Question

A random sample of n 1 equals 135n1=135 individuals results in x 1 equals 40x1=40 successes. An independent sample of n 2 equals 150n2=150 individuals results in x 2 equals 60x2=60 successes. Does this represent sufficient evidence to conclude that p 1 less than p 2p1

Answer 1

`p_v =P(Z<-1.837)=0.033`

If we compare the p value with any significance level for example
`\alpha=0.05,0.1` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the true proportion 1 is less than the true proportion 2, at 5% or 10% of significance .

Explanation:

1) Data given and notation

`X_(1)=40` represent the number of successes for 1

`X_(2)=60` represent the number of successes for 2

`n_(1)=135` sample of 1 selected

`n_(2)=150` sample of 2 selected

`\hat p_(1)=(40)/(135)=0.296` represent the sample proportion for 1

`\hat p_(2)=(60)/(150)=0.40` represent the sample proportion 2

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the proportion 1 is less than the proportion 2, the system of hypothesis would be:

Null hypothesis:
`p_(1) \geq p_(2)`

Alternative hypothesis:
`p_(1) < p_(2)`

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

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

Where
`\hat p=(X_(1)+X_(2))/(n_(1)+n_(2))=(40+60)/(135+150)=0.3509`

3) Calculate the statistic

Replacing in formula (1) the values obtained we got this:

`z=\frac{0.296-0.40}{\sqrt{0.3509(1-0.3509)((1)/(135)+(1)/(150))}}=-1.837`

4) Statistical decision

For this case we don't have a significance level provided
`\alpha`, but we can calculate the p value for this test.

Since is a one left tailed test the p value would be:

`p_v =P(Z<-1.837)=0.033`

If we compare the p value with any significance level for example
`\alpha=0.05,0.1` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the true proportion 1 is less than the true proportion 2, at 5% or 10% of significance .