Millions of North Americans including Canadians now arrange

Millions of North Americans including Canadians now arrange their travel plans on the Web. According to an article in USA Today, 77% of travelers purchased plane tickets online.

Solution

a)

Here,

mean = 0.77
standard deviation = sqrt(p (1-p)/n) = sqrt(0.77*(1-0.77)/200) = 0.029757352

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    0.75      
x2 = upper bound =    0.8      
u = mean =    0.77      
          
s = standard deviation =    0.029757352      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.672102813      
z2 = upper z score = (x2 - u) / s =    1.00815422      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.250759123      
P(z < z2) =    0.843309784      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.592550661   [ANSWER]

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b)

It is like asking the probability of p < 170/200 = 0.85.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.85      
u = mean =    0.77      
          
s = standard deviation =    0.029757352      
          
Thus,          
          
z = (x - u) / s =    2.688411254      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.688411254   ) =    0.003589646 [ANSWER]

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c)

The probability in B is unlikely if the proportion stayed at 77%. Thus, this suggests that the increase in service fee may have influenced the increase in online buying of tickets.

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d)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.85          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.025248762          
              
Now, for the critical z,              
alpha/2 =   0.01          
Thus, z(alpha/2) =    2.326347874          
Thus,              
              
lower bound = p^ - z(alpha/2) * sp =   0.791262595          
upper bound = p^ + z(alpha/2) * sp =    0.908737405          
              
Thus, the confidence interval is              
              
(   0.791262595   ,   0.908737405   ) [ANSWER]

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