Marketers want to know about the differences between men\'s and women\'s use of

Question

Marketers want to know about the differences between men's and women's use of the Internet. A research poll in April 2009 from a random sample of adults found that 2378 of 3048 men use the Internet, while 2351 of 3193 women did. a). Find the proportions of men and women who said they use the Internet at least occasionally. b). What is the difference in proportions? C). What is the standard error of the difference? d). Find a 95% confidence interval for the difference between percentages of usage by men and women nationwide

Explanation / Answer

a)

Here,

x1 = 2378
n1 = 3048
x2 = 2351
n2 = 3193

Getting p1^ and p2^,          
          
p1^ = x1/n1 = 2378/3048 =   0.780183727      
p2 = x2/n2 = 2351/3193 =   0.736298152   [ANSWER]

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b)

Hence,

p1^ - p2^ =   0.043885575   [ANSWER]

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c)

Also, the standard error of the difference is                      
sd = sqrt[ p1 (1 - p1) / n1 + p2 (1 - p2) / n2] =    0.010820095   [ANSWER]  
  
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d)

For the   95%   confidence level, then  
          
alpha/2 = (1 - confidence level)/2 =    0.025      
z(alpha/2) =    1.959963985      
Margin of error = z(alpha/2)*sd =    0.021206997      
lower bound = p1^ - p2^ - z(alpha/2) * sd =    0.022678578      
upper bound = p1^ - p2^ + z(alpha/2) * sd =    0.065092572      
          
Thus, the confidence interval is          
          
(   0.022678578   ,   0.065092572 ) [ANSWER]

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