Question

Explain how to perform a two-sample z-test for the difference between two population means using independent samples with

Answer 1

Answer:

Z= (X1`-X2`)- (u1-u2)/$\sqrt{ \frac{ s_1^2}{n_1^2} +\frac{ s_2^2}{n_2}$   (here s1=σ1 and s2= σ2)

Step-by-step explanation:

Let  X1` be the mean of the first random sample of size n1 from a normal population with a mean u1 and known standard deviation σ1.Let  X2` be the mean of the second random sample of size n2 from another normal population with a mean u2 and known standard deviation σ2. Then the sampling distribution of the difference X1`-X2` is normally distributed with a mean of u1-u2 and a standard deviation of

$\sqrt{ \frac{ s_1^2}{n_1^2} +\frac{ s_2^2}{n_2}$ .  (here s1=σ1 and s2= σ2)  In other words the variable

Z= (X1`-X2`)- (u1-u2)/$\sqrt{ \frac{ s_1^2}{n_1^2} +\frac{ s_2^2}{n_2}$   (here s1=σ1 and s2= σ2)

is exactly standard normal no matter how small sample sizes are . Hence it is used as a test statistic for testing hypotheses about the difference between two population means.

The procedure is stated below.

1) Formulate the null and alternative hypotheses.

2) Decide on significance level ∝

3) Use the test statistic Z= (X1`-X2`)- (u1-u2)/$\sqrt{ \frac{ s_1^2}{n_1^2} +\frac{ s_2^2}{n_2}$   (here s1=σ1 and s2= σ2)

4) Find the rejection region

5)Compute the value of Z from the sample data

6) Rehect H0 if Z falls in the critical region, accept H0 , otherwise.