Respuesta :
Answer:
[tex]t=\frac{(1085 -1034)-(0)}{57.646\sqrt{\frac{1}{10}+\frac{1}{15}}}=2.167[/tex]
[tex]p_v =2*P(t_{23}<-2.167) =0.0408[/tex]
We see that the p value is lower than the significance level provided of 0.05 so then we have enough evidence to conclude that the true means are different at 5% of significance.
Step-by-step explanation:
Data given
[tex]n_1 =10[/tex] represent the sample size for group 1
[tex]n_2 =15[/tex] represent the sample size for group 2
[tex]\bar X_1 =1085[/tex] represent the sample mean for the group 1
[tex]\bar X_2 =1034[/tex] represent the sample mean for the group 2
[tex]s_1=52[/tex] represent the sample standard deviation for group 1
[tex]s_2=61[/tex] represent the sample standard deviation for group 2
We are assuming that the two samples are normally distributed with equal variances and that means:
[tex]\sigma^2_1 =\sigma^2_2 =\sigma^2[/tex]
System of hypothesis
Null hypothesis: [tex]\mu_1 = \mu_2[/tex]
Alternative hypothesis: [tex]\mu_1 \neq \mu_2[/tex]
The statistic 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 degrees of freedom are given by:
[tex]n_1+n_2 -2[/tex]
And 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]
Replacing we got:
[tex]S^2_p =\frac{(10-1)(52)^2 +(15 -1)(61)^2}{10 +15 -2}=3323.043[/tex]
And the deviation would be:
[tex]S_p=57.646[/tex]
The degrees of freedom are:
[tex]df=10+15-2=23[/tex]
The statistic would be:
[tex]t=\frac{(1085 -1034)-(0)}{57.646\sqrt{\frac{1}{10}+\frac{1}{15}}}=2.167[/tex]
The p value would be
[tex]p_v =2*P(t_{23}<-2.167) =0.0408[/tex]
We see that the p value is lower than the significance level provided of 0.05 so then we have enough evidence to conclude that the true means are different at 5% of significance.