Two Means - Known Variance: Videos & Practice Problems

When population standard deviations are known, hypothesis testing for two means uses a $z-test$ instead of a $t-test$. The test compares sample means with known population standard deviations, applying the formula $z=\frac{\mu }{}$₁ - μ₂σ₁2n1 + σ₂2n2. A small p-value relative to the alpha level leads to rejecting the null hypothesis, supporting the alternative. Conditions include independent random samples and normality or large sample sizes, ensuring valid inference in comparing means.

Means Known Variances Hypothesis Test Using TI-84

When conducting hypothesis tests for two means with known population standard deviations, the process closely resembles tests where these values are unknown, with two main differences. Instead of using sample standard deviations, the known population standard deviations are used, and the test statistic is a z test rather than a t test. This approach is applicable when comparing two independent sample means to determine if there is a significant difference between the population means.

Consider a scenario where a grocery chain wants to test if self-checkout lanes have shorter checkout times than cashier lanes. They collect independent random samples of 35 checkout times from each lane type. The sample means are 4.5 minutes for self-checkout and 6.4 minutes for cashier lanes. Prior knowledge provides population standard deviations of σ₁ = 1.1 and σ₂ = 1.4. The significance level (α) is set at 0.05.

The hypotheses are formulated as follows: the null hypothesis (H₀) states that the population means are equal, $\mu ₁\; =\; \mu ₂$, while the alternative hypothesis (H₁) claims that the mean checkout time for self-checkout lanes is less than that for cashier lanes, .

To perform the test, the z test statistic is calculated using the formula:

$$z\; =\; \backslash frac\{\backslash bar\{x\}\_1\; -\; \backslash bar\{x\}\_2\}\{\backslash sqrt\{\backslash frac\{\backslash sigma\_1^2\}\{n\_1\}\; +\; \backslash frac\{\backslash sigma\_2^2\}\{n\_2\}\}\}$$

Here, $\backslash bar\{x\}\_1$ and $\backslash bar\{x\}\_2$ are the sample means, $\backslash sigma\_1$ and $\backslash sigma\_2$ are the known population standard deviations, and $n\_1$ and $n\_2$ are the sample sizes.

Using the given data, the z score is approximately -6.31, indicating a substantial difference between the two means. The corresponding p-value is the probability of observing a z score less than -6.31, which is about $1.37\; \backslash times\; 10^\{-10\}$. Since this p-value is much smaller than the significance level of 0.05, the null hypothesis is rejected.

This result provides strong evidence that self-checkout lanes have significantly shorter checkout times than cashier lanes. It is important to verify that the assumptions for the z test are met: the samples must be independent and randomly selected, and the sample sizes should be sufficiently large (typically greater than 30) or the data should be approximately normally distributed. In this case, with sample sizes of 35, the conditions are satisfied.

Understanding how to perform hypothesis tests for two means with known population standard deviations is essential for accurately comparing group differences when population variability is established. This method leverages the z distribution, providing precise inference when the population standard deviations are available, and reinforces the importance of formulating clear hypotheses, calculating the test statistic correctly, and interpreting the p-value in the context of the problem.

For ${\bar{x}}_1=40,{\sigma }_1=8.3,n_1=40,{\bar{x}}_2=39,{\sigma }_2=5.8$, & $n_2=50$, perform a hypothesis test to test the claim that ${\mu }_1>{\mu }_2$ for $\alpha =0.05$.

• 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: (D)

Means Known Variances Hypothesis Test Using TI-84 Example 1

When comparing attendance between yoga and weightlifting classes, a hypothesis test can determine if yoga classes are significantly more well attended. Given sample data where weightlifting classes have a mean attendance of 20.4 with a sample size of 32, and yoga classes have a mean attendance of 21.3 with the same sample size, the test uses known population standard deviations: σ₁ = 3.6 for weightlifting and σ₂ = 4.2 for yoga.

The null hypothesis (H₀) assumes no difference in average attendance between the two class types, expressed as $\mu$₁=μ₂. The alternative hypothesis (Hₐ) posits that yoga classes have higher attendance, or $\mu$₁<μ₂, where μ₁ is the mean attendance for weightlifting and μ₂ for yoga.

Using a significance level of α = 0.1, the test statistic is calculated based on the sample means, sizes, and known standard deviations. The p-value obtained from this test is approximately 0.18, which exceeds the alpha level. Since the p-value is greater than 0.1, the null hypothesis cannot be rejected, indicating insufficient evidence to conclude that yoga classes have higher average attendance than weightlifting classes.

Consequently, the gym does not have enough statistical evidence to justify adding more yoga classes over weightlifting classes. This conclusion emphasizes the importance of hypothesis testing in making data-driven decisions about resource allocation in fitness programs.

Means Known Variances Hypothesis Test Using TI-84 Example 2

A call center conducted a hypothesis test to determine if the average number of calls received differs between the morning and afternoon. The sample data showed that the morning had a mean call volume of 19.6 with a sample size of 90, while the afternoon had a mean of 15.3 with a sample size of 80. Using prior knowledge, the population standard deviations were known: σ₁ = 7.6 for the morning and σ₂ = 6.5 for the afternoon.

The hypotheses were set up as follows: the null hypothesis (H₀) stated that the mean number of calls in the morning (μ₁) equals the mean number in the afternoon (μ₂), implying no difference in call frequency. The alternative hypothesis (Hₐ) proposed that the means are not equal (μ₁ ≠ μ₂), indicating a difference in call volumes between the two times of day.

To test this, a two-sample z-test for means with known population standard deviations was performed at a significance level of α = 0.01. The test statistic was 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 𝑥̄₁ and 𝑥̄₂ are the sample means, σ₁ and σ₂ are the population standard deviations, and n₁ and n₂ are the sample sizes for morning and afternoon, respectively.

The resulting p-value was approximately 7.03 × 10⁻⁵, which is significantly less than the alpha level of 0.01. This led to rejecting the null hypothesis, providing strong evidence that the average number of calls differs between morning and afternoon periods.

Based on these findings, the call center should adjust staffing by increasing the number of agents during their busiest times rather than distributing shifts evenly throughout the day. This approach optimizes resource allocation by aligning employee schedules with call volume patterns, enhancing operational efficiency.

Here’s what students ask on this topic:

What is the difference between a z-test and a t-test when comparing two means?

When comparing two means, the key difference between a z-test and a t-test lies in whether the population standard deviations are known. A z-test is used when the population standard deviations are known, allowing us to use the exact values in the test statistic formula. In contrast, a t-test is used when the population standard deviations are unknown, and we estimate them using sample standard deviations. Because of this estimation, the t-test uses the t-distribution, which accounts for additional uncertainty. The z-test formula for two means is given by $z=\frac{\mu }{}$₁−μ₂σ2n₁+σ2n₂, where $\mu$₁ and $\mu$₂ are sample means, $\sigma$₁ and $\sigma$₂ are population standard deviations, and $n$₁ and $n$₂ are sample sizes. The t-test uses sample standard deviations instead and the t-distribution.

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

To perform a hypothesis test for two means with known population standard deviations, follow these steps: First, state the null hypothesis ($H{0}_{}:\mu$₁=μ₂) and the alternative hypothesis based on the research question. Second, collect sample data including sample means, sample sizes, and known population standard deviations. Third, calculate the z-test statistic using the formula $z=\frac{\mu }{}$₁−μ₂σ2n₁+σ2n₂. Fourth, find the p-value corresponding to the z-score. Finally, compare the p-value to the significance level (alpha). If the p-value is less than alpha, reject the null hypothesis, indicating evidence for the alternative hypothesis.

What are the assumptions and conditions required for hypothesis testing of two means with known population standard deviations?

When conducting hypothesis tests for two means with known population standard deviations, certain assumptions must be met to ensure valid results. First, the samples must be independent random samples from their respective populations. Second, the populations should be normally distributed, or if the sample sizes are large (typically greater than 30), the Central Limit Theorem allows us to proceed even if normality is not strictly met. These conditions help justify using the z-test and ensure the sampling distribution of the difference between means is approximately normal. Checking these assumptions is crucial before interpreting the test results.

How do you interpret the p-value in a two means hypothesis test with known variances?

The p-value in a two means hypothesis test with known variances represents the probability of observing a test statistic as extreme as, or more extreme than, the one calculated from the sample data, assuming the null hypothesis is true. A small p-value (less than the chosen significance level, often 0.05) indicates strong evidence against the null hypothesis, leading us to reject it in favor of the alternative hypothesis. Conversely, a large p-value suggests insufficient evidence to reject the null hypothesis. For example, if testing whether one mean is less than another, a p-value smaller than 0.05 supports the claim that the first mean is indeed less than the second.

Can you explain the formula for the z-test statistic when comparing two means with known population standard deviations?

The z-test statistic for comparing two means with known population standard deviations is calculated using the formula: $z=\frac{\mu }{}$₁−μ₂σ2n₁+σ2n₂. Here, $\mu$₁ and $\mu$₂ are the sample means, $\sigma$₁ and $\sigma$₂ are the known population standard deviations, 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, assuming the null hypothesis that the population means are equal.