Answer:
Check the explanation
Step-by-step explanation:
Let [tex]\overline{x}[/tex] and [tex]\overline{y}[/tex] be sample means of white and Jesse denotes are two random variables.
Given that both samples are having normally distributed.
Assume [tex]\overline{x}[/tex] having with mean [tex]\mu_{1}[/tex] and [tex]\overline{y}[/tex] having mean [tex]\mu_{2}[/tex]
Also we have given the variance is constant
A)
We can test hypothesis as
[tex]H0: \mu_{1} = \mu_{2}H1: \mu_{1} > \mu_{2}[/tex]
For this problem
Test statistic is
[tex]T=\frac{\overline {x}-\overline {y}}{s\sqrt{\frac {1}{n1} +\frac{1}{n2}}}[/tex]
Where
[tex]s=\sqrt{\frac{(n1-1)*s1^{2}+(n2-1)*s2^{2}}{n1+n2-2}}[/tex]
We have given all information for samples
By calculations we get
s=2.41
T=2.52
Here test statistic is having t-distribution with df=(10+7-2)=15
So p-value is P(t15>2.52)=0.012
Here significance level is 0.05
Since p-value is <0.05 we are rejecting null hypothesis at 95% confidence.
We can conclude that White has significant higher mean than Jesse. This claim we can made at 95% confidence.