1. To compare a new drug for diabetes with the standard one, 36 diabetics were g

Question

1.       To compare a new drug for diabetes with the standard one, 36 diabetics were given the new drug and 49 were given the standard one. After six months, the average fasting blood sugar level of the patients who took the new drug was 120 with a standard deviation of 10. The patients who took the standard drug had a mean fasting blood sugar level of 135 with a standard deviation of 12.

a.       Write the relevant null and alternative hypotheses to determine if the new drug is more effective in reducing blood sugar.

b.       Refer to your formula sheet and write down the appropriate formula for the test statistic.

c.       A 98% confidence interval for the actual difference between mean fasting blood sugar levels with the standard one and the new one is (15.1, 20.1), Can you conclude that the new drug is more effective in reducing the fasting blood sugar?

d.       Use STATKEY to find a 98% confidence interval for the true mean fasting blood sugar after taking the new drug for six months.

Explanation / Answer

1.

Given that,
mean(x)=120
standard deviation , 1 =10
number(n1)=36
y(mean)=135
standard deviation, 2 =12
number(n2)=49
null, Ho: u1 = u2
alternate, H1: 1 < u2
level of significance, = 0.02
from standard normal table,left tailed z /2 =2.05
since our test is left-tailed
reject Ho, if zo < -2.05
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=120-135/sqrt((100/36)+(144/49))
zo =-6.274
| zo | =6.274
critical value
the value of |z | at los 0.02% is 2.05
we got |zo | =6.274 & | z | =2.05
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: left tail - Ha : ( p < -6.274 ) = 0
hence value of p0.02 > 0,here we reject Ho
ANSWERS
---------------
a.
null, Ho: u1 = u2
alternate, H1: 1 < u2
b.
test statistic: -6.274
critical value: -2.05
decision: reject Ho
p-value: 0
c.
TRADITIONAL METHOD
given that,
mean(x)=120
standard deviation , 1 =10
population size(n1)=36
y(mean)=135
standard deviation, 2 =12
population size(n2)=49
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((100/36)+(144/49))
= 2.391
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.02
from standard normal table,left tailed z /2 =2.054
since our test is left-tailed
value of z table is 2.054
margin of error = 2.054 * 2.391
= 4.911
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (120-135) ± 4.911 ]
= [-19.911 , -10.089]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=120
standard deviation , 1 =10
number(n1)=36
y(mean)=135
standard deviation, 2 =12
number(n2)=49
CI = x1 - x2 ± Z a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ ( 120-135) ±Z a/2 * Sqrt( 100/36+144/49)]
= [ (-15) ± Z a/2 * Sqrt( 5.717) ]
= [ (-15) ± 2.054 * Sqrt( 5.717) ]
= [-19.911 , -10.089]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 98% sure that the interval [-19.911 , -10.089] contains the difference between
true population mean U1 - U2
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 98% of these intervals will contains the difference between
true population mean U1 - U2
3. Since this Cl does contain a zero we can conclude at 0.02 true mean
difference is zero

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