Question

A company that manufactures small lathes is interested in establishing standards for employees. A random sample of 18 employees is selected in order to develop the standards. The data collected is below. A manager at the firm feels that assembly time is related to intelligence. She feels that employees who did well in high school (as measured by high school averages) should be able to do the job faster. Does the data support her hunch?

high school

average

Time to assemble lathe

(minutes)

50                             52

62                             53

65                             62

68                             70

71                             73

73                             78

75                             80

79                             82

80                             85

82                             86

83                             90

85                             94

88                             95

90                           106

90                           111

94                           120

94                           139

100                           145

Answer 1

x	y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
50	52	863.7068	1448.2253	1118.4105
62	53	302.3735	1373.1142	644.3549
65	62	207.0401	787.1142	403.6883
68	70	129.7068	402.2253	228.4105
71	73	70.3735	290.8920	143.0772
73	78	40.8179	145.3364	77.0216
75	80	19.2623	101.1142	44.1327
79	82	0.1512	64.8920	3.1327
80	85	0.3735	25.5586	-3.0895
82	86	6.8179	16.4475	-10.5895
83	90	13.04	0.00	-0.20
85	94	31.48	15.56	22.13
88	95	74.15	24.45	42.58
90	106	112.60	254.23	169.19
90	111	112.60	438.67	222.24
94	120	213.48	896.67	437.52
94	139	213.48	2395.56	715.13
100	145	424.82	3018.89	1132.47

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	1429.00	1621.00	2836.28	11698.94	5389.61
mean	79.39	90.06	SSxx	SSyy	SSxy

sample size ,   n =   18          
here, x̅ = Σx / n=   79.389   ,     ȳ = Σy/n =   90.056  
                  
SSxx =    Σ(x-x̅)² =    2836.2778          
SSxy=   Σ(x-x̅)(y-ȳ) =   5389.6          
                  
estimated slope , ß1 = SSxy/SSxx =   5389.6   /   2836.278   =   1.90024
                  
intercept,   ß0 = y̅-ß1* x̄ =   -60.80246          
                  
so, regression line is   Ŷ =   -60.802   +   1.900   *x
                  
SSE=   (SSxx * SSyy - SS²xy)/SSxx =    1457.3848          
                  
std error ,Se =    √(SSE/(n-2)) =    9.544          
                  
correlation coefficient ,    r = Sxy/√(Sx.Sy) =   0.9356          

===============

Ho:   ß1=   0          
H1:   ß1 >   0          
n=   18              
alpha =   0.05              
estimated std error of slope =Se(ß1) = Se/√Sxx =    9.544   /√   2836.28   =   0.1792
                  
t stat = estimated slope/std error =ß1 /Se(ß1) =    1.9002   /   0.1792   =   10.6037
                  
t-critical value=    1.7459   [Excel function: =T.INV(α,df) ]          
Degree of freedom ,df = n-2=   16              
p-value = 0.00   
decison :    p-value<α , reject Ho              

reject Ho and conclude  that linear relations exists between X and y