Suppose a group of 23 patients were selected for a study to

Suppose a group of 23 patients were selected for a study to examine the effects of a new drug in reducing blood pressure (BP). The group of patients were selected from a population of people classified with a high body mass index. The average systolic BP of the sample patients was 135 with a standard deviation of 16. Use a type I error of = 0.05 to conduct a hypothesis test (assuming the normal distribution) to determine if the population has a systolic BP of greater than 120. What is the calculated p-value?

The appropriate hypotheses are:

H0: = = 120

H1: = > 120.

Suppose a group of 23 patients were selected for a study to examine the effects of a new drug in reducing blood pressure (BP). The group of patients were selected from a population of people classified with a high body mass index. The average systolic BP of the sample patients was 135 with a standard deviation of 16. Use a type I error of = 0.05 to conduct a hypothesis test (assuming the t-distribution) to determine if the population has a systolic BP of greater than 120. What is the calculated p-value?

The appropriate hypotheses are:

H0: = = 120

H1: = > 120.

3. What are the 99% confidence bounds for the sample patients BP assuming a normal distribution?

4. What are the 99% confidence bounds for the sample patients BP assuming a t-distribution?

Solution

1)
Set Up Hypothesis
Null Hypothesis, has a systolic BP equals to 120 H0: U=120
Alternate, has a systolic BP of greater than 120 H1: U>120
Test Statistic
Population Mean(U)=120
Sample X(Mean)=135
Standard Deviation(S.D)=16
Number (n)=23
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =135-120/(16/Sqrt(22))
to =4.496
| to | =4.496
Critical Value
The Value of |t | with n-1 = 22 d.f is 1.717
We got |to| =4.496 & | t | =1.717
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value :Right Tail - Ha : ( P > 4.4961 ) = 0.00009
Hence Value of P0.05 > 0.00009,Here we Reject Ho

We conclude that, has a systolic BP of greater than

3)

Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=135
Standard deviation( sd )=16
Sample Size(n)=23
Confidence Interval = [ 135 ± Z a/2 ( 16/ Sqrt ( 23) ) ]
= [ 135 - 2.58 * (3.34) , 135 + 2.58 * (3.34) ]
= [ 126.39,143.61 ]