Two Means - Sigma Known Hypothesis Test - Excel: Videos & Practice Problems

When conducting a hypothesis test for two population means with known population standard deviations, the normal distribution and a z test are used instead of the t distribution. This approach applies when the population standard deviations, denoted as σ₁ and σ₂, are known values. Unlike the t-test, Excel’s built-in z-test function only works for one mean, so the test for two means requires manually calculating the test statistic and p-value through a step-by-step process.

Consider a scenario where a manufacturing company suspects that Machine A produces fewer widgets per batch on average than Machine B. Data is collected from 30 random batches for each machine, with population standard deviations σ₁ = 9.73 and σ₂ = 5.91, and a significance level α = 0.05. The hypotheses are formulated as follows: the null hypothesis (H₀) states that the two population means are equal (μ₁ = μ₂), while the alternative hypothesis (H₁) claims that the mean production of Machine A (μ₁) is less than that of Machine B (μ₂), reflecting a left-tailed test.

The z test statistic is calculated using the formula:

\[z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\]

where \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means, \(n_1\) and \(n_2\) are the sample sizes, and \(\sigma_1\) and \(\sigma_2\) are the known population standard deviations. To reduce errors, it is practical to compute the numerator (\(\bar{x}_1 - \bar{x}_2\)) and each component of the denominator separately before combining them.

For example, if the sample means are \(\bar{x}_1 = 42.43\) and \(\bar{x}_2 = 45.97\), the numerator is \$42.43 - 45.97 = -3.54\(. The denominator involves calculating \)\frac{\sigma_1^2}{n_1} = \frac{9.73^2}{30} \approx 3.156\( and \)\frac{\sigma_2^2}{n_2} = \frac{5.91^2}{30} \approx 1.164$. The square root of their sum is then used as the denominator.

After computing the z score (approximately -1.7 in this case), the p-value is found using the cumulative distribution function of the standard normal distribution, which gives the probability of observing a value as extreme or more extreme than the test statistic under the null hypothesis. Since the alternative hypothesis is left-tailed, the p-value corresponds to the left-tail probability:

\[p = P(Z \leq z)\]

Using Excel’s NORM.S.DIST function with the cumulative option returns this p-value, which in this example is about 0.04. Comparing the p-value to the significance level α = 0.05, since 0.04 < 0.05, the null hypothesis is rejected. This indicates sufficient evidence to support the claim that Machine A produces fewer widgets on average than Machine B.

This method highlights the importance of correctly identifying when to use a z test versus a t test based on knowledge of population standard deviations, and demonstrates how to systematically calculate the test statistic and p-value for hypothesis testing of two means using the normal distribution. Mastery of these steps enables accurate decision-making in quality control and other applications involving comparison of two population means.

When comparing the average number of volunteers between two locations, such as a local animal shelter and a food pantry, hypothesis testing can determine if one location consistently receives more volunteers than the other. Given population standard deviations (σ₁ = 5.36 for the animal shelter and σ₂ = 4.25 for the food pantry) and equal sample sizes (n₁ = n₂ = 50), a two-sample z-test is appropriate to analyze the difference in means.

The null hypothesis (H₀) assumes no difference in average volunteers between the two locations, expressed as \(μ_1 = μ_2\). The alternative hypothesis (Hₐ) reflects the claim that the animal shelter receives more volunteers on average, stated as \(μ_1 > μ_2\).

To conduct the test, first calculate the sample means: the animal shelter’s average (\(\bar{x}_1\)) is 17.1 volunteers, and the food pantry’s average (\(\bar{x}_2\)) is 15.12 volunteers. The test statistic z is computed using the formula:

\[z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\]

Substituting the values:

\[z = \frac{17.1 - 15.12}{\sqrt{\frac{5.36^2}{50} + \frac{4.25^2}{50}}} = \frac{1.98}{\sqrt{0.574 + 0.361}} = \frac{1.98}{\sqrt{0.935}} \approx \frac{1.98}{0.967} \approx 2.05\]

Since the alternative hypothesis is one-sided (\(μ_1 > μ_2\)), the p-value corresponds to the right-tail probability of the standard normal distribution beyond z = 2.05. Using the cumulative distribution function (CDF) for the standard normal distribution, the left-tail probability is found first, then subtracted from 1 to obtain the right-tail p-value:

\[p = 1 - \Phi(2.05) \approx 1 - 0.9798 = 0.0202\]

With a significance level (α) of 0.1, the p-value of approximately 0.02 is less than α, leading to rejection of the null hypothesis. This statistical evidence supports the claim that the animal shelter receives more volunteers on average than the food pantry.

This process highlights the importance of formulating clear hypotheses, calculating the test statistic accurately, and interpreting the p-value in the context of the chosen significance level to make informed conclusions about population means.

When comparing the average amounts of milk dispensed at two different locations, a hypothesis test can determine if there is a significant difference between the two means. Given population standard deviations (σ₁ = 0.46 and σ₂ = 0.55) and sample sizes (n₁ = n₂ = 50), a two-sample z-test is appropriate for this analysis. The null hypothesis (H₀) states that the mean amounts dispensed at both locations are equal, expressed as \( \mu_1 = \mu_2 \). The alternative hypothesis (Hₐ) suggests that the means are not equal, or \( \mu_1 \neq \mu_2 \), indicating a two-tailed test.

To perform the test, calculate the sample means for each location, denoted as \( \bar{x}_1 \) and \( \bar{x}_2 \). The difference between these sample means forms the numerator of the z-score formula:

\[\text{Numerator} = \bar{x}_1 - \bar{x}_2\]

The denominator involves the standard error of the difference between means, which incorporates the population standard deviations and sample sizes:

\[\text{Standard Error} = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}\]

The z-score is then calculated as:

\[z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\]

Once the z-score is obtained, the p-value is determined based on the two-tailed test. Since the alternative hypothesis is non-directional (\( \mu_1 \neq \mu_2 \)), the p-value is twice the smaller tail probability of the standard normal distribution corresponding to the calculated z-score. If the z-score is negative, the left tail probability is used; if positive, the right tail probability is considered. The p-value is computed as:

\[p = 2 \times P(Z \leq z)\]

Comparing the p-value to the significance level (α = 0.01) guides the decision. If the p-value exceeds α, there is insufficient evidence to reject the null hypothesis, indicating no significant difference in the average amounts dispensed between the two locations. Conversely, a p-value less than α would suggest a statistically significant difference.

In this scenario, the calculated z-score was approximately -0.63, leading to a p-value around 0.53, which is much greater than the 0.01 threshold. Therefore, the conclusion is to fail to reject the null hypothesis, meaning the data do not provide enough evidence to claim that the two dispensing locations pour different average amounts of milk per bottle.