Question

1. Let the test statistics Z have a standard normal distribution when H0 is true. Find the p-value for each of the following situations:
a) H1:μ>μ0,z=1.88
b) H1:μ<μ0,z=−2.75
c) H1:μ≠μ0,z=2.88
2. Let the test statistics T have t distribution when H0 is true. Find the p-value for each of the following situations (provide an interval if the exact one cannot be found using a table):
a) H1:μ>μ0,n=16,t=3.733
b) H1:μ<μ0,df=23,t=−2.500
c) H1:μ≠μ0,n=7,t=−2.250

Answer 1

Answer:

1 a  $p -value = 0.030054$

1b   $p -value = 0.0029798$

1c  $p -value = 0.0039768$  

2a  $p-value = 0.00099966$

2b  $p-value = 0.00999706$

2c  $p-value = 0.0654412$

Step-by-step explanation:

Considering question a

  The alternative hypothesis is H1:μ>μ0

   The test statistics is  z =1.88

Generally from the z-table  the  probability of   z =1.88 for a right tailed test is

    $p -value = P(Z > 1.88) = 0.030054$

Considering question b

  The alternative hypothesis is H1:μ<μ0

   The test statistics is  z=−2.75

Generally from the z-table  the  probability of   z=−2.75 for a left tailed test is

    $p -value = P(Z < -2.75) = 0.0029798$

Considering question c

  The alternative hypothesis is H1:μ≠μ0

   The test statistics is  z=2.88

Generally from the z-table  the  probability of  z=2.88 for a right  tailed test is

    $p -value = P(Z >2.88) = 0.0019884$    

Generally the p-value for the two-tailed test is

    $p -value = 2 * P(Z >2.88) = 2 * 0.0019884$    

=> $p -value = 0.0039768$  

Considering question 2a

    The alternative hypothesis is H1:μ>μ0

     The sample size is  n=16

     The  test statistic is  t =  3.733

Generally the degree of freedom is mathematically represented as

        $df = n - 1$

=>     $df = 16 - 1$

=>     $df = 15$

Generally from the t distribution table  the probability of   t =  3.733 at a degree of freedom of  $df = 15$ for a right tailed test is  

       $p-value = t_{3.733 , 15} = 0.00099966$

Considering question 2b

    The alternative hypothesis is H1:μ<μ0

     The degree of freedom is df=23

     The  test statistic is ,t= −2.500

Generally from the t distribution table  the probability of   t= −2.500 at a degree of freedom of  df=23 for a left  tailed test is  

       $p-value = t_{-2.500 , 23} = 0.00999706$

Considering question 2c

    The alternative hypothesis is H1:μ≠μ0

     The sample size is  n= 7

     The  test statistic is ,t= −2.2500

Generally the degree of freedom is mathematically represented as

        $df = n - 1$

=>     $df = 7 - 1$

=>     $df = 6$

Generally from the t distribution table  the probability of   t= −2.2500 at a degree of freedom of  $df = 6$ for a left   tailed test is  

       $t_{-2.2500 , 6} = 0.03272060$

Generally the p-value  for t= −2.2500 for a two tailed test is

     $p-value = 2 * 0.03272060 = 0.0654412$