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

When testing 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 hypotheses with the null $H{0}_{}:\mu$₁=μ₂ and alternative $H{a}_{}:\mu$₁<μ₂. Calculate the t-statistic using the pooled variance $s_p=\sqrt{\frac{s_{1_{}}²(n}{}}$₁-1)+s2²(n₂-1)n₁+n₂-2. A small p-value compared to alpha indicates rejecting the null, supporting the alternative hypothesis. Ensure samples are independent and sufficiently large for valid inference.

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 the 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 refines the t-test for two means, improving the precision of the results, although it involves more complex calculations.

Consider a practical example where a school claims that 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 a standard deviation of 4.8, while the other uses the new app with a sample mean of 82 and a standard deviation of 4.4. Assuming equal population variances, the goal is to test if the app significantly improves 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, so the mean score with the app is higher). The next step involves calculating the pooled variance using the formula:

$$s{p}_2=\frac{}{}$$

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 t-statistic:

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

Here, 𝑥̄₁ and 𝑥̄₂ are the sample means, and sₚ is the square root of the pooled variance (the pooled standard deviation). In the example, the pooled variance is approximately 21, leading to a calculated t-value near -5.43.

To determine the p-value, the degrees of freedom are calculated as df = n₁ + n₂ - 2, which equals 98 in this case. The p-value corresponds to the probability of observing a t-value less than -5.43 under the null hypothesis. This p-value is extremely small, approximately 2.05 × 10⁻⁷, indicating strong evidence against the null hypothesis.

Since the p-value is less than the significance level of 0.05, the null hypothesis is rejected, supporting the claim that the new math app improves average test scores.

It is essential to verify the assumptions before performing this test: the samples must be independent and randomly selected, and the populations should be approximately normal or the sample sizes sufficiently large (usually greater than 30). In this example, both samples have sizes of 50, satisfying the normality condition by the Central Limit Theorem.

Using technology such as a TI-84 calculator can simplify these calculations, especially when selecting the option to use the pooled standard deviation. This approach ensures accurate hypothesis testing when equal population variances are assumed, enhancing the reliability of conclusions drawn about differences between two means.

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. $P-value<\alpha$; 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. Assuming equal variances between the two populations, the pooled standard deviation method is appropriate for this two-sample t-test.

The hypotheses are formulated as follows: the null hypothesis (H₀) states that the mean wait times for both services are equal, mathematically expressed as $\backslash (\; \backslash mu\_1\; =\; \backslash mu\_2\; \backslash )$. The alternative hypothesis (H_a) claims that Service A has longer wait times than Service B, or $\backslash (\; \backslash mu\_1\; >\; \backslash mu\_2\; \backslash )$. This is a one-tailed test focusing on whether Service B offers shorter wait times.

Using the pooled variance assumption, the test statistic is calculated with the formula for the two-sample t-test:

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

where $\backslash (\; \backslash bar\{x\}\_1\; \backslash )$ and $\backslash (\; \backslash bar\{x\}\_2\; \backslash )$ are the sample means, $\backslash (\; n\_1\; \backslash )$ and $\backslash (\; n\_2\; \backslash )$ are the sample sizes, and $\backslash (\; s\_p\; \backslash )$ is the pooled standard deviation calculated as:

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

Here, $\backslash (\; s\_1\; \backslash )$ and $\backslash (\; s\_2\; \backslash )$ represent the sample standard deviations. After computing the test statistic, the p-value is obtained to assess the strength of evidence against the null hypothesis.

In this case, the p-value was found to be approximately $\backslash (\; 4.05\; \backslash times\; 10^\{-5\}\; \backslash )$, which is significantly less than the chosen significance level $\backslash (\; \backslash alpha\; =\; 0.1\; \backslash )$. This leads to rejecting the null hypothesis, providing strong evidence that Service B has shorter wait times than Service A.

Consequently, the grocery chain has sufficient statistical evidence to prefer Service B for outsourcing its home delivery service, as the data supports the claim that Service B offers faster delivery times. This conclusion is based on rigorous hypothesis testing, ensuring that the decision is data-driven and reliable.

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 variability from both samples for the test.

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

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

where \( s_1 \) and \( s_2 \) are the sample standard deviations, and \( n_1 \) and \( n_2 \) are the sample sizes. The test statistic \( t \) is then:

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

Using the given data, the calculated p-value is approximately 0.189, which is greater than the significance level of 0.01. This means there is insufficient evidence to reject the null hypothesis. Therefore, it cannot be concluded that the average number of views differs between the two platforms.

Based on these results, 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 over the other.

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 variances from two independent samples. It is used when performing hypothesis tests for two population means where the population standard deviations are unknown but assumed to be equal. By pooling the variances, we get a better estimate of the common population variance, which improves the accuracy of the test statistic. This approach replaces the individual sample standard deviations in the formula with a single pooled value, simplifying calculations and providing more reliable results. The formula for the pooled variance $s_p²=\frac{{\mathrm{(n}}_1}{-}{s}_1²+\frac{{\mathrm{(n}}_2}{-}{s}_2²)/{\mathrm{(n}}_1+n_2-2$ shows how it combines the variances weighted by their degrees of freedom. This method is essential for accurate t-tests when variances are assumed equal but unknown.

How do you perform a hypothesis test for two means assuming equal variances using a TI-84 calculator?

To perform a hypothesis test for two means with equal variances on a TI-84 calculator, first press the STAT button, then scroll right to the TESTS menu. Select option 4, "2-SampTTest." Choose "Stats" to enter summary statistics. Input the sample means, standard deviations, and sample sizes for both groups. Set the alternative hypothesis (e.g., $\mu 1_{}<\mu 2_{}$ if testing if the second mean is greater). Then, select "Yes" for the pooled option to indicate equal variances. Finally, scroll down and select "Calculate" to get the t-value and p-value. This process uses the pooled standard deviation to improve test accuracy when variances are assumed equal but unknown.

What are the assumptions and conditions required for conducting a two-sample t-test with equal variances?

When conducting a two-sample t-test assuming equal variances, several key assumptions must be met: (1) The two samples must be independent random samples from their respective populations. (2) The populations should have equal variances, which justifies using the pooled standard deviation. (3) The data in each group should be approximately normally distributed, especially important for small sample sizes. However, if each sample size is greater than 30, the Central Limit Theorem allows us to proceed even if normality is questionable. Meeting these conditions ensures the validity of the t-test results and the accuracy of conclusions drawn from the hypothesis test.

How do you calculate the degrees of freedom for a two-sample t-test assuming equal variances?

For a two-sample t-test assuming equal variances, the degrees of freedom (df) are calculated by adding the degrees of freedom from each sample. Specifically, $\mathrm{df}=(n_1-1)+(n_2-1)$, where $n_1$ and $n_2$ are the sample sizes of the two groups. This sum accounts for the total number of independent pieces of information used to estimate the pooled variance. For example, if both samples have 50 observations, the degrees of freedom would be 98. This df value is used to find critical t-values or p-values from the t-distribution during hypothesis testing.

What is the formula for the test statistic in a two-sample t-test assuming equal variances?

The test statistic for a two-sample t-test assuming equal variances is calculated using the formula: $t=\frac{{\mu }_1-{\mu }_2}{s_p}$. Here, ${\mu }_1$ and ${\mu }_2$ are the sample means, $s_p$ is the pooled standard deviation, and $n_1$ and $n_2$ are the sample sizes. This formula measures how many standard errors the difference between sample means is away from zero, helping determine if the observed difference is statistically significant.