You have a coin and p is the probability of heads.You throw the coin until you obtain heads for the first time.You want to test H 0 : p = 1 / 2 against H 1 : p &lt; 1 / 2. You reject H 0 in favor of H 1 if T ≥ k, where T is the number of the throw which yielded heads for the first time and k an integer.Determine the smallest value of k corresponding to the level of significance α=0.01.Answer k=8I can't find the correct solution for this question, how it's possible to have H 1 : p &lt; 1 / 2 and then they say that we reject H 0 in favor of H 1 if T ≥ k, shouldn't they be in the same direction?I have to compute P ( T &gt; k | H 0 ) = P ( T &gt; k | p = 1 / 2 ) right? How can I compute it?

Question

You have a coin and p is the probability of heads.You throw the coin until you obtain heads for the first time.You want to test       H    0    :  p  =  1      /    2 against       H    1    :  p  &amp;lt;  1      /    2. You reject       H    0   in favor of       H    1   if   T  ≥  k, where T is the number of the throw which yielded heads for the first time and k an integer.Determine the smallest value of k corresponding to the level of significance α=0.01.Answer k=8I can&#039;t find the correct solution for this question, how it&#039;s possible to have       H    1    :  p  &amp;lt;  1      /    2 and then they say that we reject       H    0   in favor of       H    1   if   T  ≥  k, shouldn&#039;t they be in the same direction?I have to compute   P  (  T  &amp;gt;  k      |        H    0    )  =  P  (  T  &amp;gt;  k      |    p  =  1      /    2  ) right? How can I compute it?

bibliothecaqz · Accepted Answer

You are in the right direction. You need to solve the inequality   P  (  T  ≥  k      |    p  =      1    2    )  ≤  0.01(I hope this makes sense for you, as this is just the definition of level of the test. Some authors also call it size. But there is a distinction between them. For e.g. look here. To be more precise, if you are given the size for the test of a simple hypothesis, we would have used equality sign above. )T being the total no. of throws required for obtaining the first head, its probability distribution can be computed as mentioned in the comment above. I would apply rather an intuitive method to do the same.                    P        (        T        ≥        k                  |                p        =                  1          2                )                            =        1        −        P        (        T        &amp;lt;        k        )        =        1        −                  [          P          (          T          =          1          )          +          P          (          T          =          2          )          +          …                                  +                    P          (          T          =          k          −          1          )          ]                                                  =        1        −                  [                      1            2                    +                                    (                              1                2                            )                        2                    ⋯          +                                    (                              1                2                            )                                      k              −              1                                ]                                                    (When we say that                       T            =            i            ,            i            &amp;gt;            1                    , we mean that the first                       (            i            −            1            )                     results in tails.                                                             Head is obtained only on the                                     i                              t                h                                               toss. Since the trials are independent, this                                                             probability is just                                                     (                                  1                  2                                )                                            i                −                1                                      .                          1              2                        =                                          (                                  1                  2                                )                                            i                                                    )                                          =                              (                          1              2                        )                                k            −            1                          ≤        0.01            The least value of k for which this holds is 8.

Answered question

Answer & Explanation

New Questions in College Statistics