Two Means - Known Variance: Videos & Practice Problems

Performing hypothesis tests for two means with known population standard deviations involves using a $z$-test instead of a $t$-test. The test compares sample means using the formula for the $z=\frac{\mu }{}$₁−μ₂σ₁21n₁+σ₂21n₂. A small p-value compared to the significance level $\alpha$ leads to rejecting the null hypothesis, supporting the alternative hypothesis. Conditions include independent random samples and normality or large sample sizes, ensuring valid inferential statistics.

Means Known Variances Hypothesis Test Using TI-84

When performing 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 follows a z distribution rather than a t distribution. This allows for a two-sample z-test to compare the means.

Consider a scenario where a grocery chain wants to determine 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 goal is to test the claim that self-checkout times are shorter, using a significance level of α = 0.05.

The hypothesis test begins by stating the null hypothesis H₀: μ₁ = μ₂, indicating no difference in mean checkout times, and the alternative hypothesis Hₐ: μ₁ < μ₂, suggesting self-checkout lanes are faster.

Next, the test statistic z is calculated using the formula:

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

Here, 𝑥̄₁ and 𝑥̄₂ are the sample means, σ₁ and σ₂ are the known population standard deviations, and n₁ and n₂ are the sample sizes. Substituting the values:

\[ z = \frac{4.5 - 6.4}{\sqrt{\frac{1.1^2}{35} + \frac{1.4^2}{35}}} \approx -6.31 \]

The resulting z-score indicates how many standard errors the observed difference in means is from zero under the null hypothesis.

To determine the p-value, which represents the probability of observing a z-score as extreme as -6.31 or more under the null hypothesis, one looks up the cumulative probability for z < -6.31. This p-value is approximately 1.37 × 10^-10, an extremely small value.

Since the p-value is much less than the significance level α = 0.05, the null hypothesis is rejected. This provides strong evidence that the mean checkout time for self-checkout lanes is indeed shorter than for cashier lanes.

Before finalizing conclusions, it is essential to verify the assumptions of the test: the samples must be independent and randomly selected, and the sampling distribution of the difference in means should be approximately normal. With sample sizes greater than 30, the Central Limit Theorem ensures normality, satisfying these conditions.

Understanding how to conduct a two-sample z-test with known population standard deviations is crucial for accurately comparing means when such information is available. This method provides a reliable way to test claims about differences between two population means, leveraging the z distribution for hypothesis testing.

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 $\mu ₁\; =\; \mu ₂$. The alternative hypothesis (H₁) posits that yoga classes have higher attendance, or , where μ₁ is the mean attendance for weightlifting and μ₂ for yoga.

Using a significance level of α = 0.1, a two-sample z-test for means with known standard deviations is appropriate. The test statistic formula is:

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

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

Calculating the p-value from the z-test indicates the probability of observing the data assuming the null hypothesis is true. In this case, the p-value is approximately 0.18, which exceeds the significance level of 0.1. Since the p-value is greater than α, the null hypothesis cannot be rejected. This means there is insufficient statistical evidence to conclude that yoga classes have higher average attendance than weightlifting classes.

Consequently, the gym should not base decisions on adding more yoga classes over weightlifting classes solely on attendance data, as the evidence does not support a significant difference in attendance rates. This approach highlights the importance of hypothesis testing in making data-driven decisions in business and education settings.

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. Prior knowledge provided population standard deviations of σ₁ = 7.6 for the morning and σ₂ = 6.5 for the afternoon. The significance level (α) was set at 0.01.

The hypotheses were formulated as follows: the null hypothesis (H₀) stated that the mean number of calls in the morning (μ₁) equals that in the afternoon (μ₂), implying no difference in call frequency. The alternative hypothesis (H₁) proposed that the means are not equal (μ₁ ≠ μ₂), indicating a difference in call volumes between the two time periods.

Using a two-sample z-test for means with known population standard deviations, the test statistic was calculated based on the sample means, sizes, and standard deviations. 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.

Given these results, the call center should adjust staffing by increasing the number of agents during the busier time periods rather than distributing shifts evenly. This approach optimizes resource allocation based on the observed variation in call frequency throughout the day.

Here’s what students ask on this topic:

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

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 ($\sigma$) 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 ($s$). Because of this estimation, the t-test uses the t-distribution, which accounts for extra uncertainty, especially with smaller sample sizes. The z-test formula for the test statistic is $z=\frac{\mathrm{̄x}}{}$₁ - ̄x₂σ2n1 + σ2n2. Using the z-test is appropriate when sample sizes are large or when population variances are known, ensuring more precise hypothesis testing.

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 ($H{0}_{}$) and alternative hypothesis ($H{a}_{}$). For example, $H{0}_{}:\; \mu$₁ = μ₂ and $H{a}_{}:\; \mu$₁ < μ₂. Next, collect sample means ($\mathrm{̄x}$), population standard deviations ($\sigma$), and sample sizes ($n$). Calculate the z-score using the formula $z=\frac{\mathrm{̄x}}{}$₁ - ̄x₂σ2n1 + σ2n2. Then, 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 sufficient evidence for the alternative hypothesis.

What are the assumptions and conditions required for a two-sample z-test for means with known variances?

The two-sample z-test for means with known variances requires several assumptions to ensure valid results. First, the samples must be independent and randomly selected from their respective populations. Second, the populations should be normally distributed, especially if the sample sizes are small. However, if the sample sizes are large (usually $n\; >\; 30$), the Central Limit Theorem allows us to proceed even if the populations are not perfectly normal. Third, the population standard deviations ($\sigma$₁ and $\sigma$₂) must be known and used in the test statistic calculation. Meeting these conditions ensures that the z-test is appropriate and that the p-values and conclusions drawn from the test are reliable.

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

The p-value in a two-sample z-test 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 significance level $\alpha$, often 0.05) indicates strong evidence against the null hypothesis, leading us to reject it. This suggests that there is sufficient evidence to support the alternative hypothesis. Conversely, a large p-value means there is not enough evidence to reject the null hypothesis, so we fail to reject it. For example, if the p-value is 0.000000000137, which is much smaller than 0.05, we reject the null hypothesis and conclude that the means are significantly different in the direction specified by the alternative hypothesis.

Can you explain the formula for the z-score in a two-sample test for means with known population standard deviations?

The z-score formula for a two-sample test comparing means with known population standard deviations is: $z=\frac{\mathrm{̄x}}{}$₁ - ̄x₂σ2n1 + σ2n2. Here, $̄x$₁ and $̄x$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. The denominator calculates the standard error of the difference between the two means, combining the variability from both samples. This z-score measures how many standard errors the observed difference in sample means is away from the hypothesized difference (usually zero under the null hypothesis). A large absolute z-score indicates a significant difference between the means.