Question

The table below shows the data of the new type of virus disease (COVID-19) from 11 March 2020, the day when the first case occurred in our country, until 21 April 2020, when the virus peaked. In response to these data, the number of patients recovering within the same time frame is given. Find a 2nd order polynomial equation (Ŷ = a0 + a₁ x + a₂ x²) that fits these data. Then calculate the correlation coefficient. Using the parabola equation, find a regression curve for the number of patients recovering based on the number of cases. Also, estimate when the number of cases will end according to these data.

	DAILY CASE NUMBER(Y)	DAILY HEALING PATIENTS(x)	x^2
11-Mar	0	1	1
12-Mar	0	0	0
13-Mar	0	4	16
14-Mar	0	1	1
15-Mar	1	12	144
16-Mar	1	29	841
17-Mar	2	41	1681
18-Mar	3	93	8649
19-Mar	4	168	28224
20-Mar	9	311	96721
21-Mar	21	277	76729
22-Mar	30	289	83521
23-Mar	37	293	85849
24-Mar	44	343	117649
25-Mar	59	561	314721
26-Mar	75	1196	1430416
27-Mar	92	2069	4280761
28-Mar	108	1704	2903616
29-Mar	131	1815	3294225
30-Mar	168	1610	2592100
31-Mar	214	2704	7311616
1-Apr	277	2148	4613904
2-Apr	356	2456	6031936
3-Apr	425	2786	7761796
4-Apr	501	3013	9078169
5-Apr	574	3135	9828225
6-Apr	649	3148	9909904
7-Apr	725	3892	15147664
8-Apr	812	4117	16949689
9-Apr	908	4056	16451136
10-Apr	1006	4747	22534009
11-Apr	1101	5138	26399044
12-Apr	1198	4789	22934521
13-Apr	1296	4093	16752649
14-Apr	1403	4062	16499844
15-Apr	1518	4281	18326961
16-Apr	1643	4801	23049601
17-Apr	1769	4353	18948609
18-Apr	1890	3783	14311089
19-Apr	2017	3977	15816529
20-Apr	2140	4674	21846276
21-Apr	2259	4611	21261321

i know how to do with excel so please dont do this with excel

Answer 1

We have used R software to fit 2nd order polynomial equation (Ŷ = a₀ + a₁X+ a₂ X²) to data of the new type of virus disease (COVID-19) from 11 March 2020, the day when the first case occurred in our country, until 21 April 2020, when the virus peaked. Y is the “Daily Case Number” and X is the “Daily Healing Patients”. The model output is given below

The fitted Ŷ = a₀ + a₁X+ a₂ X²) is

# Fitted quadratic model is Model
Y= -1.246e+01+ (8.925e-03)*X+(7.039e-05)*X^2

Note that we have given coefficients in scientific notation, if we convert them to numer form, model is given as

Y= -12.46+0.0089*X+0.00007*X²

Daily Healing Patients (X)= 363

The model shows that

Daily Case Number (Y)= -12.46+0.0089*363+0.00007*363²

Daily Case Number (Y)=0 (approx.)

Thus, the number of cases will end, when Daily Healing Patients (X) reaches to 363

The fitted values given by the model are

The correlation coefficient between Y and Y fitted values is 0.886, which is highly positive and the fitted model is good.

####################### R code employed for data analysis ######################

data<-read.csv("data.csv",sep=",",header=T)
head(data)
Model<-lm(DAILY.CASE.NUMBER.Y~DAILY.HEALING.PATIENTS.X+DAILY.HEALING.PATIENTS.X_Square,data=data)
summary(Model)
fitted.values(Model)# Fitted quadratic model is Model
Y= -1.246e+01+ (8.925e-03)*X+(7.039e-05)*X^2

X=362

Y= -1.246e+01+ (8.925e-03)*X+(7.039e-05)*X^2

Y<-round(Y,2)
Y

X=363
Y= -12.46+0.0089*X+0.00007*X^2
Y

cor(fitted.values(Model), data$DAILY.CASE.NUMBER.Y)