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.
Second option: down on the left, down on the right.
Clearly when x is bigger than one the term -x^38 will dominate the result.
When absolute value of x grows, x^38 becomes a larger positive number and - x^38 a larger negative number. Then both ends of the function tend to negative infinity.
Ummm let me see so what u have to do is
I’m can’t freakin stay up I need to sleep and I need to get 4 papers done for my portfolio I can’t help but I wish you luck
