# Z test for proportion of ear infections.

Many elementary school students in a school district currently have ear infections. A random sample of children in two different schools found that 16 of 42 at one school and 21 of 36 at the other had this infection. At the .05 level of significance, is there sufficient evidence to conclude that a difference exists between the proportion of students who have ear infections at one school and the other?

A)No, there is not sufficient information to reject the hypothesis that the proportions of students at the two schools who have ear infections are the same because the test value -1.78 is inside the acceptance region (-1.96,1.96).

B)Yes, there is sufficient information to reject the hypothesis that the proportions of students at the two schools who have ear infections are the same because the test value -2.34 is outside the acceptance region (-1.96,1.96).

C)Yes, there is sufficient information to reject the hypothesis that the proportions of students at the two schools who have ear infections are the same because the test value -8.76 is outside the acceptance region (-1.96,1.96).

D)Yes, there is sufficient information to reject the hypothesis that the proportions of students at the two schools who have ear infections are the same because the test value -15.73 is outside the acceptance region (-1.96,1.96).

https://brainmass.com/statistics/z-test/z-test-for-proportion-of-ear-infections-236251

#### Solution Summary

The solution provides step by step method for the calculation of Z test for proportion of ear infections.. Formula for the calculation and Interpretations of the results are also included.