Assume I'm thinking of investing in a small biotech company. In a phase 2 study with 120 patients sp

2d3vljtq 2022-07-08 Answered
Assume I'm thinking of investing in a small biotech company. In a phase 2 study with 120 patients split 80/40 between the new drug and old drug, the progression free survival (PFS) rate of the patients taking the new drug was greater than that of those taking the old drug at the 1% level (P = 0.01).
Now, the company is conducting a phase 3 trial with 300 patients split 200/100 between the new drug and the old drug. What is the probability that the PFS will be greater for the patients taking the new drug than for those taking the old drug at the 5% level (P = 0.05) or better?
In other words, what is the probability that the phase 3 trial will be successful so that the FDA will presumably approve the drug? I want to know if the drug company is a good investment, which of course also depends upon the market capitalization of the company and the size of the potential market, but the first question is what the probability is of the drug trial being successful? Is there any way to get a handle on that based only on the P2 results?
Edit: P.S. This is not a homework problem, though it (or something like it) would seem to be a standard and useful problem.
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (1)

vrtuljakwb
Answered 2022-07-09 Author has 13 answers
It seems reasonable to assume that the tests involved in both Phase 2 and Phase 3 are tests that compare 'survival' fractions p 1 = X 1 / n 1 (new drug) with p 2 = X 2 / n 2 (old drug) to see if the former significantly exceeds the latter.
The test statistic z = ( p 1 p 2 ) / d, where d 2 = p 1 ( 1 p 1 ) / n 1 + p 2 ( 1 p 2 ) / n 2 .. We would say that the new drug is significantly better than the old at if z>z∗, where the critical value is z∗=2.3263 for a test at the 1% level and z∗=1.6449 at the 5% level.
For the Phase 2 test, we had n 1 = 80 and n 2 = 40.. The surviving numbers X 1 and X 2 with the new and old drugs, respectively, are unknown. To start, we assume that X 2 = 20 so that p 2 = 0.5. Then we ask what X1 must have been in order to achieve significance at the 1% level. Algebra would be a little messy, but a straightforward grid search for the smallest X1 that would give the desired result is X 1 = 58 so that p 1 = 58 / 80 = 0.725 ,, z = 2.4064 > 2.3263.
Then for a the Phase 3 test (assuming the same protocol for choosing patients, drug dosages, etc.) we suppose we would be choosing X 2 B i n o m ( 100 , 0.5 ) . Finding the corresponding p 1 , p 2 , and z, we wonder what proportion of the z's would exceed 1.6449 in order to have significance at the 5% level, and (we hope) approval by the FDA. A simple simulation shows that proportion to be about 98.4%.
Fortunately, the outlook is about as promising, if we assume p 2 = 0.3 in the the Phase 2 trial; and even more promising (proportion about 99.5%), if we assume the the old drug has p2=0.8.
By making additional, possibly realistic, assumptions, this problem might be solved satisfactorily without using a computer. If this is a homework problem, I will leave that to you.
In case you are interested in my computation and simulation in R, the code is shown below. To avoid confusion in programming, I used X's and p's in Phase 2 and Y's and q's in Phase 3. (I do not claim my code is optimal or elegant.)
# Phase 2: Grid search for required p1
n1 = 80; n2 = 40; x2 = 20; z = numeric(n1)
for(x1 in 1:80) {
p1 = x1/n1; p2 = .5
den = sqrt( p1*(1-p1)/n1 + p2*(1-p2)/n2 )
z[x1] = (p1 - p2)/den }
crit = qnorm(.99)
i = 1:n1
pp1 = min(i[z > crit])/n1 ; pp1
## 0.725
Phase 3: Proportion of expts signif at 5%
m = 10^5; n1 = 200; n2 = 100
y1 = rbinom(m, n1, pp1); y2 = rbinom(m, n2, p2)
q1 = y1/n1; q2 = y2/n2
den = sqrt( q1*(1-q1)/n1 + q2*(1-q2)/n2 )
z = (q1 - q2)/den
mean(z > qnorm(.95))
## 0.98436
Did you like this example?
Subscribe for all access

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2022-06-20
An electronics retailer would like to investigate the relationship between the selling price of a digital camera and the demand for it. The table shown below gives the weekly demand for the camera in one particular market along with the corresponding price. These data have a sample correlation​ coefficient, rounded to three decimal​ places, of negative 0.888. Using alpha equals 0.10​, test if the population correlation coefficient between the selling price and the demand for the camera is less than zero.
Demand Price 17 300 20 310 15 320 11 330 9 340
What is the​ p-value?
asked 2022-06-26
H 0 : μ = 0 against H A : μ > 0.
Had the following question on my exam today and I'm just wondering if I did this correctly.
From a normally distributed population with mean μ and variance σ 2 a sample has been drawn:
X = ( 0.05 , 4.35 , 0.48 , 0.63 , 1.17 , 2.01 ) ..
a) Test H 0 : μ = 0 against H A : μ > 0 on a 5% significance level under the assumption that σ2 is unknown.
c) Do the same test, given that σ = 1.5..
Solution: I know that H 0 is to be rejected if T c ,, where T is the teststatistic and F t n 1 = 1 α ,, where α is the significance level.
So lets first compute them for a), where we need to use a t-distribution since σ is unknown, so we have that c = F t 5 1 ( 0.95 ) 2.015.. Also, the estimator for σ 2 is
s 2 = 1 n 1 ( k = 1 n X k 2 1 n ( k = 1 n X k ) 2 ) = 1 5 ( 24.961 1 6 ( 41.861 ) ) = 3.405 ,
so, s = 3.405 = 1.845.. We also have that X ¯ = 6.47 / 6 = 1.078..
The teststatistic is
T = X ¯ μ 0 s / n = 1.078 0 1.845 / 6 = 1.078 0.753 = 1.43 c ,
which means that we can not reject H0.
For b) we do an identical approach but wtith s = σ = 1.5 and we use a z-test instead. Here we have that c = Φ 1 ( 0.95 ) = 1.644. The teststatistic in this case is
Z = X ¯ μ 0 σ / n = 1.078 1.5 / 6 = 1.76 c ,
thus we reject H 0 ..
Is this correct?
What my professor does in the solutions is he just makes a 95% confidenceinterval and just arrives to the same conclusions as I've done. Is his method correct without using the teststatistics?
asked 2022-06-15
The log return X on a certain stock investment is an N(μ,σ2) random variable.
A financial analyst has claimed that the volatility σ of the log return on this stock is less than 3 units. A random sample of 11 returns on this stock gave an estimated variance of the log-returns as s2=16.
Assess the analyst's claim by using a significance test at level
α=0.05 to test
H0:σ2≤9 against H1:σ2>9.
Find a two-sided 95% confidence interval for σ2.
asked 2022-10-18
We have the Random Variable X, which is Γ ( p , λ ) distributed with the density:
f p , λ ( x ) = λ p Γ ( p ) x p 1 e λ x
with p=10 and H 0 : λ = 2 or H 1 : λ = 4 and α = 0.001
I want to apply the Lemma of Neyman Pearson which states:
Be c>0 fixed and chosen in the way that A ( c ) = { x B : f 0 ( x ) f 1 ( x ) c } such that P H 0 ( X A ( c ) ) = α
Then the test with the region A(c) among all tests with significance level α is the most powerful.
I am now trying to calculate A(c), but got stuck. I have:
A ( c ) f 0 ( x ) d x = A ( c ) λ p Γ ( p ) x p 1 e λ x d x = α .
But I don't know how to get A(c) from this integral...
f 0 ( x ) f 1 ( x ) = 1 1024 e 2 x
asked 2022-10-24
A chi square test is conducted to check whether a person's ability in Mathematics has an impact on his/her interest in Statistic.
The test statistic is 13.277 under the tested null hypothesis. write a recommended null hypothesis and an alternative hypothesis. Briefly describe your conclusion on this test at the 0.01 significance level.
asked 2021-06-18

Although tea is the world’s most widely consumed beverage after water, little is known about its nutritional value. Folacin is the only B vitamin present in any significant amount in tea, and recent advances in assay methods have made accurate determination of folacin content feasible. Consider the accompanying data on folacin content for randomly selected specimens of the four leading brands of green tea. Brand 1234amp; Observations amp;7.9amp;5.7amp;6.8amp;6.4amp;6.2amp;7.5amp;7.5amp;7.1amp;6.6amp;9.8amp;5.0amp;7.9amp;8.6amp;6.1amp;7.4amp;4.5amp;8.9amp;8.4amp;5.3amp;5.0amp;10.1amp;amp;6.1amp;4.0amp;9.6amp;amp;amp;
(Data is based on “Folacin Content of Tea,” J. Amer. Dietetic Assoc., 1983: 627–632.) Does this data suggest that true average folacin content is the same for all brands? a. Carry out a test using α:=.05 via the P-value method. b. Assess the plausibility of any assumptions required for your analysis in part (a). c. Perform a multiple comparisons analysis to identify significant differences among brands.

asked 2021-11-19
Your math test scores are 68, 78, 90, and 91.
What is the lowest score you can earn on the next test and still achieve an average of at least 85?
Getting an Answer Solve your inequality to find the lowest score you can earn on the next test and still achieve an average of at least 85.
What score do you need to earn?

New questions