A survey company has introduced a more modern way of conducting surveys online, and the company wants to test its effectiveness. The numbers of the surveys carried out in a week by a random sample of agents are: 47 53 50 49 56 53 60 54 58 67 59 60 57 58 55 53 48 62 65 51 Note: The calculations obtained from the sample must be worked to 4 decimal places. (a) (7pts) Define the random variable of interest involved (b) (31pts) With a significance level of 10%, it can be concluded that the average number of surveys carried out by the agents is greater than 53 surveys per week. (c) (31pts) With a significance level of 5%, it can be concluded that the variance of the surveys carried out by the agents is different from 14 (surveys per week)2 (d) (31pts) With a significance level of 1%, it can be concluded that less than 30% of the agents carry out more than 59 surveys per week

Let's go through each part of the question step by step: (a) Define the random variable of interest involved: The random variable of interest here is the number of surveys carried out by a random sample of agents in a week. (b) With a significance level of 10%, it can be concluded that the average number of surveys carried out by the agents is greater than 53 surveys per week: To test this hypothesis, we'll perform a one-sample t-test. The null hypothesis (H0) is that the average number of surveys carried out is not greater than 53 surveys per week, and the alternative hypothesis (H1) is that the average is greater than 53 surveys per week. H0: μ ≤ 53 H1: μ > 53 We'll calculate the t-statistic and compare it to the critical value at a 10% significance level. First, calculate the sample mean (x̄) and sample standard deviation (s) of the data: x̄ = (47 + 53 + 50 + 49 + 56 + 53 + 60 + 54 + 58 + 67 + 59 + 60 + 57 + 58 + 55 + 53 + 48 + 62 + 65 + 51) / 20 = 56.5 s = √[(Σ(xi - x̄)²) / (n - 1)] = √[(Σ(47-56.5)² + (53-56.5)² + ... + (51-56.5)²) / 19] ≈ 5.7868 Now, calculate the t-statistic: t = (x̄ - μ) / (s / √n) t = (56.5 - 53) / (5.7868 / √20) t ≈ 1.2959 Now, find the critical value for a one-tailed test at a 10% significance level and 19 degrees of freedom (n-1). You can consult a t-table or use a t-distribution calculator. For a 10% significance level and one tail, the critical t-value is approximately 1.7291. Since the calculated t-statistic (1.2959) is less than the critical t-value (1.7291), we fail to reject the null hypothesis. There is not enough evidence to conclude that the average number of surveys carried out by the agents is greater than 53 surveys per week at a 10% significance level. (c) With a significance level of 5%, it can be concluded that the variance of the surveys carried out by the agents is different from 14 (surveys per week)^2: To test this hypothesis, we'll perform a chi-squared test for variance. The null hypothesis (H0) is that the variance is equal to 14^2, and the alternative hypothesis (H1) is that the variance is different. H0: σ^2 = 14^2 H1: σ^2 ≠ 14^2 We'll calculate the chi-squared statistic and compare it to the critical chi-squared values at a 5% significance level and (n-1) degrees of freedom, where n is the sample size. First, calculate the sample variance (s²): s² = [(Σ(xi - x̄)²) / (n - 1)] = [(Σ(47-56.5)² + (53-56.5)² + ... + (51-56.5)²) / 19] ≈ 33.5368 Now, calculate the chi-squared statistic: χ² = (n - 1) * (s² / σ²) χ² = 19 * (33.5368 / (14^2)) χ² ≈ 17.4626 Find the critical chi-squared values for a 5% significance level and (n-1) degrees of freedom. You can consult a chi-squared distribution table or use a calculator. For a two-tailed test at a 5% significance level and 19 degrees of freedom, the critical chi-squared values are approximately 33.9244 and 8.9065. Since the calculated chi-squared statistic (17.4626) does not fall in the rejection region defined by the critical values (8.9065 to 33.9244), we fail to reject the null hypothesis. There is not enough evidence to conclude that the variance of the surveys carried out by the agents is different from 14 (surveys per week)^2 at a 5% significance level. (d) With a significance level of 1%, it can be concluded that less than 30% of the agents carry out more than 59 surveys per week: To test this hypothesis, we'll perform a one-sample proportion z-test. The null hypothesis (H0) is that the proportion is greater than or equal to 30%, and the alternative hypothesis (H1) is that the proportion is less than 30%. H0: p ≥ 0.30 H1: p < 0.30 We'll calculate the z-statistic and compare it to the critical value at a 1% significance level. First, calculate the sample proportion (p̂) of agents carrying out more than 59 surveys per week: p̂ = (number of agents carrying out more than 59 surveys) / n Count the number of agents carrying out more than 59 surveys: There are 7 agents. p̂ = 7 / 20 = 0.35 Now, calculate the standard error of the proportion: SE = √[(p̂ * (1 - p̂)) / n] SE = √[(0.35 * (1 - 0.35)) / 20] SE ≈ 0.1491 Now, calculate the z-statistic: z = (p̂ - p) / SE z = (0.35 - 0.30) / 0.1491 z ≈ 3.3564 Find the critical z-value for a one-tailed test at a 1% significance level. You can consult a z-table or use a z-distribution calculator. For a one-tailed test at a 1% significance level, the critical z-value is approximately -2.33 (since we are testing if the proportion is less than 30%). Since the calculated z-statistic (3.3564) is greater than the critical z-value (-2.33), we reject the null hypothesis. There is enough evidence to conclude that less than 30% of the agents carry out more than 59 surveys per week at a 1% significance level.