Question

Instead, suppose the public health authority chooses to leave the park open until it is proven, at a ten percent significance level, that MORE THAN two percent of mosquitoes in the park carry the virus. For a random sample of 1250 mosquitos, what is the unstandardized critical value? In other words, how large of a fraction of virus-carrying mosquitoes would need to be in the random sample to close the park given the desired ten percent significance level? (Answer as a proportion, not a percent. Record your answer accurate to at least the nearest THIRD decimal place with standard rounding.)

Answer 1

The public health authority chooses to leave the park open until it is proven, at a ten percent significance level, that MORE THAN two percent of mosquitoes in the park carry the virus.

So we need to find right tailed critical value ( unstandardized ) of proportion of  mosquitoes in the park carry the virus.

Under null hypothesis p = proportion of  mosquitoes in the park carry the virus = 0.02

n = sample size = 1250

For 10% level of significance the critical z value is as follow:

z_c = "=NORMSINV(0.9)" = 1.281552 ...{ by using excel command

The formula of unstandardized critical value of p is as follow:

So final answer is 0.025