Respuesta :
Answer:
The answer is "(0.7322,0.8234)".
Step-by-step explanation:
please find the complete question in the attached file.
The sample has 225 persons, thus n = 225. Of these 175 people, the sample proportion was decreased through depression
[tex]\to \hat{p}=\frac{175}{225}=0.7778[/tex]
So z-value is 1.645 for an interval of 90%. The 90% trust interval will be
[tex]\to \hat{p}\pm 1.645\sqrt{\frac{\hat{p}\left ( 1-\hat{p} \right )}{n}}=0.7778\pm 1.645\sqrt{\frac{0.7778\left ( 1-0.7778 \right )}{225}}[/tex]
[tex]=0.7778\pm 1.645\cdot 0.0277\\\\=0.7778\pm 0.0456[/tex]
The confidence interval therefore is 90% (0.7322,0.8234).
The 90% confidence interval's lower limit is 0.7322.
The 90% confidence interval is 0.8234 at its upper limit.
