Question

The null, Ho, indicates that there is either no relationship or a positive relationship between Amazon’s growth over the past ten years and number of Best Buy brick and mortar locations.  The alternative, H1, seeks to prove that there is a negative relationship between the variables, Amazon and Best Buy brick and mortar locations. In other words, Amazon’s growth is negatively impacting Best Buy by forcing store location closures. Using a 95% confidence interval, construct a test of hypothesis using the following data:

Total number of Best Buy stores worldwide 2010-2019
2010	1,565
2011	1,550
2012	1,711
2013	1,779
2014	1,779
2015	1,732
2016	1,632
2017	1,581
2018	1,514
2019	1,238

Answer 1

X	Y	XY	X²	Y²
2010	1565	3145650	4040100	2449225
2011	1550	3117050	4044121	2402500
2012	1711	3442532	4048144	2927521
2013	1779	3581127	4052169	3164841
2014	1779	3582906	4056196	3164841
2015	1732	3489980	4060225	2999824
2016	1632	3290112	4064256	2663424
2017	1581	3188877	4068289	2499561
2018	1514	3055252	4072324	2292196
2019	1238	2499522	4076361	1532644
Ʃx =	Ʃy =	Ʃxy =	Ʃx² =	Ʃy² =
20145	16081	32393008	40582185	26096577

Sample size, n =	10
x̅ = Ʃx/n = 20145/10 =	2014.5
y̅ = Ʃy/n = 16081/10 =	1608.1
SSxx = Ʃx² - (Ʃx)²/n = 40582185 - (20145)²/10 =	82.5
SSyy = Ʃy² - (Ʃy)²/n = 26096577 - (16081)²/10 =	236720.9
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 32393008 - (20145)(16081)/10 =	-2166.5

Slope, b = SSxy/SSxx = -2166.5/82.5 =    -26.26060606

y-intercept, a = y̅ -b* x̅ = 1608.1 - (-26.26061)*2014.5 =    54510.09091

Regression equation :   

ŷ = 54510.0909 + (-26.2606) x  

Sum of Square error, SSE = SSyy -SSxy²/SSxx = 236720.9 - (-2166.5)²/82.5 =    179827.297

Standard error, se = √(SSE/(n-2)) = √(179827.29697/(10-2)) =    149.92802

Standard error for slope, se(b1) = se/√SSxx = 149.92802/√82.5 =    16.50653208

Null and alternative hypothesis:  

Ho: β₁ = 0 ; Ha: β₁ < 0  

Test statistic:  

t = b1/se(b1) = -26.2606/16.5065 = -1.5909

df = n-2 = 8

p-value = T.DIST.2T(ABS(-1.5909), 8) = 0.0751

Conclusion:  

p-value > α , Fail to reject the null hypothesis.  

---------

Critical value, t_c = T.INV.2T(0.05, 8) = 2.306  

95% Confidence interval for slope:  

Lower limit = b1 - tc*se(b1) = -26.2606 - 2.306*16.5065 =    -64.3247

Upper limit = b1 + tc*se(b1) = -26.2606 + 2.306*16.5065 =    11.8035