Question

Assume that you plan to use a significance level of alpha α equals =0.05 to test the claim that p1 equals = p2. Use the given sample sizes and numbers of successes to find the pooled estimate p. Round your answer to the nearest thousandth. n1 equals =​100, n2 equals =100 x1 equals =​42, x2 equals =45.

Answer 1

`z=\frac{0.42-0.45}{\sqrt{(0.42(1-0.42))/(100)+(0.45(1-0.45))/(100)}}=-0.428`

`p_v =2*P(Z<-0.428)=0.669`

And we can use the following R code to find it: "2*pnorm(-0.428)"

The p value is a very high value and using any significance given
`\alpha=0.05` always
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the two proportions are not significantly different.

Explanation:

1) Data given and notation

`X_1 =42` people with some characteristic desired on sample 1

`X_2 =45` people with some characteristic desired on sample 2

`n_(1)=100` sample selected 1

`n_(2)=100` sample selected 2

`p_(1)=(42)/(100)=0.42` represent the proportion estimated for 1

`p_(2)=(45)/(100)=0.45` represent the proportion estimated for 2

`\alpha=0.05` represent the significance level

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 for the two samples are different , the system of hypothesis would be:

Null hypothesis:
`p_(1) = p_(2)`

Alternative hypothesis:
`p_(1) \\eq 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{(p_1 (1-p_1))/(n_(1))+(p_2 (1-p_2))/(n_(2))}}` (1)

3) Calculate the statistic

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

`z=\frac{0.42-0.45}{\sqrt{(0.42(1-0.42))/(100)+(0.45(1-0.45))/(100)}}=-0.428`

4) Statistical decision

We can calculate the p value for this test.

Since is a two tailed test the p value would be:

`p_v =2*P(Z<-0.428)=0.669`

And we can use the following R code to find it: "2*pnorm(-0.428)"

The p value is a very high value and using any significance given
`\alpha=0.05` always
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the two proportions are not significantly different.