Question

A new online test preparation company compared 3,025 students who had not used its program with 2,150 students who had. Of those students who did not use the online test preparation program, 1,513 increased their scores on the SAT examination compared with 1,100 who did use the program. A significance test was conducted to determine whether there is evidence that the online test preparation company's students were more likely to increase their scores on the SAT exam. What is the p-value for an appropriate hypothesis test

Answer 1

0.2082

Explanation:

I took the test and got it right

Answer 2

`z=\frac{0.512-0.500}{\sqrt{0.505(1-0.505)((1)/(3025)+(1)/(2150))}}=0.851`

`p_v =P(Z>0.851)= 0.197`

Comparing the p value with the significance level assumed
`\alpha=0.05` we see that
`p_v>\alpha` 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.

Explanation:

Data given and notation

`X_(1)=1513` represent the number of students who had not used the program and increase the scores

`X_(2)=1100` represent the number of students who had used the program and increase the scores

`n_(1)=3025` sample of students who not ue the program

`n_(2)=2150` sample of students who use the program

`p_(1)=(1513)/(3025)=0.500` represent the proportion of students who had not used the program and increase the scores

`p_(2)=(1100)/(2150)=0.512` represent the proportion of students who had used the program and increase the scores

z would represent the statistic (variable of interest)

`p_v` represent the value for the test (variable of interest)

`\alpha=0.05` 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:
`p_(2) - p_(1) \leq 0`

Alternative hypothesis:
`p_(2) - \mu_(1) > 0`

We need to apply a z test to compare proportions, and the statistic is given by:

`z=\frac{p_(2)-p_(1)}{\sqrt{\hat p (1-\hat p)((1)/(n_(2))+(1)/(n_(1)))}}` (1)

Where
`\hat p=(X_(1)+X_(2))/(n_(1)+n_(2))=(1513+1100)/(3025+2150)=0.505`

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:

`z=\frac{0.512-0.500}{\sqrt{0.505(1-0.505)((1)/(3025)+(1)/(2150))}}=0.851`

Statistical decision

Since is a right tailed side test the p value would be:

`p_v =P(Z>0.851)= 0.197`

Comparing the p value with the significance level assumed
`\alpha=0.05` we see that
`p_v>\alpha` 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.