關於operation mangement功課

求助...關於operation mangement功課(regression..要用excel)

圖片參考:http://i.uwants.com/u/images/default/emailtofriend...

[按此打開] [隱藏] graduateGPA(independent variable)starting salary(dependent variable)13.26$33,80022.6$29,80033.35$33,50042.86$30,40053.82$36,40062.21$27,60073.47$35,300a.construct a scatter plot od the above dataBy inspection,is there any potential relationship between GPA and Staring Salary?

b.Using linear regression to model the data.What is the Regression Equation?express the equation in the form:y=a+bx

c.what is the correlation coefficient(r) and the coefficient of determination(r2) of your model?what does they tell?

d.what is the range of the predicted starting salary is a student's GPA is around 3.50 to 3.70?

我知道上述q要用excel做~要利用excel的regression把圖表列出!但我不知道怎麼回答!!求名大大 一齊討論!thx

1 個解答

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  • TamTam
    Lv 7
    10 年前
    最愛解答

    圖片參考:http://imgcld.yimg.com/8/n/HA00004060/o/7010122102...

    The formulae for the cells:

    A10: =SLOPE(B1:B7,A1:A7)

    B10: =INTERCEPT(B1:B7,A1:A7)

    C10: =CORREL(B1:B7,A1:A7)

    (a) The scatter plot is drawn above. From the plot, there is a potential relationship between GPA and starting salary.

    (b) a=y-intercept=14815.6

    b=slope=5706.6

    Regression equation:

    y= 14815.6 + 5706.6x

    (c) correlation coefficient r = 0.9882

    coefficient of determination r^2 = 0.9882^2=0.9765

    (d) Regression equation:

    y= 14815.6 + 5706.6x

    For the GPA is around 3.50 to 3.70,

    y= 14815.6 + 5706.6(3.50) = 34789

    y= 14815.6 + 5706.6(3.70) = 35930

    $34789 < predicted salary < $35930

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