The owner of a restaurant conducted two statistical studies to determine the proportion of female customers, one by his workers and the other by two marketing professionals. According to the sampling data, of the study of the workers of 30 clients, 60% were women; On the other hand, in the sampling carried out by professionals of 40 clients, 30% were men. With the results of the two studies, you want to know if there will be a significant difference between the proportions of female clients, at a significance level of 0.045.

To determine if there is a significant difference between the proportions of female clients based on the two studies, we will perform a hypothesis test.

Let's define the null hypothesis $$H_0$$ as the proportion of female clients being the same in the two studies. The alternative hypothesis $$H_1$$ will be that there is a significant difference in the proportions of female clients.

Let's denote $$p_1$$ as the proportion of female clients in the study conducted by the restaurant workers and $$p_2$$ as the proportion of female clients in the study conducted by the marketing professionals.

We can use the z-test for comparing proportions to test the null hypothesis. The formula for calculating the z-test statistic is:

$$Z = \frac{(p_1 - p_2) - 0}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}$$

where $$n_1$$ and $$n_2$$ are the sample sizes of the two studies.

Given that the sample size of the study conducted by the workers is 30 with a proportion of female clients of 60%, we have:

$$p_1 = 0.6$$ and $$n_1 = 30$$

Similarly, the sample size of the study conducted by the marketing professionals is 40 with a proportion of male clients of 30%. Since we are interested in the proportion of female clients, we can calculate the proportion of female clients as:

$$p_2 = 1 - 0.3 = 0.7$$ and $$n_2 = 40$$

Now, we can plug these values into the formula:

$$Z = \frac{(0.6 - 0.7) - 0}{\sqrt{\frac{0.6(1-0.6)}{30} + \frac{0.7(1-0.7)}{40}}}$$



Calculating the z-test statistic:

$$Z \approx -0. 78$$

To determine if there is a significant difference at a significance level of 0.045, we compare the absolute value of the calculated z-test statistic with the critical value of the standard normal distribution.

The critical value for a two-tailed test at a significance level of 0.045 is approximately $$\pm 2.0047$$

Since the absolute value of the calculated z-test statistic (-0.78)is less than the critical value (-2.0047,2.0047), we fail to reject the null hypothesis.

Therefore, based on the given data and significance level of 0.045, there is not enough evidence to conclude that there is a significant difference between the proportions of female clients in the two studies.

Answer: There is no significant difference between the proportions of female clients in the two studies.