Respuesta :
Answer:
a. Null hypothesis: [tex]\mu_1 \leq \mu_2[/tex]
Alternative hypothesis: [tex]\mu_1 >\mu_2[/tex]
b. [tex]t=\frac{(92.88 -86.90)-(0)}{\sqrt{\frac{15.34^2}{78}}+\frac{8.99^2}{21}}=2.282[/tex]
c. [tex]p_v =P(t_{97}>2.287) =0.0122[/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 reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).
Step-by-step explanation:
a. State and label the null and alternative hypotheses.
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]
Or equivalently:
Null hypothesis: [tex]\mu_1 - \mu_2 \leq 0[/tex]
Alternative hypothesis: [tex]\mu_1 -\mu_2>0[/tex]
Our notation on this case :
[tex]n_1 =78[/tex] represent the sample size for group 1
[tex]n_2 =21[/tex] represent the sample size for group 2
[tex]\bar X_1 =92.88[/tex] represent the sample mean for the group 1
[tex]\bar X_2 =86.90[/tex] represent the sample mean for the group 2
[tex]s_1=15.34[/tex] represent the sample standard deviation for group 1
[tex]s_2=8.99[/tex] represent the sample standard deviation for group 2
b. State the value of the test statistic.
And 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]
Where t follows a t distribution with [tex]n_1+n_2 -2[/tex] degrees of freedom. If we replace the values given we have:
[tex]t=\frac{(92.88 -86.90)-(0)}{\sqrt{\frac{15.34^2}{78}}+\frac{8.99^2}{21}}=2.282[/tex]
Now we can calculate the degrees of freedom given by:
[tex]df=78+21-2=97[/tex]
c. Find either the critical value(s) and draw a picture of the critical region(s) or find the P-value for this test. Indicate which method you are using: ( CIRCLE ONE: Critical value / P-value )
Method used: P value
And now we can calculate the p value using the altenative hypothesis, since it's a right tail test the p value is given by:
[tex]p_v =P(t_{97}>2.287) =0.0122[/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 reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).
