Answer:
a. H0:μ1≥μ2
Ha:μ1<μ2
b. t=-3.076
c. Rejection region=[tcalculated<−1.717]
Reject H0
Step-by-step explanation:
a)
As the score for group 1 is lower than group 2,
Null hypothesis: H0:μ1≥μ2
Alternative hypothesis: H1:μ1<μ2
b) t test statistic for equal variances
t=(xbar1-xbar2)-(μ1-μ2)/sqrt[{1/n1+1/n2}*{((n1-1)s1²+(n2-1)s2²)/n1+n2-2}
t=63.3-70.2/sqrt[{1/11+1/13}*{((11-1)3.7²+(13-1)6.6²)/11+13-2}
t=-6.9/sqrt[{0.091+0.077}{136.9+522.72/22}]
t=-3.076
c. α=0.05, df=22
t(0.05,22)=-1.717
The rejection region is t calculated<t critical value
t<-1.717
We can see that the calculated value of t-statistic falls in rejection region and so we reject the null hypothesis at 5% significance level.
We can solve this problem by using the distance formula. The distance formula is: 
We can now put in values and solve.
I believe the answer would be 18,618
25,600/11= 2,327
2327*8= 18,618
hope this helped
Answer:
B
Step-by-step explanation:
Given 2 quantities that vary directly, then the graph must pass through the origin.
Nikiya's graph is the only one to do this ⇒ B
