Answer: No, this is not necessarily evidence that the proportion of Americans who are afraid to fly has increased.
Explanation:
Since we have given that
n = 1400
x = 154
So,
![\hat{p}=(154)/(1400)=0.11](https://img.qammunity.org/2020/formulas/mathematics/college/mayn8monbrsbd5qzz096z4qr6az07r6svd.png)
and p = 0.10
So, hypothesis would be
![H_0:\hat{p}=p\\\\H_a:\hat{p}>p](https://img.qammunity.org/2020/formulas/mathematics/college/grzmf9oaz8eu4ab1xp5eiurn42q36v32s0.png)
So, the test statistic value would be
![z=\frac{\hat{p}-p}{\sqrt{(p(1-p))/(n)}}\\\\z=\frac{0.11-0.10}{\sqrt{(0.1* 0.9)/(1400)}}\\\\z=(0.01)/(0.008)\\\\z=1.25](https://img.qammunity.org/2020/formulas/mathematics/college/dcfoncfj6x3p26ubza7qirtx6wm6rxm859.png)
At 5% level of significance,
critical value would be 1.96.
Since 1.96>1.25.
so, we will accept the null hypothesis.
Hence, no, this is not necessarily evidence that the proportion of Americans who are afraid to fly has increased.