2. You run the advertising department for a large hotel chain. You want to estim

Question

2. You run the advertising department for a large hotel chain. You want to estimate with 98% confidence the proportion of your guests who make reservations using your hotel’s website. You wish for the estimate to be within 5% of the true proportion.

A. With no known prior proportion to go by, calculate minimum sample size required.

B. Assuming the data was gathered via survey methods that had approximately a 40% response rate, what would be the actual sample size needed to produce the minimum size required in Part A?

C. A study was done that included responses from 600 randomly sampled quests, of which 55% said they register using the hotel website. Calculate the 98% confidence interval for the population of interest.

D. What is a proper statistical interpretation of the confidence interval you calculated in Part B?

E. If the more common 95% confidence were calculated, assuming the same sample proportion and sample size from Part C, would the resulting interval be wider or narrower? Do not recalculate the interval, but instead explain why it would be wider or narrower.

F. Using the 98% confidence interval you calculated in Part C, would it be reasonable to conclude that more than 50% of the hotel guests register using the hotel website?   Using your answer to Part E, what would be your answer if the interval were at 95% confidence?

Explanation / Answer

a)

Note that      
      
n = z(alpha/2)^2 p (1 - p) / E^2      
      
where      
      
alpha/2 =    0.01  
As there is no previous estimate for p, we set p = 0.5.      
      
Using a table/technology,      
      
z(alpha/2) =    2.326347874  
      
Also,      
      
E =    0.05  
p =    0.5  
      
Thus,      
      
n =    541.1894431  
      
Rounding up,      
      
n =    542   [ANSWER]

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b)

Note that      
      
n = z(alpha/2)^2 p (1 - p) / E^2      
      
where      
      
alpha/2 =    0.01  
       
      
Using a table/technology,      
      
z(alpha/2) =    2.326347874  
      
Also,      
      
E =    0.05  
p =    0.4  
      
Thus,      
      
n =    519.5418654  
      
Rounding up,      
      
n =    520   [ANSWER]

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c)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.55          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.020310096          
              
Now, for the critical z,              
alpha/2 =   0.01          
Thus, z(alpha/2) =    2.326347874          
Thus,              
Margin of error = z(alpha/2)*sp =    0.047248349          
lower bound = p^ - z(alpha/2) * sp =   0.502751651          
upper bound = p^ + z(alpha/2) * sp =    0.597248349          
              
Thus, the confidence interval is              
              
(   0.502751651   ,   0.597248349   ) [ANSWER]

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d)

We are 98% confident that the true population proportion of those who register using the hotel website is between 0.50275 and 0.59725. [ANSWER]

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Hi! Please submit the next part as a separate question. That way we can continue helping you! Please indicate which parts are not yet solved when you submit. Please still include the background information. Thanks!

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