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Ellena Pizza Restaurant is facing stiff competition from a new competing pizza restaurant guaranteeing pizza deliveries within 30 minutes or the pizza is given free. To answer this challenge, Ellena wants to offer a 28-minute delivery guarantee. After a careful cost analysis, Ellena has determined that such a guarantee would require an average delivery time of less than 26 minutes. She thought that this would limit the percentage of ‘free pizzas’ under the guarantee to less than 10.5% 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, Ellena 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 24.9 minutes with an 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 28 minutes to deliver. Ellena is using a non-conventional confidence level of 93% in her estimates. Please answer the following questions using the information given in this problem: a. Please calculate and interpret the 93% confidence interval for the mean total delivery time of pizzas for the Ellena Restaurant. Please show the necessary steps. b. Please also calculate and interpret the 93% confidence interval for the proportion of ‘free pizzas’ expected under Ellena’s proposed promotion scheme. Please show the necessary steps. c. Based on the results for the two confidence intervals you calculated above, do you think Ellena’s proposed promotion scheme is viable? Please carefully justify your answer.

Answer :

To solve this problem, we need to calculate two confidence intervals: one for the mean total delivery time and another for the proportion of ‘free pizzas’. Both analyses will help us determine whether Ellena’s proposed promotion scheme is viable.

a. Confidence Interval for the Mean Total Delivery Time:

To calculate the confidence interval for the mean, we'll use the formula for the confidence interval for a population mean:

[tex]CI = \bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right)[/tex]

Where:

  • [tex]\bar{x} = 24.9[/tex] minutes (sample mean)
  • [tex]\sigma = 1.2[/tex] minutes (population standard deviation)
  • [tex]n = 81[/tex] (sample size)
  • [tex]z[/tex] is the z-value for a 93% confidence level

First, we need to find the z-value that corresponds to a 93% confidence level. In a standard normal distribution table, this is approximately [tex]z = 1.81[/tex].

Plugging the values into the formula:

[tex]CI = 24.9 \pm 1.81 \times \left( \frac{1.2}{\sqrt{81}} \right)[/tex]

[tex]CI = 24.9 \pm 1.81 \times 0.1333[/tex]

[tex]CI = 24.9 \pm 0.2413[/tex]

So, the 93% confidence interval for the mean total delivery time is [tex][24.6587, 25.1413][/tex] minutes.

Interpretation: This interval suggests that we are 93% confident that the true mean delivery time is between 24.66 and 25.14 minutes, which is less than 26 minutes required by Ellena.

b. Confidence Interval for the Proportion of ‘Free Pizzas’:

For the proportion, we'll use the formula for the confidence interval for a population proportion:

[tex]CI = \hat{p} \pm z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}[/tex]

Where:

  • [tex]\hat{p} = \frac{4}{81} \approx 0.0494[/tex] (sample proportion of free pizzas)
  • [tex]n = 81[/tex]
  • [tex]z = 1.81[/tex] (z-value as before)

Plugging the values into the formula:

[tex]CI = 0.0494 \pm 1.81 \sqrt{\frac{0.0494(1-0.0494)}{81}}[/tex]

[tex]CI = 0.0494 \pm 1.81 \sqrt{\frac{0.0494 \times 0.9506}{81}}[/tex]

[tex]CI = 0.0494 \pm 1.81 \sqrt{0.000578}[/tex]

[tex]CI = 0.0494 \pm 1.81 \times 0.02404[/tex]

[tex]CI = 0.0494 \pm 0.0435[/tex]

So, the 93% confidence interval for the proportion of free pizzas is [tex][0.0059, 0.0929][/tex].

Interpretation: We are 93% confident that the true proportion of free pizzas will be between approximately 0.59% and 9.29%, which is below the 10.5% threshold Ellena set.

c. Viability of the Proposed Promotion Scheme:

Both confidence intervals indicate favorable results for Ellena's restaurant:

  1. The mean delivery time is confidently below the required 26 minutes.
  2. The proportion of free pizzas is below the break-even point of 10.5%.

Based on this analysis, Ellena's proposed promotion scheme appears viable.

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