Respuesta :
Answer:
b) independent samples t-test
[tex]t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154[/tex]
[tex]p_v =2*P(t_{8}>0.154) =0.881[/tex]
So with the p value obtained and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).
Step-by-step explanation:
The system of hypothesis on this case are:
Null hypothesis: [tex]\mu_1 = \mu_2[/tex]
Alternative hypothesis: [tex]\mu_1 \neq \mu_2[/tex]
Or equivalently:
Null hypothesis: [tex]\mu_1 - \mu_2 = 0[/tex]
Alternative hypothesis: [tex]\mu_1 -\mu_2\neq 0[/tex]
Our notation on this case :
[tex]n_1 =5[/tex] represent the sample size for group 1
[tex]n_2 =5[/tex] represent the sample size for group 2
We can calculate the mean and deviation for eaach group with the following formulas:
[tex]\bar X= \frac{\sum_{i=1}^n X_i}{n}[/tex]
[tex] s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]
[tex]\bar X_1 =5.92[/tex] represent the sample mean for the group 1
[tex]\bar X_2 =5.74[/tex] represent the sample mean for the group 2
[tex]s_1=1.88[/tex] represent the sample standard deviation for group 1
[tex]s_2=1.81[/tex] represent the sample standard deviation for group 2
If we see the alternative hypothesis we see that we are conducting a bilateral test and with independnet samples t test.
b) independent samples t-test
The statistic is given by this formula:
[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{S^2_2}{n_2}}[/tex]
And now we can calculate the statistic:
[tex]t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154[/tex]
The degrees of freedom are given by:
[tex]df=5+5-2=8[/tex]
And now we can calculate the p value using the altenative hypothesis:
[tex]p_v =2*P(t_{8}>0.154) =0.881[/tex]
So with the p value obtained and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).
