In statistics, hypothesis testing is a common method used to determine if there is a significant difference between a sample and a known population. Another commonly used tests in hypothesis testing is the two-sample t-test and z-test. In this lesson, we will explore the differences between the two tests and how they are used in practice.
The Two-Sample t-Test is a statistical test used to compare the means of two independent samples. It is a hypothesis test that determines whether there is a significant difference between the means of the two populations from which the samples are drawn.
Assumptions
Before performing a Two-Sample t-Test, it is important to check that the following assumptions are met:
If these assumptions are not met, then a different statistical test may be more appropriate.
Example
Suppose we want to compare the mean heights of male and female students at a university. We collect a random sample of 20 male students and a random sample of 20 female students and measure their heights in inches. The data is summarized below:
Mean Height (in) | Standard Deviation | |
---|---|---|
Males | 70.5 | 2.5 |
Females | 64.8 | 3.2 |
To test whether there is a significant difference in the mean heights of male and female students, we can perform a Two-Sample t-Test. Assuming the above assumptions are met, we can use a significance level of 0.05.
The null hypothesis is that there is no significant difference between the mean heights of male and female students. The alternative hypothesis is that there is a significant difference.
We can calculate the test statistic as follows:
t = (x̄₁ - x̄₂) / (sᵢ * sqrt(1/n₁ + 1/n₂))
where x̄₁ and x̄₂ are the sample means, sᵢ is the pooled standard deviation, n₁ and n₂ are the sample sizes.
The pooled standard deviation is calculated as follows:
sᵢ = sqrt(((n₁ - 1) * s₁² + (n₂ - 1) * s₂²) / (n₁ + n₂ - 2))
where s₁ and s₂ are the sample standard deviations.
Plugging in the values, we get:
t = (70.5 - 64.8) / (sqrt(((20 - 1) * 2.5² + (20 - 1) * 3.2²) / (20 + 20 - 2))
* sqrt(1/20 + 1/20))
t = 6.08
The degrees of freedom for this test is (n₁ + n₂ - 2) = 38.
Using a t-distribution table or a statistical software, we can find the p-value for the test. The p-value is the probability of getting a test statistic as extreme as the one calculated, assuming the null hypothesis is true.
Assuming a significance level of 0.05, the critical t-value (two-tailed) is ±2.024.
The p-value is calculated to be 3.79e-07, which is much smaller than 0.05. Therefore, we reject the null hypothesis and conclude that there is a significant difference between the mean heights of male and female students.
The z-Test is a statistical test used to determine whether a sample mean is significantly different from a known population mean when the population standard deviation is known. It is based on the normal distribution and is used when the sample size is large (usually more than 30).
Assumptions
Before performing a z-Test, it is important to check that the following assumptions are met:
If these assumptions are not met, then a different statistical test may be more appropriate.
Example
Suppose a company produces light bulbs and claims that the average lifespan of its bulbs is 5000 hours, with a standard deviation of 1000 hours. To test this claim, a random sample of 50 bulbs is selected and their lifespans are measured. The sample mean is found to be 4800 hours.
To determine whether this sample mean is significantly different from the population mean, we can perform a z-Test. Assuming the above assumptions are met, we can use a significance level of 0.05.
The null hypothesis is that there is no significant difference between the sample mean and the population mean. The alternative hypothesis is that there is a significant difference.
Solution
First, we need to calculate the z-score:
z = (sample mean - population mean) /
(population standard deviation /sqrt(sample size))
= (4800 - 5000) / (1000 / sqrt(50)) = -2.24
Next, we can use a standard normal distribution table or a calculator to find the p-value associated with the z-score. In this case, the p-value is approximately 0.0125.
Since the p-value is less than 0.05, we reject the null hypothesis and conclude that the sample mean is significantly different from the population mean at a significance level of 0.05.
In conclusion, the Two-Sample t-Test and z-Test are important statistical tests used to compare means and determine whether a sample mean is significantly different from a known population mean. The Two-Sample t-Test is used when comparing the means of two independent samples, while the z-Test is used when the population standard deviation is known and the sample size is large. It is important to check the assumptions for each test before performing the test to ensure its validity. The results of these tests can provide valuable insights and help make informed decisions in various fields such as science, business, and social sciences.
Here are the key takeaways from the above text:
1. Which statistical test is used to compare the means of two independent samples?
Answer: b) Two-Sample t-Test
2. What is the null hypothesis for a Two-Sample t-Test?
Answer: b. There is no significant difference between the means of the two populations.
3. Which assumption is necessary for a z-Test?
Answer: c. The population standard deviation is known.
4. When is a z-Test appropriate?
Answer: d. When the population standard deviation is known and the sample size is large.
4. What is the alternative hypothesis for a Two-Sample t-Test?
Answer: a. There is a significant difference between the means of the two populations.
Top Tutorials
Related Articles