Answer to Question #238220 in Statistics and Probability for Madimetja

Question

Given a sample of size 𝑛1=40 from a population with a standard deviation 𝜎1=20, and an independent sample of size 𝑛2= 50 from another population with a known standard deviation 𝜎2= 10, if the manager wants to test that the two population means differ, given 𝑥1= 72 and 𝑥2=66, which one of the following statements is not true:

1. The standard error is 3.4641.

2. The test statistic is 1.8532.

3. The rejection region at 5% level of significance is 𝑧 <-1.96 and 𝑧 > 1.96.

4. The statistical conclusion is we fail to reject H_O.

5. H_O: 𝜇1- 𝜇2=0 vs H_O: 𝜇1- 𝜇2≠0 

Answer 1

Population 1

$n_1=40\\\bar x_1=72\\\sigma_1=20$

Population 2

$n_2=50\\\bar x_2=66\\\sigma_2=10$

The hypotheses to be tested are,

$H_:\mu_1-\mu_2=0\\vs\\H_1:\mu_1-\mu_2\not=0$

Since the population variances are known and the sample sizes for both populations are large $(\gt 30)$, we shall apply the standard normal distribution to perform this hypothesis test as follows.

The test statistic is given as,

$Z_c={(\bar x_1-\bar x_2)\over SE(\bar x_1-\bar x_2)}$ where $SE(\bar x_1-\bar x _2)$ is the standard error of difference of the means given as,$SE(\bar x_1-\bar x_2)=\sqrt{{\sigma^2_1\over n_1}+{\sigma^2_2\over n_2}}=\sqrt{{400\over40}+{100\over50}}=\sqrt{10+2}=\sqrt{12}=3.4641(4dp)$

The test statistic is therefore,

$Z_c={72-66\over3.4641}={6\over 3.4641}=1.7321(4dp)$

The critical value is the standard normal table value at $\alpha=0.05$ given as,

$Z_{0.05\over 2}=Z_{0.025}=1.96$

Since we are performing a two-sided test, we shall have two critical values namely, $-1.96$ and $1.96.$ Therefore, the rejection region is, $Z\lt-1.96$ and $Z\gt1.96$.

The null hypothesis is rejected if, $|Z_c|\gt Z_{0.025}$

For this case, $|Z_c|=1.7321\lt Z_{0.025}=1.96$, we fail to reject the null hypothesis and we conclude that there is no sufficient evidence to show that the means differ at 5% level of significance.

From the choices given above, the choice number (2) is not true.