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

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`

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`