A special diet is intended to reduce systolic blood pressure

A special diet is intended to reduce systolic blood pressure. If the diet is effective, the target is to have the average systolic blood pressure of this group be below 150. After six months on the diet, an SRS of 28 patients with high blood pressure had an average cholesterol of x = 143, with standard deviation s = 21. Is this sufficient evidence that the diet is effective in meeting the target? Assume the distribution of the cholesterol for patients in this group in approximately Normal with mean mu...

Solution

t-test For Single Mean
Set Up Hypothesis
Null, H0: U=150
Alternate, H1: U<150
Test Statistic
Population Mean(U)=150
Sample X(Mean)=143
Standard Deviation(S.D)=21
Number (n)=28
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =143-150/(21/Sqrt(27))
to =-1.764
| to | =1.764
Critical Value
The Value of |t | with n-1 = 27 d.f is 1.521
We got |to| =1.764 & | t | =1.521
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value :Left Tail -Ha : ( P < -1.7638 ) = 0.04454
Hence Value of P0.07 > 0.04454,Here we Reject Ho

E.
We are Reject Ho, It is TYPE I error

F.
Confidence Interval
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)=143
Standard deviation( sd )=21
Sample Size(n)=28
Confidence Interval = [ 143 ± t a/2 ( 21/ Sqrt ( 28) ) ]
= [ 143 - 2.052 * (3.969) , 143 + 2.052 * (3.969) ]
= [ 134.856,151.144 ]