In: Statistics and Probability
A company wants to know how job performance relates to IQ. They collect data on 30 employees, resulting in the following table. Enter the data into SPSS OR EXCEL with appropriate variable labels. Show all syntax and output. (5pts)
Name |
Job Performance |
IQ |
Henry |
85 |
109 |
Riley |
84 |
106 |
Alexis |
87 |
125 |
Evelyn |
69 |
84 |
Blake |
69 |
89 |
Dominic |
81 |
109 |
Jose |
71 |
121 |
Tristan |
76 |
102 |
Kayden |
77 |
111 |
Makayla |
76 |
106 |
Ella |
90 |
107 |
Piper |
74 |
97 |
Jonathan |
74 |
133 |
Joshua |
65 |
96 |
Brooklyn |
66 |
97 |
Connor |
73 |
116 |
Sadie |
80 |
108 |
Zoe |
96 |
102 |
Cameron |
77 |
94 |
Jason |
73 |
98 |
Logan |
70 |
87 |
Olivia |
68 |
104 |
Madison |
66 |
85 |
Lucas |
86 |
145 |
Tyler |
88 |
105 |
Madeline |
82 |
96 |
Michael |
85 |
103 |
Mason |
78 |
115 |
Andrew |
87 |
135 |
Joseph |
72 |
104 |
Using an excel tool of data analysis of linear regression we get
SUMMARY OUTPUT | ||||||
Regression Statistics | ||||||
Multiple R | 0.440888 | |||||
R Square | 0.194382 | |||||
Adjusted R Square | 0.16561 | |||||
Standard Error | 7.412171 | |||||
Observations | 30 | |||||
ANOVA | ||||||
df | SS | MS | F | Significance F | ||
Regression | 1 | 371.1723 | 371.1723 | 6.75592438 | 0.014742959 | |
Residual | 28 | 1538.328 | 54.94027 | |||
Total | 29 | 1909.5 | ||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | |
Intercept | 51.27493 | 10.17996 | 5.036851 | 2.50752E-05 | 30.42223226 | 72.12763 |
IQ | 0.246708 | 0.094916 | 2.599216 | 0.014742959 | 0.052280775 | 0.441135 |
a) Since Intercept=51.2749 and Slope =0.2467
hence equation will be
y=0.2467*IQ+51.2749
b) The strength of the relationship is the R-value which is 0.4409
c) The hypothesis is
Ho: Slope=0
Ha: Slope 0
Since the p-value calculated for the test is 0.015 at 95 % confidence level which is less than the level of significance (0.05) hence we reject the null hypothesis and conclude that the slope is non zero value.