Answer:
Step-by-step explanation:
Given that:
the sample proportion p = 0.39
sample size = 100
Then np = 39
Using normal approximation
The sampling distribution from the sample proportion is approximately normal.
Thus, mean ![\mu _{\hat p} = p = 0.39](https://tex.z-dn.net/?f=%5Cmu%20_%7B%5Chat%20p%7D%20%3D%20p%20%3D%200.39)
The standard deviation;
![\sigma = \sqrt{\dfrac{p(1-p)}{n} }](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7B%5Cdfrac%7Bp%281-p%29%7D%7Bn%7D%20%7D)
![\sigma = \sqrt{\dfrac{0.39(1-0.39)}{100} }](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7B%5Cdfrac%7B0.39%281-0.39%29%7D%7B100%7D%20%7D)
![\sigma = 0.048](https://tex.z-dn.net/?f=%5Csigma%20%3D%200.048)
The test statistics can be computed as:
![Z = \dfrac{{\hat _{p}} - \mu_{_ {\hat p}} }{\sigma_{\hat p}}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cdfrac%7B%7B%5Chat%20_%7Bp%7D%7D%20-%20%5Cmu_%7B_%20%7B%5Chat%20p%7D%7D%20%7D%7B%5Csigma_%7B%5Chat%20p%7D%7D)
![Z = \dfrac{0.3 - 0.39 }{0.0488}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cdfrac%7B0.3%20-%200.39%20%7D%7B0.0488%7D)
![Z = -1. 8 4](https://tex.z-dn.net/?f=Z%20%3D%20-1.%208%204)
From the z - tables;
![P (\hat p \le 0.3 ) = P(z \le -1.84)](https://tex.z-dn.net/?f=P%20%28%5Chat%20p%20%5Cle%200.3%20%29%20%3D%20P%28z%20%5Cle%20-1.84%29)
![\mathbf{P (\hat p \le 0.3 ) = 0.0329}](https://tex.z-dn.net/?f=%5Cmathbf%7BP%20%28%5Chat%20p%20%5Cle%200.3%20%29%20%3D%200.0329%7D)
(b)
Here;
the sample proportion = 0.39
the sample size n = 400
Since np = 400 * 0.39 = 156
Thus, using normal approximation.
From the sample proportion, the sampling distribution is approximate to the mean ![\mu_{\hat p} = p = 0.39](https://tex.z-dn.net/?f=%5Cmu_%7B%5Chat%20p%7D%20%3D%20%20p%20%3D%200.39)
the standard deviation ![\sigma_{\hat p} = \sqrt{\dfrac{p(1-p)}{n} }](https://tex.z-dn.net/?f=%5Csigma_%7B%5Chat%20p%7D%20%3D%20%5Csqrt%7B%5Cdfrac%7Bp%281-p%29%7D%7Bn%7D%20%7D)
![\sigma_{\hat p} = \sqrt{\dfrac{0.39 (1-0.39)}{400} }](https://tex.z-dn.net/?f=%5Csigma_%7B%5Chat%20p%7D%20%3D%20%5Csqrt%7B%5Cdfrac%7B0.39%20%281-0.39%29%7D%7B400%7D%20%7D)
![\sigma_{\hat p} =0.0244](https://tex.z-dn.net/?f=%5Csigma_%7B%5Chat%20p%7D%20%3D0.0244)
The test statistics can be computed as:
![Z = \dfrac{{\hat _{p}} - \mu_{_ {\hat p}} }{\sigma_{\hat p}}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cdfrac%7B%7B%5Chat%20_%7Bp%7D%7D%20-%20%5Cmu_%7B_%20%7B%5Chat%20p%7D%7D%20%7D%7B%5Csigma_%7B%5Chat%20p%7D%7D)
![Z = \dfrac{0.3 - 0.39 }{0.0244}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cdfrac%7B0.3%20-%200.39%20%7D%7B0.0244%7D)
![Z = -3.69](https://tex.z-dn.net/?f=Z%20%3D%20-3.69)
From the z - tables;
![P (\hat p \le 0.3 ) = P(z \le -3.69)](https://tex.z-dn.net/?f=P%20%28%5Chat%20p%20%5Cle%200.3%20%29%20%3D%20P%28z%20%5Cle%20-3.69%29)
![\mathbf{P (\hat p \le 0.3 ) = 0.0001}](https://tex.z-dn.net/?f=%5Cmathbf%7BP%20%28%5Chat%20p%20%5Cle%200.3%20%29%20%3D%200.0001%7D)
(c) The effect of the sample size on the sampling distribution is that:
As sample size builds up, the standard deviation of the sampling distribution decreases.
In addition to that, reduction in the standard deviation resulted in increases in the Z score, and the probability of having a sample proportion that is less than 30% also decreases.