Question

A furniture store has maintained monthly sales records for the past 20​ months, with the results shown below.

Month	Sales
1	2360
2	1820
3	1760
4	1560
5	1950
6	1950
7	3360
8	1740
9	3780
10	2400
11	2160
12	2760
13	3570
14	2820
15	2800
16	1890
17	2500
18	3630
19	2530
20	3270

Assume you have determined there is NO SEASONALITY in this time series.​ Therefore, you want to fit a linear trend model​ (that is, trend​ only) to the data.

Calculate the linear trend equation.​ (Round coefficients to the nearest whole​ number.)

y Subscript t Baseline equalsyt=nothing​+

nothing​*t

What are the test statistic and​ p-value to test for a significant trend. Round both to two decimal places.

T​ = nothing

​p-value = nothing

Is the trend significant using a​ 10% significance​ level?

Yes

No

What is the value of​ R-squared? (Round to two​ decimals.)

nothing

Forecast the sales for the next month​ (t =​ 21). (Round to the nearest whole​ number.)

Upper F 21 equalsF21=nothing

Based on the​ R-squared value, how confident are you in this​ forecast? (That​ is, how accurate do you think the forecasts will​ be?)

A.

Not confident at all because the​ R-squared value is so low

B.

Very confident because the​ R-squared value is high

C.

Somewhat confident because the​ R-squared value is moderate​ (not extremely high but not extemely​ low)

Answer 1

x	y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
1	2360	90.25	29070.25	1619.75
2	1820	72.25	504810.25	6039.25
3	1760	56.25	593670.25	5778.75
4	1560	42.25	941870.25	6308.25
5	1950	30.25	336980.25	3192.75
6	1950	20.25	336980.25	2612.25
7	3360	12.25	688070.25	-2903.25
8	1740	6.25	624890.25	1976.25
9	3780	2.25	1561250.25	-1874.25
10	2400	0.25	17030.25	65.25
11	2160	0.25	137270.25	-185.25
12	2760	2.25	52670.25	344.25
13	3570	6.25	1080560.25	2598.75
14	2820	12.25	83810.25	1013.25
15	2800	20.25	72630.25	1212.75
16	1890	30.25	410240.25	-3522.75
17	2500	42.25	930.25	-198.25
18	3630	56.25	1208900.25	8246.25
19	2530	72.25	0.25	-4.25
20	3270	90.25	546860.25	7025.25

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	210	50610	665	9228495.0	39345.00
mean	10.50	2530.50	SSxx	SSyy	SSxy

sample size ,   n =   20          
here, x̅ = Σx / n=   10.50   ,     ȳ = Σy/n =   2530.50  
                  
SSxx =    Σ(x-x̅)² =    665.0000          
SSxy=   Σ(x-x̅)(y-ȳ) =   39345.0          
                  
estimated slope , ß1 = SSxy/SSxx =   39345.0   /   665.000   =   59.16541
                  
intercept,   ß0 = y̅-ß1* x̄ =   1909.26316          
                  
so, regression line is   Ŷ =   1909 +   59*x

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estimated std error of slope =Se(ß1) = Se/√Sxx =    619.168   /√   665.00   =   24.0103
                  
t stat = estimated slope/std error =ß1 /Se(ß1) =    59.1654   /   24.0103   =   2.46

p-value =    0.024

Is the trend significant using a​ 10% significance​ level?

Yes

---------------

R² =    (Sxy)²/(Sx.Sy) =    0.25

Predicted Y at X=   21   is                  
F21  =   1909.2632   +   59.1654   *   21   =   3152

-----------

Not confident at all because the​ R-squared value is so low