Hypothesis testing and p-values | Inferential statistics | Probability and Statistics | Khan Academy
TLDRIn the provided transcript, a neurologist investigates the impact of a drug on the response time of rats by conducting an experiment on 100 rats. The rats are injected with the drug and then subjected to a neurological stimulus, with their response times recorded. The experiment compares the mean response time of the injected rats (1.05 seconds) with the known mean response time of non-injected rats (1.2 seconds). The neurologist sets up a null hypothesis stating that the drug has no effect, and an alternative hypothesis suggesting it does. Using statistical analysis, the neurologist calculates a Z-score to determine how many standard deviations the sample mean is from the population mean. The result is 3 standard deviations, which corresponds to a P-value of 0.003, indicating a less than 1 in 300 chance of observing such an extreme result if the null hypothesis were true. This leads to the rejection of the null hypothesis in favor of the alternative, suggesting that the drug significantly affects the rats' response time.
Takeaways
- 🧪 The neurologist conducted an experiment by injecting 100 rats with a drug to test its effect on response time.
- ⏱️ The mean response time for rats not injected with the drug is 1.2 seconds, while the mean for the injected rats is 1.05 seconds.
- 📊 A sample standard deviation of 0.5 seconds was observed for the injected rats' response times.
- 🔍 Two hypotheses were set up: the null hypothesis (drug has no effect) and the alternative hypothesis (drug has an effect).
- ⚖️ The null hypothesis assumes the mean response time remains at 1.2 seconds even with the drug, while the alternative suggests a different mean.
- 🤔 The decision to accept or reject the null hypothesis is based on the probability of obtaining the observed results if the null hypothesis were true.
- 📉 A sampling distribution with a mean of 1.2 seconds and a standard deviation of 0.05 seconds (based on the sample standard deviation divided by the square root of the sample size) was considered.
- 📊 The Z-score for the sample mean of 1.05 seconds was calculated to be 3, indicating it is 3 standard deviations away from the mean under the null hypothesis.
- 📊 According to the empirical rule, only 0.3% of the data falls more than 3 standard deviations from the mean, which corresponds to a P-value of 0.003.
- 🔎 A P-value of less than 0.003 suggests a very low probability of observing the sample results if the null hypothesis were true, leading to its rejection.
- 🧐 The neurologist concludes that the drug likely has an effect on response time, as the results strongly favor the alternative hypothesis.
Q & A
What is the null hypothesis in the context of the neurologist's experiment?
-The null hypothesis is the assumption that the drug has no effect on the response time of the rats. It suggests that the mean response time for rats, even when injected with the drug, remains the same as the control group, which is 1.2 seconds.
What is the alternative hypothesis in this scenario?
-The alternative hypothesis is the opposite of the null hypothesis. It posits that the drug does have an effect on the response time, meaning the mean response time of the rats that received the drug would be different from the control group's mean of 1.2 seconds.
What is the mean response time for the rats that were injected with the drug?
-The mean response time for the rats that were injected with the drug is 1.05 seconds.
What is the sample standard deviation of the response times for the injected rats?
-The sample standard deviation of the response times for the injected rats is 0.5 seconds.
How many rats were used in the experiment?
-A total of 100 rats were used in the experiment.
What is the significance of calculating the Z-score in this context?
-The Z-score is used to determine how many standard deviations away from the mean the sample mean of 1.05 seconds is. This helps in understanding the likelihood of observing such a sample mean if the null hypothesis were true.
What is the Z-score for the sample mean of 1.05 seconds?
-The Z-score is calculated as (1.05 - 1.2) / 0.05, which equals 3. This means the sample mean is 3 standard deviations below the mean response time under the null hypothesis.
What does the empirical rule state about the distribution of data?
-The empirical rule states that approximately 99.7% of the data falls within 3 standard deviations of the mean in a normal distribution.
What is the P-value in this experiment, and what does it indicate?
-The P-value is 0.003, which represents the probability of obtaining a sample mean as extreme as 1.05 seconds or more extreme if the null hypothesis were true. A P-value this low indicates strong evidence against the null hypothesis.
Why would a researcher reject the null hypothesis in this case?
-A researcher would reject the null hypothesis because the P-value (0.003) is less than the typical threshold for statistical significance, such as 0.05 or 0.01. This suggests that the observed effect is unlikely to have occurred by chance alone, indicating that the drug likely has an effect on response time.
What is the standard deviation of the sampling distribution if the null hypothesis is true?
-The standard deviation of the sampling distribution, under the assumption of the null hypothesis, is the sample standard deviation (0.5 seconds) divided by the square root of the sample size (100), which equals 0.05 seconds.
What is the general approach in scientific research when testing a hypothesis?
-The general approach is to assume the null hypothesis is true and then calculate the probability of obtaining the observed results or more extreme results under this assumption. If this probability is very low, the null hypothesis is likely incorrect, and the alternative hypothesis is favored.
Outlines
🧪 Testing Drug Effects on Rat Response Time
In this paragraph, a neurologist conducts an experiment to determine if a drug has an effect on the response time of rats. The experiment involves injecting 100 rats with a unit dose of the drug and then measuring their response times after a neurological stimulus. The neurologist compares the mean response time of the injected rats (1.05 seconds) with the known mean for non-injected rats (1.2 seconds). To test for a significant effect, two hypotheses are set up: the null hypothesis (the drug has no effect, and the mean remains 1.2 seconds) and the alternative hypothesis (the drug does have an effect, and the mean is not 1.2 seconds). The goal is to calculate the probability of observing the sample mean of 1.05 seconds assuming the null hypothesis is true, using the sample standard deviation and the central limit theorem to estimate the standard error.
📊 Calculating the Probability of Observing the Sample Mean
This paragraph delves into the statistical analysis of the experiment's results. The standard deviation of the sampling distribution is calculated by dividing the sample standard deviation (0.5 seconds) by the square root of the sample size (10), resulting in 0.05. The Z-score, which indicates how many standard deviations the sample mean is from the population mean under the null hypothesis, is then computed. The sample mean of 1.05 seconds is found to be 3 standard deviations away from the hypothesized population mean of 1.2 seconds. Using the empirical rule, the probability of observing a result this extreme or more (3 or more standard deviations away) is determined to be 0.3%, or 0.003 in decimal form. This very low probability suggests that the null hypothesis is likely false, and the alternative hypothesis that the drug has an effect is favored.
🔍 Interpreting the Results and Hypothesis Decision
In the final paragraph, the results of the statistical analysis are interpreted. Given that the probability of observing the sample mean of 1.05 seconds if the null hypothesis were true is only 0.3%, it is concluded that the null hypothesis can be rejected in favor of the alternative hypothesis. This decision is supported by the fact that the observed result is extremely unlikely under the assumption of no drug effect. The term P-value is introduced as the probability of obtaining a result more extreme than the observed sample mean, should the null hypothesis be true. A common threshold for rejecting the null hypothesis is a P-value less than 5%, but with a P-value of 0.003, the evidence is much stronger than this threshold, indicating that the drug likely has an effect on the response time of rats.
Mindmap
Keywords
💡neurologist
💡response time
💡drug effect
💡null hypothesis
💡alternative hypothesis
💡sampling distribution
💡standard deviation
💡Z-score
💡P-value
💡empirical rule
💡hypothesis testing
Highlights
A neurologist is testing the effect of a drug on response time by injecting 100 rats with a unit dose of the drug and recording their response times.
The mean response time for rats not injected with the drug is known to be 1.2 seconds.
The mean response time for the 100 injected rats is 1.05 seconds with a sample standard deviation of 0.5 seconds.
Two hypotheses are set up: the null hypothesis stating the drug has no effect, and the alternative hypothesis suggesting it does.
The null hypothesis assumes the mean response time remains 1.2 seconds even with the drug.
The alternative hypothesis posits that the mean response time will differ from 1.2 seconds when the drug is administered.
The probability of obtaining the sample results under the null hypothesis is calculated to test its validity.
A sampling distribution with a mean of 1.2 seconds and a standard deviation of 0.05 seconds is assumed under the null hypothesis.
The sample mean of 1.05 seconds is 3 standard deviations away from the mean of the assumed sampling distribution.
The Z-score for the sample mean is calculated to be 3, indicating a significant deviation from the null hypothesis mean.
According to the empirical rule, only 0.3% of results fall more than 3 standard deviations from the mean.
The probability of obtaining a result this extreme or more by chance under the null hypothesis is only 0.3%.
The neurologist rejects the null hypothesis in favor of the alternative hypothesis due to the low probability of the observed results.
The P-value, representing the probability of obtaining a result more extreme given the null hypothesis, is 0.003.
A P-value less than 5% typically leads to the rejection of the null hypothesis, and in this case, it is much lower.
The results strongly indicate that the drug has an effect on response time, as the observed effects are unlikely under the null hypothesis.
The study demonstrates the use of hypothesis testing and statistical significance in determining the efficacy of a drug.
The neurologist's findings could have practical applications for the development and approval of new pharmaceuticals.
Transcripts
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