Two Means - Unknown, Equal Variance: Videos & Practice Problems

When performing hypothesis tests for two population means with unknown standard deviations but assumed equal variances, use the pooled standard deviation, a weighted average of sample variances, to improve accuracy. Formulate null ($H{0}_{}:\mu$₁=μ₂) and alternative hypotheses based on the claim. Calculate the t-statistic using the pooled variance and degrees of freedom ($\mathrm{df}=n$₁+n₂-2). Compare the p-value to alpha ($\alpha$) to decide on rejecting $H{0}_{}$, supporting conclusions about population means.

Means Unknown Equal Variances Hypothesis Test Using TI-84

When conducting a hypothesis test for two population means without knowing the population standard deviations, an important scenario arises if we can assume the populations have equal variances. This assumption allows us to use a pooled standard deviation, which is a weighted average of the sample standard deviations, providing a more accurate estimate of the common population standard deviation. Using the pooled standard deviation improves the precision of the test but involves more complex calculations, often facilitated by technology such as the TI-84 calculator.

Consider a practical example where a school claims a new math app improves test scores. Two independent random samples of 50 students each are taken: one group uses the traditional method with a sample mean of 77 and standard deviation of 4.8, while the other uses the new app with a sample mean of 82 and standard deviation of 4.4. Assuming equal population variances, the goal is to test if the app significantly increases scores at a significance level of α = 0.05.

The hypothesis test begins by stating the null hypothesis H₀: μ₁ = μ₂ (no difference in means) and the alternative hypothesis Hₐ: μ₁ < μ₂ (the app improves scores). The pooled variance s_p² is calculated using the formula:

\[ s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2} \]

where n₁ and n₂ are the sample sizes, and s₁ and s₂ are the sample standard deviations. This pooled variance is then used to compute the test statistic t:

\[ t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \]

In the example, the pooled variance is approximately 21, and the calculated t-value is about -5.43. The degrees of freedom for this test are df = n_1 + n_2 - 2 = 98. Using either a t-distribution table or a calculator, the p-value corresponding to this t-value is approximately 2.05 × 10⁻⁷, which is much smaller than the significance level of 0.05.

Since the p-value is less than α, the null hypothesis is rejected, providing strong evidence that the new math app improves average test scores. This conclusion is valid under the conditions that the samples are independent and either the populations are normally distributed or the sample sizes are sufficiently large (greater than 30), both of which are satisfied in this case.

Understanding how to perform a two-sample t-test with pooled variance is essential when the assumption of equal population variances is reasonable. It enhances the accuracy of hypothesis testing for comparing means, especially when population standard deviations are unknown. Utilizing technology like the TI-84 calculator can simplify the computational process, but knowing the underlying formulas and assumptions remains crucial for interpreting results correctly.

For ${\bar{x}}_1=41,S_{x1}=8.3,n_1=40,{\bar{x}}_2=39,S_{x2}=5.8$, & $n_2=50$, perform a hypothesis test to test the claim that ${\mu }_1>{\mu }_2$, assuming ${\sigma }_1={\sigma }_2$ for $\alpha =0.01$.

• A. $P-value>\alpha$; fail to reject $H_0$ since there is not enough evidence to suggest $\mu_1>\mu_2$.
• B. $P-value>\alpha$; reject $H_0$ since there is enough evidence to suggest $\mu_1>\mu_2$.
• C. ; reject $H_0$ since there is enough evidence to suggest $\mu_1>\mu_2$.
• D. $P-value<\alpha$; fail to reject $H_0$ since there is not enough evidence to suggest $\mu_1>\mu_2$.

Answer: (A)

Means Unknown Equal Variances Hypothesis Test Using TI-84 Example 1

When comparing the average wait times of two delivery services, it is essential to perform a hypothesis test to determine if one service offers significantly shorter wait times than the other. In this scenario, two samples were collected: Service A (population one) with a sample mean of 18.4 minutes, a standard deviation of 3.2, and a sample size of 35; and Service B (population two) with a sample mean of 14.6 minutes, a standard deviation of 4.3, and a sample size of 35. The goal is to test, at a significance level of α = 0.1, whether Service B has shorter wait times than Service A, assuming equal variances between the two populations.

The hypothesis test begins by setting the null hypothesis (H₀) as the equality of the population means:

H₀: μ₁ = μ₂

which states that the average wait times for both services are the same. The alternative hypothesis (H₁) reflects the claim that Service B has shorter wait times, meaning Service A's mean wait time is greater:

H₁: μ₁ > μ₂

Since the variances are assumed equal, the pooled standard deviation is used to calculate the test statistic. The pooled standard deviation (s_p) combines the variability of both samples and is calculated as:

$$s{p}_{}=\; \backslash sqrt\{\backslash frac\{(n\_1\; -\; 1)s\_1^2\; +\; (n\_2\; -\; 1)s\_2^2\}\{n\_1\; +\; n\_2\; -\; 2\}\}$$

where n₁ and n₂ are the sample sizes, and s₁ and s₂ are the sample standard deviations.

The test statistic t is then computed as:

$$t\; =\; \backslash frac\{\backslash bar\{x\}\_1\; -\; \backslash bar\{x\}\_2\}\{s\_p\; \backslash sqrt\{\backslash frac\{1\}\{n\_1\}\; +\; \backslash frac\{1\}\{n\_2\}\}\}$$

where 𝑥̄₁ and 𝑥̄₂ are the sample means.

Using statistical software or a calculator, the p-value associated with the test statistic was found to be approximately 4.05 × 10⁻⁵, which is significantly less than the alpha level of 0.1. This leads to rejecting the null hypothesis, providing strong evidence that Service A has longer wait times than Service B. Consequently, the grocery chain has sufficient statistical evidence to prefer Service B for home delivery due to its shorter wait times.

Means Unknown Equal Variances Hypothesis Test Using TI-84 Example 2

When comparing the average number of views on posts between two major social media platforms, a hypothesis test can determine if there is a significant difference in traffic. Given two samples—Platform A with a mean of 2.5 million views, a standard deviation of 0.57, and a sample size of 90, and Platform B with a mean of 2.4 million views, a standard deviation of 0.39, and a sample size of 80—the goal is to assess whether the difference in means is statistically significant at a 1% significance level (α = 0.01).

The null hypothesis (H₀) assumes that the population means are equal (μ₁ = μ₂), indicating no difference in average views between the platforms. The alternative hypothesis (H_a) posits that the means are not equal (μ₁ ≠ μ₂), reflecting the manager’s concern that the platforms receive different traffic levels. Since the population variances are assumed equal, a pooled standard deviation is used to combine the sample variances for the test.

The test statistic for comparing two means with equal variances is calculated using the formula:

\[ t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \]

where the pooled standard deviation \( s_p \) is:

\[ s_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}} \]

Here, \( \bar{x}_1 \) and \( \bar{x}_2 \) are the sample means, \( s_1 \) and \( s_2 \) are the sample standard deviations, and \( n_1 \) and \( n_2 \) are the sample sizes.

After performing the hypothesis test, the p-value obtained is approximately 0.189, which is greater than the significance level of 0.01. Since the p-value exceeds α, the null hypothesis cannot be rejected. This indicates insufficient evidence to conclude that the average number of views differs between the two platforms.

Consequently, the company should continue allocating equal time to posting on both platforms, as the data does not support a difference in audience engagement. This approach ensures resources are distributed effectively without bias toward one platform based on unsubstantiated assumptions about traffic disparities.

Here’s what students ask on this topic:

What is the pooled standard deviation and why is it used in hypothesis testing for two means with unknown but equal variances?

The pooled standard deviation is a weighted average of the sample standard deviations from two independent samples. It is used when we assume the population variances are equal but unknown. By combining the variances, the pooled standard deviation provides a better estimate of the common population variance, improving the accuracy of the hypothesis test. Mathematically, the pooled variance $s_p²$ is calculated as $\frac{(n_1-1)s_1²+(n_2-1)s_2²}{n_1+n_2-2}$. This value replaces individual sample standard deviations in the t-test formula, leading to more reliable results when the equal variance assumption holds.

How do you perform a hypothesis test for two means assuming unknown but equal variances?

To perform a hypothesis test for two means with unknown but equal variances, follow these steps: First, state the null hypothesis $H{0}_{}:\mu$₁=μ₂ and the alternative hypothesis depending on the claim (e.g., , $\mu \_1\; >\; \mu \_2$, or $\mu \_1\; \ne \; \mu \_2$). Next, calculate the pooled standard deviation using the formula for $s\_p^2$. Then compute the t-statistic with $t\; =\; rac\{ar\{x\}\_1\; -\; ar\{x\}\_2\}\{s\_p\; imes\; oot\{2\}rac\{1\}\{n\_1\}\; +\; rac\{1\}\{n\_2\}\}$. Determine the degrees of freedom as $df\; =\; n\_1\; +\; n\_2\; -\; 2$. Finally, find the p-value from the t-distribution and compare it to the significance level $\alpha$. If the p-value is less than $\alpha$, reject the null hypothesis, indicating sufficient evidence for the alternative.

What are the assumptions required for using the two means t-test with unknown but equal variances?

The key assumptions for the two means t-test with unknown but equal variances are: (1) The two samples are independent random samples from their respective populations. (2) The populations have equal variances, which justifies using the pooled standard deviation. (3) The populations are approximately normally distributed, or the sample sizes are large (usually greater than 30) to invoke the Central Limit Theorem. Meeting these assumptions ensures the validity of the t-test results and accurate inference about the population means.

How do you calculate the degrees of freedom when performing a two means t-test assuming equal variances?

When assuming equal variances in a two means t-test, the degrees of freedom (df) are calculated by adding the sample sizes of both groups and subtracting 2. Mathematically, $\mathrm{df}=n_1+n_2-\; 2$. For example, if both samples have 50 observations, then $df\; =\; 50\; +\; 50\; -\; 2\; =\; 98$. This df value is used to find critical values or p-values from the t-distribution during hypothesis testing.

How do you use a TI-84 calculator to perform a two means t-test assuming equal variances?

To perform a two means t-test with equal variances on a TI-84 calculator, press the STAT button, scroll right to the TESTS menu, and select option 4: 2-SampTTest. Choose the Stats input method if you have summary statistics. Enter the sample means, standard deviations, and sample sizes for both groups. Set the alternative hypothesis (e.g., , $\mu \_1\; >\; \mu \_2$, or $\mu \_1\; \ne \; \mu \_2$). Then, select Yes for the Pooled option to indicate equal variances. Finally, scroll down to Calculate and press ENTER. The calculator will display the t-statistic, degrees of freedom, and p-value for your test.