Your advertising campaign has worked out so well for your pi

Your advertising campaign has worked out so well for your pizza deliveries that you would now like to promote your restaurants submarine sandwiches. So you are considering offering a $1 off coupon on subs to increase the amount of submarine sandwiches you are selling each week. However you are unable find the total number of subs sold each week anywhere on your computer and no way for you to total them up without going through the sales records one by one. Never fear! Since you have now learned about confidence intervals in your Statistics class you can get an estimate for the mean number of subs sold each week by taking a random sample of 36 weeks and calculating the number of subs sold during those weeks. Your random sample resulted in a mean of 7.25 sub sandwiches per week with a standard deviation of 2.5.

a) Construct a 95% confidence interval for mean number of submarines sandwiches sold each week. Round 3 decimal.

b) Interpret the interval.

c) Construct a 98% confidence interval for the mean number of submarine sandwiches sold each week. Round 3 decimals.

d) What effect did increasing the level of confidence have on the interval?

e) How many weeks should you include in your sample if you want your estimate for the mean number of submarine sandwiches sold each week to be within 0.5 sub sandwiches with 99% confidence?

Solution

a)

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    7.25          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    2.5          
n = sample size =    36          
              
Thus,              
Margin of Error E =    0.81665166          
Lower bound =    6.43334834          
Upper bound =    8.06665166          
              
Thus, the confidence interval is              
              
(   6.43334834   ,   8.06665166   ) [ANSWER]

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b)

We are 95% confident that the true mean number of subs sold per week is between 6.4333 and 6.0667. [ANSWER]

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c)

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.01          
X = sample mean =    7.25          
z(alpha/2) = critical z for the confidence interval =    2.326347874          
s = sample standard deviation =    2.5          
n = sample size =    36          
              
Thus,              
Margin of Error E =    0.969311614          
Lower bound =    6.280688386          
Upper bound =    8.219311614          
              
Thus, the confidence interval is              
              
(   6.280688386   ,   8.219311614   ) [ANSWER]

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d)

It became wider because of the increase in critical z.

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e)

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.005  
      
Using a table/technology,      
      
z(alpha/2) =    2.575829304  
      
Also,      
      
s = sample standard deviation =    2.5  
E = margin of error =    0.5  
      
Thus,      
      
n =    165.872415  
      
Rounding up,      
      
n =    166   [ANSWER]