Question

A random sample of n 1n1equals=139139 individuals results in x 1x1equals=3737 successes. An independent sample of n 2n2equals=147147 individuals results in x 2x2equals=5858 successes. Does this represent sufficient evidence to conclude that p 1 less than p 2p1

Answer 1

`z=-2.32`

`p_v =P(Z<-2.32)= 0.010`

If we compare the p value and using any significance level for example
`\alpha=0.05` 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 proportion 1 is significant lower than the proportion 2 at 5% of significance.

Explanation:

1) Data given and notation

`X_(1)=37` represent the number of people with characteristic 1

`X_(2)=58` represent the number of people with characteristic 2

`n_(1)=139` sample 1 selected

`n_(2)=147` sample 2 selected

`p_(1)=(37)/(139)=0.266` represent the proportion of people with characteristic 1

`p_(2)=(58)/(147)=0.395` represent the proportion of people with characteristic 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{p_(1)-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))=(37+58)/(139+147)=0.332`

3) Calculate the statistic

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

`z=\frac{0.266-0.395}{\sqrt{0.332(1-0.332)((1)/(139)+(1)/(147))}}=-2.32`

4) Statistical decision

For this case we don't have a significance level provided
`\alpha` we can assuem it 0.05, and 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<-2.32)= 0.010`

So if we compare the p value and using any significance level for example
`\alpha=0.05` 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 proportion 1 is significant lower than the proportion 2 at 5% of significance.