A new drug claims to reduce the number of migraines an individual suffers. In a study on the effectiveness of the new drug, 50 s
ubjects were given the new drug and another 50 subjects were given a placebo. At the end of the 6-month study, the group given the new drug reported an average of 1.3 migraines a week. The group given the placebo reported an average of 2.5 migraines a week. The data from the study are rerandomized 10 times. The difference of the means from rerandomization are 1.2, 3.4, 1.8, 1.3, 1.1, 0.9, 2.5, 0.7, 1.1, and 0.9. What is the most appropriate conclusion about the new drug to draw from this information? The new drug appears to reduce the number of migraines since the rerandomized mean differences are close to the mean difference found between the two groups studied. The new drug appears to reduce the number of migraines since the rerandomized mean differences are considerably less than the mean difference found between the two groups studied. The new drug does not appear to reduce the number of migraines since the rerandomized mean differences are close to the mean difference found between the two groups studied. The new drug does not appear to reduce the number of migraines since the rerandomized mean differences are considerably greater than the mean difference found between the two groups studied.
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Answer:
0.5 = 50% probability that a random sample of 100 independent persons will cause an overload
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For the sum of n values of a distribution, the mean is and the standard deviation is
An expert in ski lifts thinks that the weights of individuals using the lift have expected weight of 200 pounds and standard deviation of 30 pounds. 100 individuals.
This means that
If the expert is right, what is the probability that a random sample of 100 independent persons will cause an overload
Total load of more than 20,000 pounds, which is 1 subtracted by the pvalue of Z when X = 20000. So
has a pvalue of 0.5
1 - 0.5 = 0.5
0.5 = 50% probability that a random sample of 100 independent persons will cause an overload