A Google research study asked 5093 smartphone users about ho

A Google research study asked 5093 smartphone users about how they used the phonos in response to a question about purchases. 2654 reported that they purchased an item after using the smanpnono to search for information about the item. Find ( 0.001) SE beta the standard error of beta? A Pew Internet poll asked cell phone owners about how they used their cell phones. One question asked whether or not during the past 30 days they had used their phone while in a store to call a fiend or family member for advice about a purchase they were considering. The poll surveyed 1030 adults Irving m the United States by telephone. Of these. 462 responded that they had used their cell phone while m a store within the last 30 days to call a friend or family member for advice about a purchase they were conceding. Find the margin of error for a 95% confidence interval. Margin of error beta = A Pew Internet poll asked cell phone owners about how they used their cell phones. One question ask

Solution

1.

a)

Here,

p^ = 2654/5093 = 0.521107402

As

SE(p^) = sqrt(p^ (1-p^) / n) = sqrt(0.521107402*(1-0.521107402)/5093)

SE(p^) = 0.006999965 [ANSWER]

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b)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.521107402          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.006999965          
              
Now, for the critical z,              
alpha/2 =   0.025          
Thus, z(alpha/2) =    1.959963985          
Thus,              
Margin of error = z(alpha/2)*sp =    0.013719679          
lower bound = p^ - z(alpha/2) * sp =   0.507387724          
upper bound = p^ + z(alpha/2) * sp =    0.534827081          
              
Thus, the confidence interval is              
              
(   50.7387724%   ,   53.4827081%   ) [ANSWER]