Large Sample Test

• Z-test: Difference between two means of large samples with unknown population variance

A Z-test for the difference between two means of large samples is used when comparing the means of two independent populations to determine whether there is a significant difference between them. This test is applied under certain conditions, especially when the population variances are unknown, but sample sizes are large enough (typically $n > 30$ ) for the Central Limit Theorem to apply.

Here’s how it works:

1. Assumptions:

Both samples are large ( $n_1 > 30$ and $n_2 > 30$ ).
The population variances are unknown, but since the samples are large, we assume the sample variances approximate the population variances.
The samples are independent of each other.
The data from both samples follow an approximately normal distribution due to the large sample size.

2. Test Statistic (Z):

The Z-test statistic for the difference between two means is calculated as:

$\frac{(\bar{X}_1 – \bar{X}_2) – (\mu_1 – \mu_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$

Where:

$Xˉ1\bar{X}_1$ and $Xˉ2\bar{X}_2$ are the sample means of sample 1 and sample 2, respectively.
$n_1$ and $n_2$ are the sizes of sample 1 and sample 2.
$s_1^2$ and $s_2^2$ are the sample variances of sample 1 and sample 2.
$μ1−μ2\mu_1 – \mu_2$ is the hypothesized difference between the two population means (usually 0 when testing for no difference).

3. Steps in Performing the Z-Test:

State the hypotheses:
- Null hypothesis ( $H_0$ ): $μ1=μ2\mu_1 = \mu_2$ (no difference in population means).
- Alternative hypothesis ( $H_1$ ): $μ1≠μ2\mu_1 \neq \mu_2$ , $μ1>μ2\mu_1 > \mu_2$ , or $μ1<μ2\mu_1 < \mu_2$ (depending on whether it’s a two-tailed, left-tailed, or right-tailed test).
Calculate the Z-statistic using the formula mentioned above.
Determine the critical value from the standard normal distribution table corresponding to your chosen significance level (e.g., 1.96 for a 5% significance level in a two-tailed test).
Compare the calculated Z-value with the critical value:
- If $∣ Z ∣$ is greater than the critical value, you reject the null hypothesis.
- If $∣ Z ∣$ is less than the critical value, you fail to reject the null hypothesis.
Draw conclusions based on the comparison.

4. Example:

Suppose two independent samples are taken from two populations, and we want to test whether there is a difference in their means.

Sample 1: $Xˉ1=100\bar{X}_1 = 100$ , $s_1 = 15$ , $n_1 = 40$
Sample 2: $Xˉ2=95\bar{X}_2 = 95$ , $s_2 = 10$ , $n_2 = 50$

We want to test if there is a significant difference between the two population means ( $μ1\mu_1$ and $μ2\mu_2$ ) at a 5% significance level.

Steps:

Hypotheses:
- $H_0$ : $μ1=μ2\mu_1 = \mu_2$
- $H_1$ : $μ1≠μ2\mu_1 \neq \mu_2$ (two-tailed test)
Calculate the Z-statistic:

$\frac{(100 – 95)}{\sqrt{\frac{15^2}{40} + \frac{10^2}{50}}} = \frac{5}{\sqrt{\frac{225}{40} + \frac{100}{50}}} = \frac{5}{\sqrt{5.625 + 2}} = \frac{5}{\sqrt{7.625}} = \frac{5}{2.76} \approx 1.81$

Critical Value: At a 5% significance level (two-tailed), the critical value from the Z-table is ±1.96.
Compare: Since 1.81 is less than 1.96, we fail to reject the null hypothesis.
Conclusion: There is not enough evidence to conclude that the two population means are significantly different.

Unit VII Large Sample Test