Question

Maldini Pizza Restaurant is facing a stiff competition from the new competing pizza

restaurant guaranteeing pizza deliveries within 32 minutes or the pizza is given free. To

answer this challenge, Maldini wants to offer a 30-minute delivery guarantee. After a

careful cost analysis, Maldini has determined that such a guarantee would require an

average delivery time of less than 27 minutes. She thought that this would limit the

percentage of ‘free pizzas’ under the guarantee to less than 8% of all the deliveries, which

she had figured to be the break-even point for such a promotion. To find out if the

restaurant can meet these requirements, Maldini collected data on the total delivery times

and ‘free pizzas’ for a random sample of 81 orders within a month. The total delivery

time includes the preparation-time, the wait-time, and the travel-time for the drivers. The

sample data show an average total delivery time of 26.7 minutes with estimated

population standard deviation of 1.2 minutes. The data also show that 4 out of the 81

orders resulted in ‘free pizza’ deliveries because they took longer than 30 minutes to

deliver. Maldini is using a non-conventional confidence level of 97% in her estimates.

Please answer the following questions using the information given in this problem.

a.

Please calculate and interpret the 97% confidence interval for the mean total delivery

time of pizzas for the Maldini Restaurant. Please show the necessary steps

.

[2 points]

b. Please also calculate and interpret the 97% confidence interval for the proportion of

‘free pizzas’ expected under Maldini’s proposed promotion scheme. Please show the

necessary steps

. [2 points]

c. Based on the results for the two confidence intervals you calculated above, do you

think Maldini’s proposed promotion scheme is viable? Please carefully justify your

answer.

Answer 1

a.
TRADITIONAL METHOD
given that,
standard deviation, σ =1.2
sample mean, x =26.7
population size (n)=81
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 1.2/ sqrt ( 81) )
= 0.133
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.03
from standard normal table, two tailed z α/2 =2.17
since our test is two-tailed
value of z table is 2.17
margin of error = 2.17 * 0.133
= 0.289
III.
CI = x ± margin of error
confidence interval = [ 26.7 ± 0.289 ]
= [ 26.411,26.989 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =1.2
sample mean, x =26.7
population size (n)=81
level of significance, α = 0.03
from standard normal table, two tailed z α/2 =2.17
since our test is two-tailed
value of z table is 2.17
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 26.7 ± Z a/2 ( 1.2/ Sqrt ( 81) ) ]
= [ 26.7 - 2.17 * (0.133) , 26.7 + 2.17 * (0.133) ]
= [ 26.411,26.989 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 97% sure that the interval [26.411 , 26.989 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 97% of these intervals will contains the true population mean

b.
TRADITIONAL METHOD
given that,
possible chances (x)=4
sample size(n)=81
success rate ( p )= x/n = 0.049
I.
sample proportion = 0.049
standard error = Sqrt ( (0.049*0.951) /81) )
= 0.024
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.03
from standard normal table, two tailed z α/2 =2.17
margin of error = 2.17 * 0.024
= 0.052
III.
CI = [ p ± margin of error ]
confidence interval = [0.049 ± 0.052]
= [ -0.003 , 0.102]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possible chances (x)=4
sample size(n)=81
success rate ( p )= x/n = 0.049
CI = confidence interval
confidence interval = [ 0.049 ± 2.17 * Sqrt ( (0.049*0.951) /81) ) ]
= [0.049 - 2.17 * Sqrt ( (0.049*0.951) /81) , 0.049 + 2.17 * Sqrt ( (0.049*0.951) /81) ]
= [-0.003 , 0.102]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 97% sure that the interval [ -0.003 , 0.102] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 97% of these intervals will contains the true population proportion
c.
Maldini has determined that such a guarantee would require an
average delivery time of less than 27 minutes.
97% sure that the interval [26.411 , 26.989 ] ,

She thought that this would limit the
percentage of ‘free pizzas’ under the guarantee to less than 8% of all the deliveries.
97% sure that the interval [ -0.003 , 0.102] = -0.3%,10.2%
so that,
Based on the results for the two confidence intervals,
Maldini’s proposed promotion scheme is viable.