X 527 sigma 48 n 25 H0 mu 50 versus Ha mu 50 alpha 005

X = 52.7, sigma = 4.8, n = 25 H_0: mu = 50 versus H_a: mu 50, alpha = 0.05 If mu is really 53, what is the power of the test? You might get some numbers outside the normal probability charts (CPT-18,19). In that case, probability is either 0 or 1. H_0: mu = 200 versus H_a: mu 200 x = 206.73, sigma = 6.35, n = 40 What is the P-value?

Solution

I)
Set Up Hypothesis
Null Hypothesis H0: U=50
Alternate Hypothesis H1: U!=50
Test Statistic
Population Mean(U)=50
Sample X(Mean)=52.7
Standard Deviation(S.D)=4.8
Number (n)=25
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =52.7-50/(4.8/Sqrt(24))
to =2.813
| to | =2.813
Critical Value
The Value of |t | with n-1 = 24 d.f is 2.064
We got |to| =2.813 & | t | =2.064
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value :Two Tailed ( double the one tail ) -Ha : ( P != 2.8125 ) = 0.0096
Hence Value of P0.05 > 0.0096,Here we Reject Ho

Power of test = 100 * ( 1 - 0.0096 ) = 1

II)
t-test For Single Mean
Set Up Hypothesis
Null Hypothesis H0: U=200
Alternate Hypothesis H1: U!=200
Test Statistic
Population Mean(U)=200
Sample X(Mean)=206.73
Standard Deviation(S.D)=6.35
Number (n)=40
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =206.73-200/(6.35/Sqrt(39))
to =6.703
| to | =6.703
Critical Value
The Value of |t | with n-1 = 39 d.f is 2.023
We got |to| =6.703 & | t | =2.023
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value :Two Tailed ( double the one tail ) -Ha : ( P != 6.703 ) = 0
Hence Value of P0.05 > 0,Here we Reject Ho