o

calculation of test statistics - 99-1 (I for mean diftunce (-26.505 LMO 244.505). The t-statistics = 0.94. P-value = 0.3836 C. P = 0.383620.01). finally we conclude that 161=o.auLtc = 3:707. at is then calculated that the null hypothesis is not rejected.