Module 14 – 1 Sample T-Test

One-Sample T-Tests

Introduction

• Z-tests require knowing the mean and standard deviation (SD) of the parent population, which is often unrealistic.
• One-sample t-tests are used when the mean of the population is known, but the SD is not.

Scenario

• Assume the average American watches 5 hours of video each day.
• We want to determine if college students watch a statistically different amount of video.
• A sample of 25 students is surveyed, and it's found they watch an average of 6 hours of video each day.
• The variance or standard deviation of the population is unknown.

Why T-Tests?

• Without the population SD, a z-test cannot be conducted.
• The one-sample t-test is appropriate in this scenario.

One-Sample T-Tests vs. Z-Tests

• Similar to z-tests.
• Z score calculation: Subtract the mean of the comparison distribution from the mean of the sample, then divide by the standard error of the comparison distribution.
• With the population variance, divide it by $n$ and take the square root to get the standard error of the comparison distribution.
• With the population standard deviation, divide it by $\sqrt{n}$ to get the standard error of the comparison distribution.

Estimating Variance

• $s^2$ is the symbol for the variance of the sample.
• A sampling distribution of $s^2$ is used to estimate the variance of the comparison distribution.
• Imagine taking multiple samples and calculating their variances.

Issues with Sampling Distributions

• Sampling distributions have less variability than the population.
• The sampling distribution of $s^2$ is skewed because most scores fall below the variance of the population.
• Example: A distribution where 50,000 samples were drawn from a population with $\mu = 5$ and $\sigma^2 = 50$. Most scores fall below the variance of the population, resulting in a skewed distribution.

The T Distribution

• The distribution of $s^2$ is skewed, so a Z-score and Z-table cannot be used.
• In 1908, an employee of the Guinness Brewing Company created a table using the sampling distribution of $s^2$, called Student’s t distribution.
• Solve for $t$ and look up the value on the table.

Example: College Students' Media Consumption

• Goal: Determine if the average college student's media consumption differs from the rest of Americans.
• The average American watches 5 hours of video each day.
• Survey 25 college students and find they watch 6 hours of video each day with a standard deviation of 1.
• Given:
  • Population mean: $\mu = 5$
  • Sample mean: $\bar{X} = 6$
  • Sample standard deviation: $s = 1$
  • Sample size: $n = 25$

Steps for Hypothesis Testing

Step 1: State the Research and Null Hypotheses

• Research hypothesis: The media viewing habits of college students are different from the average American (non-directional, two-tailed test).
• Null hypothesis: The media viewing habits of college students are the same as the average American.

Step 2: Determine the Characteristics of the Comparison Distribution

• For a one-sample t-test, the mean of the comparison distribution is the mean of the population: $\mu_M = \mu = 5$
• Calculate standard deviation:
  • If given $s_M$, you're done.
  • If given $s^2$, divide by $n$ and take the square root.
  • If given $s$, divide it by $\sqrt{n}$.

Sampling Distributions

• Different sampling distributions exist for every $n$. This is more relevant for t-tests than z-tests.
• The t distribution changes based on the number of subjects.

Degrees of Freedom

• The shape of the t distribution is described by degrees of freedom (df).
• For a one-sample t-test: $df = n - 1$
• One degree of freedom is lost because the sample is used to calculate the variance.

Calculating Distribution Characteristics

• Determine the mean, standard deviation, and degrees of freedom of the comparison distribution.
  • $\mu_M = \mu = 5$
  • $s_M = \frac{s}{\sqrt{n}} = \frac{1}{\sqrt{25}} = \frac{1}{5} = 0.2$
  • $df = n - 1 = 24$

Step 3: Determine the Cutoff Sample Score

• The cutoff score is $\pm 2.064$

Step 4: Determine the Sample’s Score on the Comparison Distribution

• Plug the numbers into the one-sample t formula.
  • Given: $\bar{X} = 6$
  • Calculated: $\mu<em>M = 5$ and $s</em>M = 0.2$
  • $t = \frac{6 - 5}{0.2} = 5$

Step 5: Decide Whether to Reject the Null Hypothesis

• Cutoff: $\pm 2.064$
• Sample’s t score: $+5$
• Reject the null hypothesis because the sample’s t score is more extreme than the cutoff.

Similarity to Z-Tests

• This process is similar to the null hypothesis test used for z-tests.
• The core process remains the same: a new test will be introduced, formulas will be adjusted, and steps 2, 3, and 4 will be modified as a result of that new formula but the process remains
• Understanding the basic concepts is crucial; the rest involves knowing which formula to use and how to use it.

Finding the Cutoff Score

• A t table is used to find the cutoff score.
• The table shows different cutoffs depending on the number of tails used (one-tailed vs. two-tailed) and the degrees of freedom.
• If alpha is set at 0.05, the appropriate column is selected based on whether it is a one-tailed or two-tailed test.
• The row is selected based on the degrees of freedom.
• Example: With 4 df, the cutoff would be 2.132 for a one-tailed test or 2.776 for a two-tailed test.

T-Table Practice

• Use the t-table to determine the cutoff score.

Example: Moon Illusion

• The moon appears larger near the horizon than high in the sky.
• Psychologists used an artificial moon and asked participants to adjust the size of the moon in different locations.
• Data was collected from 10 subjects and recorded as a ratio of the diameter of the moon at the horizon and zenith.
  • Sample mean: $\bar{X} = 1.463$
  • Sample standard deviation: $s = 0.341$
• A ratio of 1 indicates no difference; 1.5 indicates the horizon moon is 1.5 times larger than the zenith moon; 0.5 indicates it is half the size.
• If the null hypothesis is that there is no difference, a ratio of 1 is used as the population mean: $\mu = 1$

Given:

• Population mean: $\mu = 1$
• Sample mean: $\bar{X} = 1.463$
• Sample standard deviation: $s = 0.341$
• Sample size: $n = 10$

Step 1: State the Research and Null Hypotheses

• Research hypothesis: People perceive the horizon moon to be a different size than the zenith moon.
• Null hypothesis: People perceive the moon to be the same size regardless of its location.

Step 2: Determine the Characteristics of the Comparison Distribution

• This is a one-sample t-test because the mean of the population is known, but the SD is not.
• The mean of the comparison distribution is the mean of the population: $\mu_M = \mu = 1$
• Calculate the standard deviation:
  • $s_M = \frac{s}{\sqrt{n}} = \frac{0.341}{\sqrt{10}} = 0.108$
• Degrees of freedom: $df = n - 1 = 10 - 1 = 9$

Step 3: Determine the Cutoff Sample Score

• Using a t-table we find the cutoff score.

Step 4: Determine the Sample’s Score on the Comparison Distribution

• Plug the numbers into the one-sample t formula.
  • Given: $\bar{X} = 1.463$
  • Calculated: $\mu<em>M = 1$ and $s</em>M = 0.108$
  • $t = \frac{1.463 - 1}{0.108} = 4.29$

Step 5: Decide Whether to Reject the Null Hypothesis

• Cutoff: $\pm 2.262$
• The t value is 4.29, which is more extreme than $\pm 2.262$, so reject the null hypothesis.
• People reported that the moon was significantly larger on the horizon than at its zenith.
• If a directional one-tailed test had been used (believing the moon would be bigger), the cutoff would have been +1.833 instead of +2.262 for the two-tailed test.
• Two-tailed tests are often used even when a one-tailed test would be appropriate because if the numbers come out the other way (the moon was smaller on the horizon), it would still be something important to know.

Example 3: College Students' Sleep

• The typical young adult sleeps 8 hours each night.
• Determine if the average hours of sleep for college students differ from the rest of the population.
• Survey 49 students and find they sleep an average of 6.7 hours with a standard deviation of 0.4 hours.

Given

• Population mean: $\mu = 8$
• Sample mean: $\bar{X} = 6.7$
• Sample standard deviation: $s = 0.4$
• Sample size: $n = 49$

Step 1: State the Research and Null Hypotheses

• Research hypothesis: College students don’t get the same amount of sleep as other young adults.
• Null hypothesis: College students get the same amount of sleep as the rest of the population.

Step 2: Determine the Characteristics of the Comparison Distribution

• Because we have the mean of the population, but not the SD, this is definitely a one-sample t-test
• The mean of the comparison distribution is the mean of the population: $\mu_M = \mu = 8$
• Calculate the standard deviation:
  • $s_M = \frac{s}{\sqrt{n}} = \frac{0.4}{7} = 0.057$
• Degrees of freedom: $df = n - 1 = 49 - 1 = 48$

Step 3: Determine the Cutoff Sample Score

• With 48 df, but the table goes from 40 to 50.
• The cutoff doesn’t change much from 40 to 50 then from 50 to 100.
• 50 can be used because it is close to 48. If the result is close to the cutoff, the exact value can be looked up to be safe.

Step 4: Determine the Sample’s Score on the Comparison Distribution

• Plug the numbers into the one-sample t formula.
  • Given: $\bar{X} = 6.7$
  • Calculated: $\mu<em>M = 8$ and $s</em>M = 0.057$
  • $t = \frac{6.7 - 8}{0.057} = -22.81$

Step 5: Decide Whether to Reject the Null Hypothesis

• Cutoff: $\pm 2.009$
• The t value is -22.81, which is much more extreme than $\pm 2.009$, so reject the null hypothesis.
• College students reported significantly fewer hours of sleep on average than the typical young adult.