MATH SOLVE

4 months ago

Q:
# A chemical laboratory in Peru has discovered certain substances to lose weight in a short time to replace existing medications. The proportion of people in which the substance is effective is 0.7, while currently existing drugs are 60% effective a random sample of 100 people were given the new substance While the old drug was used in a sample of 150 What is the probability of observing a p1-p2 value less than 5%?

Accepted Solution

A:

To begin with, we need to calculate the sample proportions for both the new substance and the old drug:
For the new substance:
Sample size (n1) = 100
Proportion of people for whom the substance is effective (p1) = 0.7
Sample proportion (p̂1) = x1/n1, where x1 is the number of people in the sample for whom the substance was effective.
We don't know the value of x1, but we can use the given proportion to estimate it:
x1 = p1 * n1 = 0.7 * 100 = 70
Sample proportion (p̂1) = x1/n1 = 70/100 = 0.7
For the old drug:
Sample size (n2) = 150
Proportion of people for whom the drug is effective (p2) = 0.6
Sample proportion (p̂2) = x2/n2, where x2 is the number of people in the sample for whom the drug was effective.
We don't know the value of x2, but we can use the given proportion to estimate it:
x2 = p2 * n2 = 0.6 * 150 = 90
Sample proportion (p̂2) = x2/n2 = 90/150 = 0.6
The null hypothesis is that the proportion of people for whom the new substance is effective is equal to the proportion for whom the old drug is effective, that is, H0: p1 = p2. The alternative hypothesis is that the new substance is more effective than the old drug, that is, Ha: p1 > p2.
We can use a two-sample z-test to compare the sample proportions and calculate the p-value:
Test statistic: z = (p̂1 - p̂2) / sqrt(p̂(1-p̂)(1/n1 + 1/n2))
where p̂ = (x1 + x2) / (n1 + n2) is the pooled sample proportion.
z = (0.7 - 0.6) / sqrt(0.63 * 0.37 * (1/100 + 1/150)) = 2.22 (rounded to two decimal places)
p-value = P(Z > 2.22) = 0.0139 (using a standard normal distribution table or a calculator)
The p-value is less than 5%, so we reject the null hypothesis at the 5% level of significance. We can conclude that there is strong evidence to suggest that the new substance is more effective than the old drug for weight loss.