Answer:
1a) p-hat=0.38
1b) P=0.08
1c) The null hypothesis is not rejected
2a) p-hat=0.64
2b) P=0.0027
2c) The null hypothesis is rejected
3a) p-hat=0.55
3b) P=0.153
3c) The null hypothesis is not rejected
Step-by-step explanation:
(1) H0: p = 0.3 vs Ha: p ≠ 0.3. In their survey, they had a count of 38 using a sample size n=100.
1a) The p-hat is p-hat=38/100=0.38.
1b) The standard deviation is
![\sigma=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.3*0.7}{100}}=0.046](https://tex.z-dn.net/?f=%5Csigma%3D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%3D%5Csqrt%7B%5Cfrac%7B0.3%2A0.7%7D%7B100%7D%7D%3D0.046)
The sample size is n=100.
The z-value is:
![z=\frac{\hat{p}-p}{\sigma}=\frac{0.38-0.3}{0.046}=\frac{0.08}{0.046}= 1.74](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B%5Chat%7Bp%7D-p%7D%7B%5Csigma%7D%3D%5Cfrac%7B0.38-0.3%7D%7B0.046%7D%3D%5Cfrac%7B0.08%7D%7B0.046%7D%3D%201.74)
As it is a two-sided test, the p-value considers both tails of the distribution.
The p-value for this |z|=1.74 is P=0.08.
1c) The null hypothesis is not rejected.
(2) H0: p = 0.7 vs Ha: p ≠ 0.7. In their survey, they had a count of 320 using a sample size n=500.
2a) The p-hat is p-hat=320/500=0.64.
2b) The standard deviation is
![\sigma=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.7*0.3}{500}}=0.02](https://tex.z-dn.net/?f=%5Csigma%3D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%3D%5Csqrt%7B%5Cfrac%7B0.7%2A0.3%7D%7B500%7D%7D%3D0.02)
The sample size is n=500.
The z-value is:
![z=\frac{\hat{p}-p}{\sigma}=\frac{0.64-0.7}{0.02}=\frac{-0.06}{0.02}=-3](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B%5Chat%7Bp%7D-p%7D%7B%5Csigma%7D%3D%5Cfrac%7B0.64-0.7%7D%7B0.02%7D%3D%5Cfrac%7B-0.06%7D%7B0.02%7D%3D-3)
As it is a two-sided test, the p-value considers both tails of the distribution.
The p-value for this |z|=3 is P=0.0027.
2c) The null hypothesis is rejected.
(3) H0: p = 0.6 vs Ha: p < 0.6. In their survey, they had a count of 110 using a sample size n=200.
2a) The p-hat is p-hat=110/200=0.55.
2b) The standard deviation is
![\sigma=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.6*0.4}{200}}=0.035](https://tex.z-dn.net/?f=%5Csigma%3D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%3D%5Csqrt%7B%5Cfrac%7B0.6%2A0.4%7D%7B200%7D%7D%3D0.035)
The sample size is n=200.
The z-value is:
![z=\frac{\hat{p}-p}{\sigma}=\frac{0.55-0.6}{0.035}=\frac{-0.05}{0.035}=-1.43](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B%5Chat%7Bp%7D-p%7D%7B%5Csigma%7D%3D%5Cfrac%7B0.55-0.6%7D%7B0.035%7D%3D%5Cfrac%7B-0.05%7D%7B0.035%7D%3D-1.43)
As it is a two-sided test, the p-value considers both tails of the distribution.
The p-value for this |z|=1.43 is P=0.153.
2c) The null hypothesis is not rejected.