Question

To examine the effect of a drug on blood pressure, 10 diabetics before and after the blood
pressures are measured and the following results are obtained.
 a) For this type of mass diabetics, what is the average difference between blood pressure before and after the drug?
You can tell?
 b) Do you have an estimate of the difference between the mean and before blood pressure?
 c) 90% confidence interval for the mean difference in blood pressure of the mass before and after the drug
Found it.
 d) What is the Spacing Length? How could the interval length be if there were 50 diabetics?

X(before) Y(after)

123 128

136 144

136 138

120 128

125 128

132 136

126 122

132 126

116 120

133 126

Answer 1

a) In this question we have two samples in which observations in one sample(Blood pressure before the drug x_i ) can be paired with observations in the other sample(Blood pressure after the drug y_i). So we can use a paired t-test.

The difference between blood pressure before and after the drug

d_i = y_i − x_i

The difference calculated using excel is shown below.

Y(after)	X(before)	Difference
128	123	5
144	136	8
138	136	2
128	120	8
128	125	3
136	132	4
122	126	-4
126	132	-6
120	116	4
126	133	-7
Mean Difference()	=(5+8+2+8+3+4-4-6+4-7)/10=1.7

Hypothesis

H₀: True mean difference is zero

H₁: True mean difference is not zero.

standard deviation of the difference(s_d) = 5.478.

Therefore, Standard Error( ) = = = 1.732

So, we have test statistic t = mean difference/standard error = 1.7/1.732= 0.981 on 9 degress of freedom

Looking this up in t table gives p = .352086..

The result is insignificant as p > .05. Which means we cant reject the null hypothesis

So we can say that there is no evidence to show that there is difference in the blood pressure before and after the drug at 0.05 significance level.

As we were not able to reject the null hypothis, the difference between the blood pressure should be considered to be 0

b)Mean difference is 0 as per the above paired t test.

c)

confidence interval for the mean difference

t value for 90% confidence at 9 degrees of freedom= 1.833113(from t table)

S_d= 5.478, n= 10

Therefore confidence interval for  = (-3.175633,3.175633)

d) Interval if there were 50 samples,

Z value for 90% confidence is 1.65

S_d= 5.478, n= 50

Therefore confidence interval for  = (-1.28,1.28)