Answer:
0.339 < p < 0.461
Explanation:
Given data:
confidence interval is 92%
Randomly selected adults = 329
Total number of adults is 763
[tex]\alpha = 1 - 0.92 = 0.08[/tex]
[tex]\frac{\alpha}{2} = \frac{0.08}{2} = 0.04[/tex]
for alpha = 0.04
z value is = 1.75
[tex]p = \frac{329}{763} = 0.43[/tex]
[tex]= p \pm z \times \sqrt{\frac{p \times (1-p)}{N}[/tex]
[tex]= 0.43 \pm 1.75 \tiimes \sqrt{\frac{0.43 \times0.57)}{763}[/tex]
[tex]=0.43 \pm 0.031[/tex]
0.339 < p < 0.461
