Mathematics, 21.10.2020 16:01 amelvin41

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

Answers: 2

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Answer from: klu65

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

Answer from: Quest

just find the gcf and solve the gcf's until you get each answer i have a really good feeling somethings missing xd (like the dimensions of the fence) *cough *cough *hint *hint

step-by-step explanation:

Answer from: Quest

fdvs

step-by-step explanation: vsfdvsfvd

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