A company is considering offering a retirement plan to its employees. Since a pension contribution has to be deducted both from the employer and the employee for the employee to be eligible, the company requires a sufficiently large number of employees to be interested for the plan to be cost-effective. In a random sample of 300 employees, 192 indicated interest in the plan (i.e. 64%).
(i) Obtain a 95% confidence interval for the proportion of all employees who would participate in the retirement plan.
(ii) Younger employees (below 35 years of age) were less interested in the retirement plan. A recent company report claimed average annual income of all employees was 48,000 Euro, with a standard deviation of 8,500 Euro. A random sample of 30 younger employees from the total of 650 employees below 35 years of age had mean income of 43,900 Euro. If expectation is that all employees should have the same average and standard deviation of income, is there evidence that younger employee income is significantly less than 48,000 Euro? Test this one-sided hypothesis at the 0.01 (1%) level of significance.
(iii) Suppose that the company wished to explicitly test the hypothesis that the average income of employees below and above 35 years of age differed significantly. A random sample of incomes of 10 employees from each of two groups was taken. Means and variances obtained were as follows, (values in thousands of Euro for convenience):
Test the hypothesis that mean incomes differ at the 0.05 (5%) level of significance. Why is the t-distribution used here?
I'm having trouble with the third part understanding why SE = Square root of 1.294/10 + 3.899/10 = 0.7206 is used as in my notes we haven't used this statistics formula. Can anyone explain what this formula is called and is there another way of answering this part of the question and why t-distribution is being used here instead of another distribution. Thank You