Question

Question 1

Jack is a restaurant owner in San Francisco downtown. Due to the recent coronavirus pandemic, he can now accept orders only for pickups and delivery services. This change in the format of incoming orders has made him revisit his assumption for daily estimation of order arrival. The reason why he needs to make this estimation, is for him to prepare the raw materials the night before, for the following day orders.

Based on his past experience, his assumption has been to expect around 18.5 orders a day. He also knows that regardless of season or the situation, the incoming orders always follow a normal distribution. However, to re-evaluate his assumption of 18.5 orders a day, he decided to keep a track of the total number of orders per day for about 10 days. Now Jack has come to you to help him evaluate the daily demand since he knows you are a pro in supply chain analytics.

Using the information that Jack has recorded in the table below, please calculate the upper and lower bound for a 90% confidence interval.

Date	#Orders
June 29, 2020	35
June 30, 2020	22
July 1, 2020	11
July 2, 2020	17
July 3, 2020	36
July 4, 2020	55
July 5, 2020	42
July 6, 2020	28
July 7, 2020	25
July 8, 2020	19

A) 90% Confidence interval lower bound

Round your answer to 1 decimal place (ex.: 19.4 for 19.432).

Answer:

B) 90% Confidence interval upper bound

Round your answer to 1 decimal place (ex.: 19.4 for 19.432).

Answer:

Question 2

After you reported the confidence interval bounds, Jack has come back to you for more information. He wants to know if his original assumption of 18.5 orders a day is still the same after the changes in the format of incoming orders.

Using the daily orders that Jack has captured during the 10 days, please run a hypothesis testing to check whether or not his original assumption of an average of 18.5 orders per day is still in line with the actuals that he has recorded.

H 0 : The average number of orders per day is equal to to 18.5  

H 1 : The average number of orders per day is not equal to 18.5

A) What is the test statistic of this hypothesis testing?

Round your answer to 1 decimal place (ex.: 19.4 for 19.432)

Answer:

B) Based on the significance level of 10% (the same as the previous section), can we reject the null hypothesis?

No

Yes

Answer 1

Result:

A) 90% Confidence interval lower bound

Round your answer to 1 decimal place (ex.: 19.4 for 19.432).

Answer:21.4

B) 90% Confidence interval upper bound

Round your answer to 1 decimal place (ex.: 19.4 for 19.432).

Answer:36.6

n=10

Mean =29

SD=13.1825

Table value t at 10-1 = 9 DF is 1.8331

=( 21.3584, 36.6416)

Excel calculations:

Confidence Interval Estimate for the Mean
Data
Sample Standard Deviation	13.1825
Sample Mean	29
Sample Size	10
Confidence Level	90%
Intermediate Calculations
Standard Error of the Mean	4.1687
Degrees of Freedom	9
t Value	1.8331
Interval Half Width	7.6416
Confidence Interval
Interval Lower Limit	21.3584
Interval Upper Limit	36.6416

Question 2

After you reported the confidence interval bounds, Jack has come back to you for more information. He wants to know if his original assumption of 18.5 orders a day is still the same after the changes in the format of incoming orders.

Using the daily orders that Jack has captured during the 10 days, please run a hypothesis testing to check whether or not his original assumption of an average of 18.5 orders per day is still in line with the actuals that he has recorded.

H 0 : The average number of orders per day is equal to to 18.5  

H 1 : The average number of orders per day is not equal to 18.5

A) What is the test statistic of this hypothesis testing?

Round your answer to 1 decimal place (ex.: 19.4 for 19.432)

Answer: 2.5

B) Based on the significance level of 10% (the same as the previous section), can we reject the null hypothesis?

Correct option: Yes

single sample t test used.

Ho: µ = 18.5   H₁: µ ≠ 18.5

=2.5188                

Table value of t with 9 DF at 0.10 level =1.8331

Rejection Region: Reject Ho if t < -1.8331 or t > 1.8331

Calculated t =2.5188   falls in the rejection region

The null hypothesis is rejected.

Excel calculations:

t Test for Hypothesis of the Mean
Data
Null Hypothesis                m=	18.5
Level of Significance	0.1
Sample Size	10
Sample Mean	29
Sample Standard Deviation	13.1825
Intermediate Calculations
Standard Error of the Mean	4.1687
Degrees of Freedom	9
t Test Statistic	2.5188
Two-Tail Test
Lower Critical Value	-1.8331
Upper Critical Value	1.8331
p-Value	0.0328
Reject the null hypothesis