Answer:
Following are the responses to the given question:
Step-by-step explanation:
Level of Significance,[tex]\alpha = 0.05[/tex]
Sample Size, [tex]n = 120[/tex]
Sample Proportion, [tex]\hat{P} = 0.600[/tex]
[tex]z -value = Z_{\frac{\alpha}{2}} = 1.960 \ \ [excel formula =NORMSINV(\frac{\alpha}{2})][/tex]
Standard Error[tex]SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = 0.0447[/tex]
margin of error [tex]E = Z\times SE = 1.960 \times 0.0447 = 0.0877[/tex]
When [tex]95\%[/tex] Confidence Interval
Interval Lower Limit[tex]= \hat{p} - E = 0.600 - 0.0877 = 0.5123[/tex]
Interval Upper Limit [tex]=\hat{p}+ E = 0.600 + 0.0877 = 0.6877[/tex]