Hypothesis Testing (t-Test): Videos & Practice Problems

The t-test is a statistical method used to compare the means of populations, particularly when the population standard deviation is unknown. The t-score is calculated using the formula: $t=\frac{\mathrm{x̄}-\mu ₀}{s/\sqrt{n}}$. A larger t-score indicates greater differences between populations. Variations include equal variance, unequal variance, and paired data, each requiring specific calculations to determine significant differences using a t-table.

t-Test

The t-test is a statistical method used to compare the means of two populations, particularly when the population standard deviation is unknown. The t-score, which is calculated using a specific formula, helps determine the similarities or differences between these populations. The formula for the t-score is given by:

\( t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} \)

In this equation, \( \bar{x} \) represents the sample average, \( \mu_0 \) is the population average, \( s \) is the sample standard deviation, and \( n \) is the number of samples. The t-test is particularly applicable when the sample size is less than 30; for larger samples, the z-test is typically used, often requiring software for calculations.

The interpretation of the t-score is straightforward: a larger t-score indicates greater differences between the populations, while a smaller t-score suggests they are more similar. When comparing two populations, the t-test can be adapted for different scenarios, including equal variances, unequal variances, and paired data.

For equal variances, the t-calculated value is determined using the pooled standard deviation, which is calculated as follows:

\( s_{pooled} = \sqrt{\frac{s_1^2(n_1 - 1) + s_2^2(n_2 - 1)}{n_1 + n_2 - 2}} \)

Here, \( s_1 \) and \( s_2 \) are the standard deviations of the two populations, and \( n_1 \) and \( n_2 \) are their respective sample sizes. The t-calculated value for equal variances is then:

\( t = \frac{|\bar{x}_1 - \bar{x}_2|}{s_{pooled} \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \)

When the variances are unequal, a different formula is used to calculate the t-value, which also involves a more complex calculation for degrees of freedom. The degrees of freedom for unequal variances can be calculated using:

\( df = \frac{(s_1^2/n_1 + s_2^2/n_2)^2}{\frac{(s_1^2/n_1)^2}{n_1 - 1} + \frac{(s_2^2/n_2)^2}{n_2 - 1}} \)

Paired data is utilized when comparing two populations measured by different methods. In this case, the t-calculated value is derived from the differences between paired observations. Regardless of the method used, once the t-calculated value is obtained, it can be compared to a t-table to determine statistical significance. If the t-calculated value exceeds the critical value from the t-table, it indicates a significant difference between the population means; if it is less, the means are considered not significantly different.

Understanding these calculations and their applications is crucial for effectively analyzing data and drawing meaningful conclusions from statistical tests.

t-Test Calculations

In analyzing the amount of arsenic in cigarettes using two different analytical methods, a student aims to determine if there is a significant difference between the results obtained from each method at a 95% confidence interval. To achieve this, the student employs a t-test, which is essential when comparing the means of two independent samples.

The first step involves calculating the means for both methods. For Method 1, the mean is calculated as:

Mean₁ = \frac{\sum X_1}{n_1} = \frac{(110.5 + 93.1 + 63.0 + 72.3 + 121.6)}{5} = 92.1

For Method 2, the mean is:

Mean₂ = \frac{\sum X_2}{n_2} = \frac{(104.7 + 95.8 + 71.2 + 69.9 + 118.7)}{5} = 92.06

Next, the standard deviations for both methods are calculated using the formula:

s = \sqrt{\frac{\sum (X - \text{Mean})^2}{n - 1}}

For Method 1, the standard deviation is found to be approximately 24.742, while for Method 2, it is about 21.27. The variances, which are the squares of the standard deviations, are then calculated as:

Variance₁ = (24.742)^2 = 612.167

t = \frac{|\text{Mean₁} - \text{Mean₂}|}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}

Substituting the calculated values, the t-calculated value is found to be approximately 0.002741.

To determine the significance of this result, the degrees of freedom must also be calculated using the formula for unequal variances:

df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}}

After performing the calculations, the degrees of freedom is determined to be approximately 8.

Referring to the t-table for 8 degrees of freedom at a 95% confidence level, the critical t-value is found to be 2.306. Comparing the t-calculated value of 0.002741 to the t-table value reveals that the t-table value is greater. This indicates that there is no significant difference between the means of the two methods.

In summary, the analysis demonstrates that the two methods yield similar results regarding arsenic content in cigarettes, as the statistical test shows no significant difference in their means at the specified confidence level.

t-Test Calculations

In this analysis, we aim to determine whether the concentrations of hydrocarbons in seawater, measured by fluorescence, differ significantly from those measured using gas chromatography with flame ionization detection (GCFID). Given that we have two sets of measurements from the same samples, this scenario involves paired data, which is essential for our statistical analysis.

To begin, we calculate the differences between the two measurement methods for each sample. For instance, if the fluorescence measurement is 100.2 µM and the GCFID measurement is 101.1 µM, the difference would be calculated as:

Difference = Fluorescence - GCFID = 100.2 - 101.1 = -0.9 µM.

We repeat this process for all seven samples to create a new column of differences.

Next, we compute the mean difference by summing all the differences and dividing by the number of measurements (n = 7). The formula for the mean difference (MD) is:

MD = (ΣDifferences) / n.

After calculating, we find the mean difference to be approximately -0.014 µM.

Following this, we calculate the standard deviation (SD) of the differences. The formula for standard deviation in this context is:

SD = √(Σ(Difference - MD)² / (n - 1)).

After performing the calculations, we determine the standard deviation to be approximately 0.47 µM.

With the mean difference and standard deviation established, we can now calculate the t statistic using the formula:

t = |MD| / (SD / √n).

Substituting our values, we find:

t = |-0.014| / (0.47 / √7) ≈ 0.08.

To assess the significance of this t statistic, we refer to the Student's t-distribution table. The degrees of freedom (DOF) for our analysis is calculated as:

DOF = n - 1 = 7 - 1 = 6.

At a 95% confidence level, the critical t value from the table for 6 degrees of freedom is approximately 2.447.

Since our calculated t value (0.08) is less than the critical t value (2.447), we conclude that there is no significant difference between the two measurement methods. This indicates that both fluorescence and GCFID methods yield similar mean concentrations for the hydrocarbon samples, reinforcing the reliability of either method for this analysis.

In summary, when comparing two different measurement methods using paired data, it is crucial to calculate the differences, mean difference, standard deviation, and t statistic, followed by a comparison with critical values from the t-distribution to draw conclusions about the significance of the results.

t-Test Calculations

To compute a 95% confidence interval for the population mean based on a sample, we start with the sample size, sample mean, and population standard deviation. In this case, the sample size \( n \) is 100, the sample mean \( \bar{x} \) is 16, and the population standard deviation \( \sigma \) is 3. Since the sample size is greater than 30, we can use the z-test instead of the t-test.

For a 95% confidence interval, we refer to the z-table to find the z-score corresponding to a 95% confidence level. With a large sample size, the degrees of freedom approach infinity, and the z-score is approximately 1.960.

The formula for the confidence interval is given by:

CI = \bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right)

Substituting the known values into the formula:

CI = 16 \pm 1.960 \left( \frac{3}{\sqrt{100}} \right)

Calculating the standard error:

\frac{3}{\sqrt{100}} = \frac{3}{10} = 0.3

Now, substituting this back into the confidence interval formula:

CI = 16 \pm 1.960 \times 0.3

Calculating the margin of error:

1.960 \times 0.3 = 0.588

Thus, the confidence interval becomes:

CI = 16 \pm 0.588

This results in the lower and upper bounds of the confidence interval:

16 - 0.588 = 15.412

16 + 0.588 = 16.588

Therefore, we can conclude that we are 95% confident that the true population mean lies between 15.412 and 16.588.

The average height of the US male is approximately 68 inches. What is the probability of selecting a group of males with average height of 72 inches or greater with a standard deviation of 5 inches?

• A. 84.61 %
• B. 78.81 %
• C. 84.85 %
• D. 79.10 %

Answer: (B)

Here’s what students ask on this topic:

What is the t-test used for in statistics?

The t-test is a statistical method used to compare the means of two populations, especially when the population standard deviation is unknown. It helps determine if there is a significant difference between the means of the populations. The t-score is calculated using the formula:

$t=\frac{{\overline{x}}_0-\mu )}{\frac{s}{\sqrt{n}}}$

A larger t-score indicates greater differences between the populations. Variations of the t-test include equal variance, unequal variance, and paired data, each requiring specific calculations to determine significant differences using a t-table.

When should you use a t-test instead of a z-test?

You should use a t-test instead of a z-test when the population standard deviation is unknown and the sample size is less than 30. The t-test is more appropriate in these cases because it accounts for the additional uncertainty in the estimate of the population standard deviation. The t-test uses the sample standard deviation (s) and adjusts for smaller sample sizes, making it more reliable for small samples. In contrast, the z-test is used when the population standard deviation is known and the sample size is large (typically n > 30).

How do you interpret the t-score in a t-test?

The t-score in a t-test indicates how much the sample mean deviates from the population mean in units of the standard error. A larger t-score suggests a greater difference between the sample and population means. To interpret the t-score, you compare it to a critical value from the t-table, which depends on the degrees of freedom and the desired confidence level. If the t-score is greater than the critical value, it indicates a significant difference between the means. If it is less, there is no significant difference.

What are the different types of t-tests and when are they used?

There are three main types of t-tests: independent t-test (equal variance), independent t-test (unequal variance), and paired t-test. The independent t-test (equal variance) is used when comparing the means of two independent groups with equal variances. The independent t-test (unequal variance) is used when the variances of the two groups are not equal. The paired t-test is used when comparing means from the same group at different times or under different conditions. Each type of t-test has specific formulas and considerations for calculating the t-score and degrees of freedom.

What is the formula for calculating the t-score in a t-test?

The formula for calculating the t-score in a t-test is:

$t=\frac{{\overline{x}}_0-\mu )}{\frac{s}{\sqrt{n}}}$

where $\overline{x}$ is the sample mean, $\mu$ is the population mean, $s$ is the sample standard deviation, and $n$ is the sample size. This formula is used when the population standard deviation is unknown and the sample size is less than 30.