What is the z Test and When is it Used?
The z Test is a statistical method used to determine if there is a significant difference between the means of two populations. This is commonly used when the sample sizes are large (typically n > 30) and the population variances are known. The z Test helps in understanding if the observed difference between the sample means is large enough to conclude that there is a meaningful difference between the population means.
What are the Assumptions for a Two-Sample z Test?
1. Independent Samples: The samples are independent of each other.2. Normal Distribution: The populations from which the samples are drawn should be normally distributed. This is usually assumed if the sample size is large due to the Central Limit Theorem.3. Known Population Variances: The variances of the two populations are known.4. Random Sampling: The samples should be randomly selected from the population.
What is the Formula for the Two-Sample z Test?
Suppose we have two sample means, ( ar{X}_1 ) and ( ar{X}_2 ), from two samples. The z Test statistic can be calculated using the formula:
z = ( ( ar{X}_1 - ar{X}_2 ) - ( Delta )) / ( SE )
Where:- ( ar{X}_1 ) = Mean of sample 1- ( ar{X}_2 ) = Mean of sample 2- ( Delta ) = Hypothesized difference between population means (commonly 0)- ( SE ) = Standard Error of the difference between the two means
The Standard Error (SE) is calculated as:
SE = sqrt((?1^2 / n1) + (?2^2 / n2))
Where:- ?1^2 = Variance of population 1- ?2^2 = Variance of population 2- n1 = Sample size of sample 1- n2 = Sample size of sample 2
Can You Provide an Example?
Sure! Let’s work through an example to make this clear:
Question:
Suppose researchers want to test if there is a difference in the average heights of men and women in a specific city. They select random samples of 50 men and 50 women. The mean height of men is 70 inches with a known population variance of 4 inches², and the mean height of women is 65 inches with a known population variance of 5 inches². Test the hypothesis that there is no difference in the means at a significance level of 0.05.
Step-by-Step Calculation:
1. State the Hypotheses: - Null Hypothesis (H0): ?1 - ?2 = 0 (no difference in means) - Alternative Hypothesis (H1): ?1 - ?2 ? 0 (there is a difference in means)
2. Calculate the Standard Error (SE): SE = sqrt((?1^2 / n1) + (?2^2 / n2)) SE = sqrt((4 / 50) + (5 / 50)) SE = sqrt(0.08 + 0.10) SE = sqrt(0.18) SE = 0.424
3. Calculate the z Statistic: z = ( ( ar{X}_1 - ar{X}_2 ) - ( Delta )) / SE z = (70 - 65 - 0) / 0.424 z = 5 / 0.424 z ? 11.79
4. Compare the z Statistic to Critical Value: For ( alpha = 0.05 ) (two-tailed test), the critical z values are ±1.96.
Since the calculated z (11.79) is well beyond 1.96, we reject the null hypothesis.
Conclusion:
Given that the calculated z-value of 11.79 is much greater than the critical value of 1.96, we reject the null hypothesis. Therefore, we have sufficient evidence to conclude that there is a significant difference in the mean heights of men and women in this city.
Summary:
The z Test is a useful statistical tool when comparing the means of two large, independent samples with known variances. By computing the test statistic and comparing it to a critical value, we can determine whether to accept or reject the null hypothesis, thus making inferences about the population means.
Commuting Times The U.S. Census Bureau reports that the average commuting time for citizens of both Baltimore, Maryland, and Miami, Florida, is appro…
Heights of 9 -Year-Olds At age 9 the average weight $(21.3 \mathrm{kg})$ and the average height $(124.5 \mathrm{cm})$ for both boys and girls are exa…
Watch the video solution with this free unlock.
EMAIL
PASSWORD