75 words or more response, Keri is the owner of a new restaurant in the downtown

Question

75 words or more response, Keri is the owner of a new restaurant in the downtown area of her hometown. To continuously improve service, she would like to know if completed dishes are being delivered to the customer’s table within one minute of being completed by the chef. A random sample of 75 completed dishes showed that 60 were delivered within one minute of completion. Calculate the 90%, 95%, and 99% confidence interval for the true population proportion. Interpret your response within the context of the situation.

Explanation / Answer

90% confidence:

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.8          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.046188022          
              
Now, for the critical z,              
alpha/2 =   0.05          
Thus, z(alpha/2) =    1.644853627          
Thus,              
Margin of error = z(alpha/2)*sp =    0.075972535          
lower bound = p^ - z(alpha/2) * sp =   0.724027465          
upper bound = p^ + z(alpha/2) * sp =    0.875972535          
              
Thus, the confidence interval is              
              
(   0.724027465   ,   0.875972535   )

Hence, we are 90% confident that the true proportion of dishes that are being delivered to the customer’s table within one minute of being completed by the chef is between 0.72403 and 0.87597. [CONCLUSION]

****************

95% confidence:

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.8          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.046188022          
              
Now, for the critical z,              
alpha/2 =   0.025          
Thus, z(alpha/2) =    1.959963985          
Thus,              
Margin of error = z(alpha/2)*sp =    0.090526859          
lower bound = p^ - z(alpha/2) * sp =   0.709473141          
upper bound = p^ + z(alpha/2) * sp =    0.890526859          
              
Thus, the confidence interval is              
              
(   0.709473141   ,   0.890526859   )

Hence, we are 95% confident that the true proportion of dishes that are being delivered to the customer’s table within one minute of being completed by the chef is between 0.70947 and 0.89053. [CONCLUSION]

*******************

99% confidence:

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.8          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.046188022          
              
Now, for the critical z,              
alpha/2 =   0.005          
Thus, z(alpha/2) =    2.575829304          
Thus,              
Margin of error = z(alpha/2)*sp =    0.118972459          
lower bound = p^ - z(alpha/2) * sp =   0.681027541          
upper bound = p^ + z(alpha/2) * sp =    0.918972459          
              
Thus, the confidence interval is              
              
(   0.681027541   ,   0.918972459   )

Hence, we are 99% confident that the true proportion of dishes that are being delivered to the customer’s table within one minute of being completed by the chef is between 0.68103 and 0.91897. [CONCLUSION]

Related Questions

Navigate

• Browse (All)
• Browse 7
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.