Significance level for a hypothesis test for a linear regressionConsider linear regression model Y i = a + b ⋅ x i + ϵ i , i = 1 , 2 , 3 , 4 , 5 where a , b ∈ R are unknown and x 1 = x 2 = 1 , x 3 = 3 , x 4 = x 5 = 5, ϵ i are iid, normally distributed with mean =0 and variance =9. Consider the hypothesis H 0 : b = 0 with the alternative H 1 : b ≠ 0, with the critical region { | b ^ | &gt; c }, where b^ is a maximum likelihood estimator and c is chosen in such a way that the significance level of a test is equal to 0,05. Calculate c.Could you help me with this exercise? It is taken from the actuary exam organized in my country. I thought that I am able solve this exercise, however my answer c=0,7528 is wrong, the correct answer is c=1,47.Edited: the exercise seems very easy, but I'm sure that my method of solution is wrong, as I've seen the similar exercise and my method results with the wrong answer as well. That's why I've decided to start a bounty, however I do not know how (I can see "question eligible for bounty in 59 minutes" only, not "start a bounty")

Question

Significance level for a hypothesis test for a linear regressionConsider linear regression model       Y    i    =  a  +  b  ⋅      x    i    +      ϵ    i  ,   i  =  1  ,  2  ,  3  ,  4  ,  5 where   a  ,  b  ∈      R   are unknown and       x    1    =      x    2    =  1  ,      x    3    =  3  ,      x    4    =      x    5    =  5,       ϵ    i   are iid, normally distributed with mean =0 and variance =9. Consider the hypothesis       H    0    :  b  =  0 with the alternative       H    1    :  b  ≠  0, with the critical region   {      |              b      ^            |    &amp;gt;  c  }, where b^ is a maximum likelihood estimator and c is chosen in such a way that the significance level of a test is equal to 0,05. Calculate c.Could you help me with this exercise? It is taken from the actuary exam organized in my country. I thought that I am able solve this exercise, however my answer c=0,7528 is wrong, the correct answer is c=1,47.Edited: the exercise seems very easy, but I&#039;m sure that my method of solution is wrong, as I&#039;ve seen the similar exercise and my method results with the wrong answer as well. That&#039;s why I&#039;ve decided to start a bounty, however I do not know how (I can see &quot;question eligible for bounty in 59 minutes&quot; only, not &quot;start a bounty&quot;)

Glenn Ware · Accepted Answer

I am having trouble making sense of this problem. In your notation, the usual regression model is      Y    i    =  a  +  b      x    i    +      ϵ    i    ,where ϵi are distributed   N  o  r  m  (  0  ,      σ    ϵ    2    ) for i=1,…,n.A 95% confidence interval for the slope is  b  ±      t    ∗        s    ϵ        1          /              S              x        x              ,where t∗=3.182 cuts off area .025 from the upper tail of Student&#039;s t distribution with n−2=3 degrees of freedom,      s          ϵ        2    =      [          ∑      i        (          Y      i        −                            Y          ^                    i              )      2        ]        /    (  n  −  2  )  ,      S          x      x        =  [      ∑    i    (      x    i    −            x      ¯            )    2    ]  , and                     Y        ^              i    =            a      ^        +            b      ^            x    i   are predicted values from the regression line. Here,       s    ϵ    2   estimates       σ    ϵ    2    ..For this model, you would reject the null hypothesis if 0 is not contained in this interval. That is, you would reject if       |              b      ^            |    &amp;gt;      t    ∗        s    ϵ        1          /              S              x        x              =  c  .The data are sufficiently simple that the computation could be done on a calculator, but some results from Minitab statistical software are shown below for verification. In particular,   b  =  1.20  ,       s          ϵ        =  0.730297  ,, and       s          ϵ            1          /              S              x        x              =  0.7303      1          /        10    =  0.2309..The regression equation isy = - 0.600 + 1.20 xPredictor     Coef  SE Coef      T      PConstant   -0.6000   0.7659  -0.78  0.491x           1.2000   0.2309   5.20  0.014S = 0.730297   The resulting value of c=3.182(0.2309)=0.735 seems close (perhaps even within rounding error) to your value 0.7528.For this usual regression model, c=1.47 cannot be correct. It would indicate that the null hypothesis       H    0    :  b  =  0 would not be rejected because b=1.20. Minitab has P-value 0.014, indicating rejection. Also, a look at the regression line through the data pretty clearly shows that a zero slope is absurd.However, yours in not a standard model because you are given that       σ    ϵ    =      9    =  3  ,, which is not anywhere near the above estimate       s    ϵ    ≈  0.73.. The only way this value of σϵ could be taken seriously would be to claim that prior experience completely overrides the current data. And in that case, what sense does it make to use the current data to estimate b?I am not sure about the distribution theory of estimating b when σϵ is known. I have experimented with several possibilities that seem reasonable, but none of them gives c=1.47.

Significance level for a hypothesis test for a linear regression Consider linear regression model

Answered question

Answer & Explanation

New Questions in College Statistics