单样本T-检验

We've already learned about the types of error and one- and two-tailed test. In this next lesson, we will learn about another statistical hypothesis test: the one sample t-test.

Making Sense of One-Sample T-Tests

Below are four steps to conducting a one-sample t-test

1) Write out hypothesis statements (null and alternative)

one-sample:

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$H_0: \mu = m_0$

$H_1: \mu \neq m_0$ (two-tailed)

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$H_0:\mu \leq m_0$

$H_1: \mu > m_0$ (upper-tailed)

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$H_0:\mu \geq m_0$

$H_1: \mu < m_0$ (lower-tailed)

Such that:

$m_0$ is the value you are comparing the mean to (aka comparison value)

and \mu is your one sample mean

2) Plug known values into one-sample t-score formula

$t = \frac{\bar{y}- m_0}{\hat{\sigma}/\sqrt{n}}$

Such that:

$\bar{y} =$ sample mean

$m_0 =$ population mean/hypothesized value

$\hat{\sigma} =$ sample standard deviation

$n =$ sample size

3) Determine the degrees of freedom ($df$)

$df = n-1$ for a one-sample t-test

4) Determine the critical value by looking up ($t_{\alpha, df}$) in a t-table

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Note: For one-sample t-tests we do not divide $\alpha$ by 2 before looking up the critical value in the t-table as we do for two-sample t-tests.

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5) T-Test Conclusion/Significance (Reject Null Hyp. or Fail to Reject Null Hyp)

If the absolute value of the t-value is greater than the critical value, you reject the null hypothesis. If the absolute value of the t-value is less than the critical value, you fail to reject the null hypothesis.

Learning More

In a one sample t-test, the test statistic compares the mean of a sample to a known population mean to determine if there is a significant difference. The standard error quantifies the variability of the sample mean around the population mean, crucial for determining the precision of the test statistic.

We can also incorporate the confidence interval. A $95\%$ confidence interval for a one-sample t-test indicates a range of values in the sample that is likely to contain the true population mean with a $95\%$ certainty.

One-Sample T-Test Example

Question: Your fishing company wants to improve their salmon sales. Past sales data indicate that the average sale for salmon was 100 dollars per transaction. After training your fish vendors (salesmen), recent sales data (taken from a sample of 25 vendors) indicates an average sale of 130 dollars, with a standard deviation of 15 dollars. Did the training significantly raise sales? Assume an alpha level of 5%.

1) Write out null and alternative hypothesis statements:

Because this question is asking whether or not we significantly raised sales, or made more than 100 dollars per transaction and only has one sample described, this is an upper-tailed, one-sample problem. Hence our hypothesis statements are as follows

$H_0:\mu \leq 100$

$H_1: \mu > 100$

2) Plug in known values into t-score formula

Known values:

$\bar{y} = sample mean = 130$

$m_0 = Population Mean/ Hypothesized Value = 100$

$\hat{\sigma} = Sample Standard Deviation = 15$

$\sqrt{n} = \sqrt{Sample Size} = \sqrt{25}$

$t = tscore = \frac{130- 100}{15/\sqrt{25}} = \frac{30}{3} = 10$

3) Figure out the degrees of freedom parameter ($df$)

for a one-sample t-test our $df = n-1$ so $df = 24$

4) Determine the critical value

Look up ($t_{0.05, 24}$) in a t-table

($t_{0.05, 24}$) = 1.710882

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Note: For one-sample t-tests we do not divide $\alpha=5$ by 2 before looking up the critical value in the t-table as we do for two-sample t-tests.

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A handy t-table to have is this one from San Jose State University!

5) Determine significance of results and whether or not to reject the null or fail to reject the null

Because the magnitude of the t-value is greater than the critical value (10 > 1.710882), we have significant evidence to reject the null hypothesis.

(Video) One Sample T-Test

by statslectures

In this video by statslectures, you will learn about the one sample t-test.