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

When population standard deviations (σ₁, σ₂) are known, hypothesis testing for two population means uses the normal distribution and a z-test. The null hypothesis assumes equal means (μ₁ = μ₂), while the alternative tests if one mean is less than the other (μ₁ < μ₂). Calculate the z-score using the formula $z=\frac{\mathrm{x}¯1_{}-\mathrm{x}¯2_{}}{\sqrt{{\sigma }^2/n_1+{\sigma }^2/n_2}}$. The p-value from the standard normal distribution determines if the null hypothesis is rejected at significance level α, supporting conclusions about population means differences.

Two Means -Sigma Known Hypothesis Test - Excel

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 function in Excel, which can directly compute p-values for two means without known standard deviations, the z-test for two means requires manually calculating the test statistic and then finding the p-value using the standard normal distribution.

Consider a scenario where a manufacturing company wants to determine if machine A produces fewer widgets per batch on average than machine B. Given sample data from 30 batches for each machine, and known population standard deviations σ₁ = 9.73 and σ₂ = 5.91, a hypothesis test at a significance level α = 0.05 can be performed.

The null hypothesis (H₀) states that the two population means are equal: \(H_0: \mu_1 = \mu_2\), where \(\mu_1\) and \(\mu_2\) represent the average widgets produced by machines A and B, respectively. The alternative hypothesis (H₁) reflects the company's concern that machine A produces less: \(H_1: \mu_1 < \mu_2\).

To calculate the z test statistic, first find the sample means \(\bar{x}_1\) and \(\bar{x}_2\) using the average function in Excel. For example, if \(\bar{x}_1 = 42.43\) and \(\bar{x}_2 = 45.97\), the numerator of the z formula is the difference between these means:

\[\bar{x}_1 - \bar{x}_2 = 42.43 - 45.97 = -3.54\]

The denominator involves the standard errors of the means, calculated using the known population standard deviations and sample sizes (\(n_1 = n_2 = 30\)):

\[\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} = \sqrt{\frac{9.73^2}{30} + \frac{5.91^2}{30}} = \sqrt{3.156 + 1.164} = \sqrt{4.32} \approx 2.08\]

Thus, the z score is:

\[z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} = \frac{-3.54}{2.08} \approx -1.70\]

Since the alternative hypothesis is one-sided (\(\mu_1 < \mu_2\)), the p-value corresponds to the cumulative probability to the left of the z score. Using Excel’s NORM.S.DIST function with the z score and cumulative set to TRUE yields a p-value of approximately 0.044.

Comparing the p-value to the significance level α = 0.05, since 0.044 < 0.05, the null hypothesis is rejected. This provides sufficient evidence to conclude that machine A produces fewer widgets on average per batch 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. It also demonstrates how breaking down the z test formula into smaller components can reduce calculation errors and improve clarity. Excel functions such as AVERAGE and NORM.S.DIST facilitate efficient computation of sample means and p-values, making hypothesis testing more accessible and accurate.

Two Means -Sigma Known Hypothesis Test - Excel Example 1

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 evaluate the claim that the animal shelter receives more volunteers on average.

The first step is to establish the hypotheses. The null hypothesis (H₀) assumes no difference in means: \(μ_1 = μ_2\), where \(μ_1\) is the mean number of volunteers at the animal shelter and \(μ_2\) is the mean at the food pantry. The alternative hypothesis (Hₐ) reflects the claim that the animal shelter has a higher average number of volunteers: \(μ_1 > μ_2\).

Next, calculate the sample means from the collected data. For example, the animal shelter’s sample mean (\(\bar{x}_1\)) might be 17.1 volunteers per week, while the food pantry’s sample mean (\(\bar{x}_2\)) is 15.12 volunteers per week. These values form the basis for the test statistic.

The z-test statistic is computed using the formula:

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

Here, the numerator is the difference between the sample means, and the denominator is the standard error of the difference, which accounts for the population variances divided by their respective sample sizes. Squaring the population standard deviations and dividing by the sample sizes yields the components inside the square root. For instance, \(\frac{\sigma_1^2}{n_1} = \frac{5.36^2}{50} \approx 0.57\) and \(\frac{\sigma_2^2}{n_2} = \frac{4.25^2}{50} \approx 0.36\). Plugging these into the formula gives a z-score of approximately 2.05.

To interpret this z-score, convert it to a p-value, which represents the probability of observing such a difference (or more extreme) if the null hypothesis were true. 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. Using the cumulative distribution function (CDF) for the standard normal distribution, the p-value is calculated as:

\[p = 1 - \Phi(z)\]

where \(\Phi(z)\) is the CDF value at the calculated z-score. For \(z = 2.05\), the p-value is approximately 0.02.

Comparing the p-value to the significance level (\(\alpha = 0.1\)), since \$0.02 < 0.1$, the null hypothesis is rejected. This statistical evidence supports the claim that the animal shelter receives more volunteers on average than the food pantry.

Understanding this process highlights the importance of formulating clear hypotheses, calculating appropriate test statistics using known population parameters, and interpreting p-values in the context of significance levels to make informed decisions based on data.

Two Means -Sigma Known Hypothesis Test - Excel Example 2

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, first 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-test statistic:

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

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

\[SE = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}\]

The z-test statistic is then computed as:

\[z = \frac{\bar{x}_1 - \bar{x}_2}{SE}\]

Once the z-score is obtained, the p-value is determined based on the two-tailed nature of the test. Since the alternative hypothesis is non-directional (\( \mu_1 \neq \mu_2 \)), the p-value is twice the smaller tail probability corresponding to the calculated z-score. This can be found using the standard normal cumulative distribution function (CDF), denoted as \( \Phi(z) \). For a negative z-score, the p-value is:

\[p = 2 \times \Phi(z)\]

Comparing the p-value to the significance level (α = 0.01) guides the decision-making process. If the p-value is greater than 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 exceeds the 0.01 threshold. Therefore, the conclusion is to fail to reject the null hypothesis, meaning the data do not provide strong enough evidence to claim that the two dispensing locations pour different average amounts of milk per bottle.

Here’s what students ask on this topic:

How do you perform a hypothesis test for two means when population standard deviations are known using Excel?

To perform a hypothesis test for two means with known population standard deviations in Excel, you first state your null hypothesis (H₀: μ₁ = μ₂) and alternative hypothesis (H₁, e.g., μ₁ < μ₂). Calculate the sample means using =AVERAGE(range). Then compute the z-score using the formula: $z=\frac{\mathrm{̅x}{1}_{}-\mathrm{̅x}{2}_{}}{\sqrt{{\sigma }^2\frac{1}{n_1}+{\sigma }^2\frac{1}{n_2}}}$. Use Excel's =NORM.S.DIST(z, TRUE) to find the p-value for a left-tailed test. Compare the p-value to your significance level α to decide whether to reject H₀. This method uses the normal distribution because population standard deviations (σ₁, σ₂) are known.

What is the difference between using a z-test and a t-test for two means in Excel?

The key difference between a z-test and a t-test for two means in Excel lies in whether the population standard deviations (σ₁, σ₂) are known. A z-test is used when these population standard deviations are known, and it relies on the normal distribution. The test statistic is calculated using the z-score formula. Excel does not have a built-in function for two-sample z-tests, so you calculate the z-score manually and then use =NORM.S.DIST to find the p-value. In contrast, a t-test is used when population standard deviations are unknown, and sample standard deviations are used instead. Excel provides the T.TEST function for this, which directly computes the p-value using the t-distribution. Thus, knowing the population standard deviations determines which test and Excel functions to use.

How do you calculate the z-score for two population means with known standard deviations in Excel?

To calculate the z-score for two population means with known standard deviations in Excel, first find the sample means using =AVERAGE(range) for each group. Then, use the formula: $z=\frac{\mathrm{̅x}{1}_{}-\mathrm{̅x}{2}_{}}{\sqrt{{\sigma }^2\frac{1}{n_1}+{\sigma }^2\frac{1}{n_2}}}$. In Excel, calculate the numerator by subtracting the two sample means, then calculate each variance term divided by sample size, sum them, and take the square root for the denominator. Finally, divide the numerator by the denominator to get the z-score. This stepwise approach reduces errors and is easy to implement in Excel cells.

How do you find the p-value from a z-score in Excel for a two means hypothesis test?

After calculating the z-score for your two means hypothesis test, you find the p-value in Excel using the =NORM.S.DIST(z, TRUE) function for a left-tailed test, which gives the cumulative probability up to the z-score. For a right-tailed test, use 1 - NORM.S.DIST(z, TRUE). For a two-tailed test, multiply the smaller tail probability by 2. The p-value indicates the probability of observing a test statistic as extreme as the z-score under the null hypothesis. Comparing this p-value to your significance level α helps you decide whether to reject the null hypothesis.

What are the steps to perform a two means hypothesis test with known sigmas in Excel?

To perform a two means hypothesis test with known population standard deviations (σ₁, σ₂) in Excel, follow these steps: (1) State your null hypothesis H₀: μ₁ = μ₂ and alternative hypothesis H₁ (e.g., μ₁ < μ₂). (2) Calculate sample means using =AVERAGE(range). (3) Calculate the z-score using the formula $z=\frac{\mathrm{̅x}{1}_{}-\mathrm{̅x}{2}_{}}{\sqrt{{\sigma }^2\frac{1}{n_1}+{\sigma }^2\frac{1}{n_2}}}$. (4) Use =NORM.S.DIST(z, TRUE) to find the p-value for a left-tailed test. (5) Compare the p-value to your significance level α to decide whether to reject H₀. This process uses Excel functions to simplify calculations and ensure accuracy.