Respuesta :
Answer:
[tex]z=\frac{0.512-0.500}{\sqrt{0.505(1-0.505)(\frac{1}{3025}+\frac{1}{2150})}}=0.851[/tex]
[tex]p_v =P(Z>0.851)= 0.197[/tex]
Comparing the p value with the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't say that the proportion of students who use the test increasse significantly the scores compared to the group who do not use the test.
Step-by-step explanation:
Data given and notation
[tex]X_{1}=1513[/tex] represent the number of students who had not used the program and increase the scores
[tex]X_{2}=1100[/tex] represent the number of students who had used the program and increase the scores
[tex]n_{1}=3025[/tex] sample of students who not ue the program
[tex]n_{2}=2150[/tex] sample of students who use the program
[tex]p_{1}=\frac{1513}{3025}=0.500[/tex] represent the proportion of students who had not used the program and increase the scores
[tex]p_{2}=\frac{1100}{2150}=0.512[/tex] represent the proportion of students who had used the program and increase the scores
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the value for the test (variable of interest)
[tex]\alpha=0.05[/tex] significance level given
Concepts and formulas to use
We need to conduct a hypothesis in order to check if the online test preparation company's students were more likely to increase their scores on the SAT exam , the system of hypothesis would be:
Null hypothesis:[tex]p_{2} - p_{1} \leq 0[/tex]
Alternative hypothesis:[tex]p_{2} - \mu_{1} > 0[/tex]
We need to apply a z test to compare proportions, and the statistic is given by:
[tex]z=\frac{p_{2}-p_{1}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{2}}+\frac{1}{n_{1}})}}[/tex] (1)
Where [tex]\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{1513+1100}{3025+2150}=0.505[/tex]
z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.
Calculate the statistic
Replacing in formula (1) the values obtained we got this:
[tex]z=\frac{0.512-0.500}{\sqrt{0.505(1-0.505)(\frac{1}{3025}+\frac{1}{2150})}}=0.851[/tex]
Statistical decision
Since is a right tailed side test the p value would be:
[tex]p_v =P(Z>0.851)= 0.197[/tex]
Comparing the p value with the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't say that the proportion of students who use the test increasse significantly the scores compared to the group who do not use the test.
Answer:
0.2082
Step-by-step explanation:
I took the test and got it right
