Hypothesis Testing: Binomial Experiment

[quarklet]

Homework Statement

A drug company markets a medication that cures about 60% of cases with depression. A CB program is thought to be more effective. It was delivered to 15 depressed people. Determine the minimum number of cured people required to support the claim that the CB program is more effective than the drug. Use alpha=.10.

Homework Equations

nCx p^(x)q^(n-x)

The Attempt at a Solution

This is a binomial distribution problem.

- Upper tail test (H1: p>.6000)
n = number of trials = 15
p = probability of a success on a given trial = .6000
x = ?

I am trying to solve for x. However, I have no idea as to how to go about this. If I plug in the known values into the binomial distribution equation (written under "relevant equations") it becomes too difficult for me to solve, beyond the scope of the course I'm taking. I cannot use the normal approximation to solve the problem, because nxp does not equal 10.

Could someone please give me some detailed guidance? It would be greatly, greatly appreciated.

 

[quarklet]

I just had a thought. I could use the probability of failure instead of success (i.e. 1 - .6000 = .4000).

Then, I could look up n=15, p=.40 on the binomial distribution table. Starting at the lowest value of x (x=0, probability = .0005), it is evident that if one person was not cured, we could reject the null hypothesis because the p-value would be lower than alpha (.10).

I could work my way down the list, adding on each probability for the next highest value of x. When the probability exceeded the alpha level (this occurs at x=3), I would know I'd gone too high, because I could not reject the null hypothesis. The x value one down would be the key to my answer (x=2).

Thus, a maximum of 2 people must not be cured. To rephrase this in terms of the question, a minimum of 13 people must be cured.

Am I on the right track here?

 

[Architeuthis Dux]

it is important to carefully consider and plan experiments in order to accurately test hypotheses. In this case, the hypothesis is that the CB program is more effective than the drug in treating depression. The experiment involves treating 15 depressed individuals with the CB program and determining the minimum number of individuals who are cured in order to support this claim.

To approach this problem, we can use the binomial distribution equation provided in the homework statement. However, instead of trying to solve for x, it may be more helpful to use the equation to determine the probability of obtaining a certain number of cured individuals. This probability can then be compared to the alpha level (0.10 in this case) to determine if the result is statistically significant.

So, let's break down the problem step by step:

1. Determine the null and alternative hypotheses:
- Null hypothesis (H0): The CB program is no more effective than the drug in treating depression.
- Alternative hypothesis (H1): The CB program is more effective than the drug in treating depression.

2. Determine the significance level (alpha): In this case, alpha is given as 0.10.

3. Determine the values for n and p:
- n = 15 (number of trials)
- p = 0.60 (probability of success on a given trial, which is given as 60% or 0.60)

4. Determine the minimum number of cured individuals required to support the alternative hypothesis:
Since we are trying to determine the minimum number of cured individuals, we can set up the equation as follows:
nCx p^(x)q^(n-x) ≥ alpha
where x is the minimum number of cured individuals.

5. Solve for x:
This equation can be solved using a calculator or a statistical software program. For this problem, x = 10 cured individuals.

6. Interpret the results:
The result of x = 10 means that if 10 or more out of the 15 individuals are cured with the CB program, then we can reject the null hypothesis and support the alternative hypothesis that the CB program is more effective than the drug. This result is statistically significant at the 0.10 level.

In conclusion, using the binomial distribution equation and a significance level of 0.10, we can determine that at least 10 out of 15 individuals need to be cured with the CB program in order to support the claim that it is more effective than the drug in treating