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廖廖酱 · 2023年02月08日

关于拒绝域

NO.PZ2015120204000015

问题如下:

Based on past research, Hansen selects the following independent variables to predict IPO initial returns:

Underwriter rank = 1–10, where 10 is highest rank

Pre-offer price adjustment (Expressed as a decimal) = (Offer price – Initial filing price)/Initial filing price

Offer size ($ millions) = Shares sold × Offer price

Fraction retained (Expressed as a decimal) = Fraction of total company shares retained by insiders

He also believes that for each 1 percent increase in pre-offer price adjustment, the initial return will increase by less than 0.5 percent, holding other variables constant. Hansen wishes to test this hypothesis at the 0.05 level of significance.

Hansen collects a sample of 1,725 recent IPOs for his regression model.

\Hansen’s Regression Results Dependent Variable: IPO Initial Return (Expressed in Decimal Form, i.e., 1% = 0.01)

Selected Values for the t-Distribution (df = ∞)

The most appropriate null hypothesis and the most appropriate conclusion regarding Hansen’s belief about the magnitude of the initial return relative to that of the pre-offer price adjustment (reflected by the coefficient bj) are:

选项:

Null Hypothesis
Conclusion about bj(0.05 Level of Significance)
A.
H0: bj=0.5
Reject H0
B.
H0: bj≥0.5
Fail to reject H0
C.
H0: bj≥0.5
Reject H0

解释:

C is correct.

C To test Hansen’s belief about the direction and magnitude of the initial return, the test should be a one-tailed test. The alternative hypothesis is H1: bj<0.5b_j<0.5, and the null hypothesis is H0:bj0.5b_j\geq0.5 . The correct test statistic is: t = (0.435-0.50)/0.0202 = -3.22, and the critical value of the t-statistic for a one-tailed test at the 0.05 level is -1.645. The test statistic is significant, and the null hypothesis can be rejected at the 0.05 level of significance.

老师您好,不太明白为什么本题的拒绝域是在左边?小于-1.645.

1 个答案
已采纳答案

星星_品职助教 · 2023年02月09日

同学你好,

1)根据题干中的“...believes that for each 1 percent increase in pre-offer price adjustment, the initial return will increase by less than 0.5 percent”,可知研究人员相信的是bj<0.5。由于“相信的”要放备择假设,所以可确定备择假设为bj<0.5。(进而得到H0为 bj≥0.5)

2)拒绝域由备择假设决定,由备择假设可知拒绝域只在“小于”的这一侧,也就是在左尾;

3)根据“at the 0.05 level of significance”,可知拒绝域的面积为0.05。根据表格值,右尾0.05对应正1.645,所以左尾对应的就是-1.645.

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