An intensive training course is given to two groups of workers. In group "A" there are 25 failed out of 180 workers who attend the course while in group "B" there are only 21 out of 170 workers. Can we conclude that group "B" takes better advantage of the instruction? Because? Use a 3% significance level.

We will check if there is a statistically significant difference between the two groups in terms of the proportion of workers who failed the course.

Let p_A be the proportion of workers who failed in group A and p_B be the proportion of workers who failed in group B.

The null hypothesis is that there is no difference in the proportions, which can be stated as:
H_0: p_A = p_B
The alternative hypothesis is that group B takes better advantage of the instruction:
H_1: p_A > p_B

The significance level given is 3%, which means \alpha = 0.03 .

Now, we will calculate the z-score and compare it to the critical z-value for rejection.

First, calculate the standard error of the difference between two sample proportions:
SE = \sqrt{p_{pool} \times (1 - p_{pool}) \times \left(\frac{1}{n_A} + \frac{1}{n_B}\right)}
where p_{pool} is the pooled sample proportion:
p_{pool} = \frac{X_A + X_B}{n_A + n_B}
X_A and X_B are the number of failures in groups A and B, respectively.

Then, calculate the z-score:
z = \frac{(p_A - p_B)}{SE}

Next, find the critical z-value at a significance level of 3%. Since it's a one-tailed test (we're checking if group B takes better advantage), the critical value is obtained by finding the z-value with a cumulative probability of 97%:
z_{\alpha} = 1.88

If the calculated z-score is greater than 1.88, we reject the null hypothesis.

Given:
- Group A: 25 failed out of 180 workers (n_A = 180, X_A = 25)
- Group B: 21 failed out of 170 workers (n_B = 170, X_B = 21)

Calculations:
p_{pool} = \frac{25 + 21}{180 + 170} = \frac{46}{350} \approx 0.1314

SE = \sqrt{0.1314 \times (1 - 0.1314) \times \left(\frac{1}{180} + \frac{1}{170}\right)} \approx 0.0335

p_A = \frac{25}{180} \approx 0.1389
p_B = \frac{21}{170} \approx 0.1235

z = \frac{0.1389 - 0.1235}{0.0335} \approx 0.459

We compare z = 0.459 to the critical z-value of 1.88. Since 0.459 < 1.88, we fail to reject the null hypothesis.

Therefore, we do not have enough evidence to conclude that group B takes better advantage of the instruction at a 3% significance level.