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小美
Lv 4
小美 asked in 教育與參考考試 · 2 decades ago

統計學的問題-假設檢定

隨機抽取24位大學畢業之社會新鮮人得其薪資所得如下: (單位:千元)

25,28,30,33,25,25,27,28,30,30,31,32,32,35,36,37,30,32,32,35,36,33,35

(1)請計算平均數、變異數及標準差

(2)請以此計算大學畢業之社會新鮮人之薪資所得之95%信賴區間。

(3)若有人宣稱大學畢業之社會新鮮人之薪資所得為35000元,請問是否可信?

2 Answers

Rating
  • Hunter
    Lv 4
    2 decades ago
    Favorite Answer

    n=23(少一個資料)

    1.樣本平均數 X bar = ΣX / n =31.174

    樣本變異數 S^2 =Σ( X - X bar )^2 / (n-1) =13.241

    樣本標準差 S=√[Σ( X - X bar )^2 / (n-1)] = 3.6388

    2.樞紐量為( X bar - μ ) / ( S / √n ) ~ T(n-1)

    μ的95%CI為X bar +- T(0.025,22) * S/√n

    =[ 31.174 - 2.074*3.6388 / √23 , 31.174 + 2.074*3.6388 / √23 ]

    =[ 29.600 , 32.748 ]

    3.對大學畢業之社會新鮮人之薪資所得為35000元做檢定

    H0:μ=35

    H1:μ=/=35 (雙尾)

    α=0.05(自己設)

    ( X bar - μ ) / ( S / √n ) ~ T(n-1)

    統計檢定量T*= ( X bar - μ0 ) / ( S / √n ) = -5.0426 < T(0.975,22) = -2.074

    所以拒絕H0

    有顯著證據說大學畢業之社會新鮮人之薪資所得不是35000

    其實從第二小題

    可以看到μ的95%CI是不包括35的

    可以很清楚的知道有顯著證據說薪資所得不是35000

    Source(s): 自己 + fx-350ms + T表
  • 小美
    Lv 4
    2 decades ago

    有認為自已的答案是正確的,歡迎提供答案,我將會選出正確的答案!

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