Respuesta :
Answer:
[tex]t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656[/tex]
[tex]df=6+5-2=9[/tex]
[tex]p_v =P(t_{9}>2.656) =0.0131[/tex]
Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.
Step-by-step explanation:
Data given
[tex]n_1 =6[/tex] represent the sample size for group 1
[tex]n_2 =5[/tex] represent the sample size for group 2
[tex]\bar X_1 =50[/tex] represent the sample mean for the group 1
[tex]\bar X_2 =38[/tex] represent the sample mean for the group 2
[tex]s_1=7[/tex] represent the sample standard deviation for group 1
[tex]s_2=8[/tex] represent the sample standard deviation for group 2
System of hypothesis
The system of hypothesis on this case are:
Null hypothesis: [tex]\mu_1 \leq \mu_2[/tex]
Alternative hypothesis: [tex]\mu_1 > \mu_2[/tex]
We are assuming that the population variances for each group are the same
[tex]\sigma^2_1 =\sigma^2_2 =\sigma^2[/tex]
The statistic for this case is given by:
[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}[/tex]
The pooled variance is:
[tex]S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}[/tex]
We can find the pooled variance:
[tex]S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67[/tex]
And the pooled deviation is:
[tex]S_p=7.46[/tex]
The statistic is given by:
[tex]t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656[/tex]
The degrees of freedom are given by:
[tex]df=6+5-2=9[/tex]
The p value is given by:
[tex]p_v =P(t_{9}>2.656) =0.0131[/tex]
Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.
