Question

1. A study is conducted to determine the security and confidentiality of personal information submitted over the Internet, including such things as social insurance and credit card numbers. A random sample of 1,000 adults were selected and asked under what circumstances they would give personal information to a company. 20% said they would never give personal data to a company, while 51% said they would if the company had strict privacy guidelines in place. Construct a 95% confidence interval for the proportion of those who would never give personal data to a company.

   Calculate the margin of error.

   	a.	0.025
   	b.	0.030
   	c.	0.016

1 points   

QUESTION 2

1. Construct a confidence interval for the proportion of those who would never give personal data to a company.

   	a.	0.465, 0.515
   	b.	0.485, 0.535
   	c.	0.175, 0.225

Answer 1

1)

Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   200          
Sample Size,   n =    1000          
                  
Sample Proportion ,    p̂ = x/n =    0.200          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0126          
margin of error , E = Z*SE =    1.960   *   0.0126   =   0.025

2)

                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.200   -   0.0248   =   0.1752
Interval Upper Limit = p̂ + E =   0.200   +   0.0248   =   0.2248
                  
95%   confidence interval is (   0.175 < p <    0.225 )

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