Question

In a study to evaluate drug efficacy, the manufacturer of a new type of generic drug for treating a
disease wants to investigate the effectiveness of this generic drug as compared to the traditional
brand name drug, At the present time, the brand name drug is the only approved treatment for the
disease. Patients diagnosed with the disease have low concentration of a specific factor in their
blood. Treatment with the brand name drug will result in an increase in the concentration of the
specific factor in the blood. An experiment is conducted in which 10 patients currently with the
disease are randomly assigned to receive either the generic drug or the brand name drug for a
period of 9 months. After 9 months have elapsed, a measure of the concentration level of the
specific factor in the blood is obtained for each patient. The results are shown in Table Q1.

Brand name drug	11	9	8	10	10
Generic drug	9	8	7	9	8

(a) At a 5 % level of significance, apply a two-sample t test to determine whether there is
any difference in the concentration levels between brand name drug and generic drug.
Comment on the results.
(15 marks)
(b) Construct a 90 % confidence interval for the difference in the mean concentration
levels between brand name drug and generic drug.
(3 marks)
(c) Suppose now the manufacturer wishes to demonstrate whether the concentration level
of brand name drug exceeds that of generic drug by more than 1, apply a two-sample t
test to analyze the problem based on 1 % level of significance.
(8 marks)
(d) For part (i) above, use R to perform the two-sample t test. Show the screenshots of the
R commands and output results.

(4 marks)

Answer 1

Soolution-A:

Ho:$\mu 1=\mu 2$

Ha:$\mu 1\neq \mu 2$

mu1--mean of brand name drug

mu2-mean of generic drug

alpha=0.05

t=x1bar-x2bar/sqrt(s1^2/n1+s2^2/n2)

For

Brandname drug

sample mean=x1bar=9.6

sample standard deviation=s1=1.140175

sample size=n1=5

For Generic drug

sample mean=x2bar=8.2

sample standard deviation=s2=0.83666

sample size=n2=5

t=(9.6-8.2)/sqrt(1.140175^2/5+0.83666^2/5)

t= 2.213595

test statistic,t= 2.213595

df=n1+n2-2=5+5-2=8

p value in excel is

=T.DIST.2T( 2.213595,8)

=0.057756258

=0.0578

p>0.05

Do not reject Ho

Accept Ho

Conclusion:

There is no suffcient statistical evidence at 5% level of significance to conclude that there is
any difference in the concentration levels between brand name drug and generic drug.

(b) Construct a 90 % confidence interval for the difference in the mean concentration
levels between brand name drug and generic drug.

Brandnamedrug 11 Generic drd=brand-gdbar d-dbar (d-bar)^2 1.4 0.6 0.36 1.4 -0.4 0.16 87 1.4 -0.4| 0.16 1 1.4 -0.4|| 0.16: 1.4 0.6 0.36 ...total 1.2

Sd=ssqrt(1.2/5-1)

Sd= 0.5477226

df=n-1=5-1=4

alpha=0.10

alpha/2=0.10/2=0.05

t critical value in excel

=T.INV(0.05,4)

=2.13185

90% confidence interval for difference in means

=dbar+-tcrit*sd/sqrt(n)

=1.4+-2.13185* 0.5477226/sqrt(5)

=1.4-(2.13185* 0.5477226)/sqrt(5),1.4+(2.13185* 0.5477226)/sqrt(5)

= 0.8778055, 1.922195

90% lower limit mean difference=0.8778055

90% upper limit mean difference=1.922195

c) Suppose now the manufacturer wishes to demonstrate whether the concentration level
of brand name drug exceeds that of generic drug by more than 1, apply a two-sample t
test to analyze the problem based on 1 % level of significance.

Ho:$\mu 1-\mu 2=1$

Ha:$\mu 1-\mu 2 >1$

alpha=0.01

t=((9.6-8.2)-1)/(sqrt(1.140175^2/5+0.83666^2/5))

t=0.6324557

df=n1+n2-2=8

pvalue in excel

==T.DIST.RT(0.6324557,8)

=0.272368598

p value=0.2724

p>0.01

Do not reject Ho

Accept Ho

There is no suffcient statistical evidence at 1% level of significance to conlcude that the concentration level
of brand name drug exceeds that of generic drug by more than 1.

(d) For part (i) above, use R to perform the two-sample t test. Show the screenshots of the
R commands and output results.

Rcode:

Brand_name_drug <- c(11,   9,   8,   10,   10)
Generic_drug <- c(   9   ,8   ,7,   9,   8)
t.test(Brand_name_drug,Generic_drug)

Output:

data: Brand_name_drug and Generic_drug
t = 2.2136, df = 7.3394, p-value = 0.06072
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.08161171 2.88161171
sample estimates:
mean of x mean of y
9.6 8.2

t=2.2136

p=0.06072

p>0.05

Do not reject Ho

There are no differences in means of Brand name drug and generic drug

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