Consider the Minitab output shown below How many degrees of

Consider the Minitab output shown below. How many degrees of freedom are there on the /-test statistic? Fill in the missing information.

Solution

a)
Standard Error= sd/ Sqrt(n)
Where,
sd = Standard Deviation
n = Sample Size

Standard deviation( sd )= Unknown
Sample Size(n)=12
Standard Error = 0.3360
0.3360 = ( s.d/ Sqrt ( 12) )
s.d = 0.3360*Sqrt ( 12) = 1.164

b)
Null, H0: U=25
Alternate, H1: U>25
Test Statistic
Population Mean(U)=25
Sample X(Mean)=25.6818
Standard Deviation(S.D)=1.164
Number (n)=12
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =25.6818-25/(1.164/Sqrt(11))
to =2.03
P-Value :Right Tail - Ha : ( P > 2.0291 ) = 0.034

c)
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=25.6818
Standard deviation( sd )=1.164
Sample Size(n)=12
Confidence Interval = [ 25.6818 ± t a/2 ( 1.164/ Sqrt ( 12) ) ]
= [ 25.6818 - 2.201 * (0.336) , 25.6818 + 2.201 * (0.336) ]
= [ 24.942,26.421 ]

ANS:
a)
The Value of |t | with n-1 = 11 d.f
b)
Std dev = 1.164
Lower bound = 24.942
T = 2.03