# Contradiction between confidence intervals and Z-test

I've got the problem with Z-test and confidence intervals. Z-test rejects null hipothesis (proportions are equal), but confidence intervals are intersect.

Let us consider 2 companies. The proportion of smokers in company A is p(A)=0.22. And the proportion of smokers in company B is p(B)=0.303. The amount of employees in company A is 403. And the amount of employees in company B is 404.

So, we have the folowing data:

p(A)=0.22
n(A)=403

p(B)=0.303
n(B)=404

Now let's apply the Z-test by the following forumula for proportions (I've took the formula from https://onlinecourses.science.psu.edu/stat414/node/268):

$$Z= \frac{p(B)-p(A)}{\sqrt{p(1-p)(\frac{1}{n(A)}+\frac{1}{n(b)})}}$$

Where p is $$p=\frac{p(A)n(A)+p(B)n(B)}{n(A)+n(B)}$$

I've made the following computation: $$p=\frac{0.22*403+0.303*404}{403+404}=0.2614$$

$$z= \frac{0.303-0.22}{\sqrt{0.2614(1-0.2614)(\frac{1}{403}+\frac{1}{404})}}=2.6828$$

So, z more than 1.96, it means that we reject the null hipothesis. The proportions p(A) and p(B) are not equal.

But then I've computed the 5% confidence intervals by the following formulas (I've took them form https://onlinecourses.science.psu.edu/stat200/node/48):

$$low=P-1.96 \sqrt{\frac{P*(1-P)}{N}}$$ $$high=P+1.96 \sqrt{\frac{P*(1-P)}{N}}$$ Where P is proportion of smokers in some company and N is amount of employees in this company.

I've computed the interval for company A: $$\text{low}(A)=p(A)-1.96 \sqrt{{p(A)(1-p(A))}{n(A)}}=0.22-1.96\sqrt{\frac{0.22\cdot 0.78}{403}}=0.1796$$ $$\text{high}(A)=p(A)+1.96 \cdot\sqrt{{p(A)(1-p(A))}{n(A)}}=0.22+1.96 \sqrt{\frac{0.22\cdot0.78}{403}}=0.2604$$ So, for $p(A)$ the CI is $(0.1796;0.2696)$.

And finally I've computed the interval for the company B: $$low(B)=p(B)-1.96*\sqrt{{p(B)(1-p(B))}{n(B)}} =0.303-1.96*sqrt{\frac{0.303*0.697}{404}}=0.2581$$ $$high(B)=p(B)+1.96*\sqrt{{p(B)(1-p(B))}{n(B)}} =0.303+1.96*sqrt{\frac{0.303*0.697}{404}}=0.3477$$ For p(B) the confidence interval is (0.2581;0.3477).

And now we see that confidence intervals are intersect, but it contradicts the computations for z-test.

Finally I've used a web service for z-test and CI computing. The service drawed intersection of CI's and rejected the null hipothesis of z-test too.