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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):

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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

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  • $\begingroup$ T distribution is very good for small sample sizes, which is what you have here with only 10 employees for part C $\endgroup$ – Klint Qinami Aug 13 '16 at 1:07
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That is just the Standard Error formula. One way to deduce the answer is to use the confidence interval

$$(\bar{x}_1 - \bar{x}_2) - t_{\alpha /2} s_p \sqrt{\frac{1}{n_1^2} + \frac{1}{n^2_2} } < \mu_1 - \mu_2 < (\bar{x}_1 - \bar{x}_2) + t_{\alpha /2} s_p \sqrt{\frac{1}{n_1^2} + \frac{1}{n^2_2} } $$

and check if 0 is within the interval. Here, $s_p = \frac{(n_1 -1)S^2_1 + (n_2 -1)S^2_2}{n_1 + n_2 -2}$ and $v = n_1 + n_2 -2$ degrees of freedom.

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If $X_o$ and $X_y$ are the random incomes of an individual in the older group and younger group, respectively, and we assume that $X_o \sim \operatorname{Normal}(\mu_o, \sigma_o^2)$, and $X_y \sim \operatorname{Normal}(\mu_y, \sigma_y^2)$, then the hypothesis being tested is $$H_0 : \mu_o = \mu_y \quad \text{vs.} \quad H_a : \mu_0 \ne \mu_y.$$ Since the true variances are unknown parameters but not of interest with respect to the desired inference, these are estimated through the sample. From the data, we find that $s_o^2 = 1.294$ is substantially smaller than $s_y^2 = 3.899$, which suggests that a model in which both $X_o$ and $X_y$ are drawn from populations with the same variance may not be appropriate to use.

Consequently, an appropriate test statistic is required. Here, we use the Welch's $t$-test: $$T = \frac{\bar x_o - \bar x_y}{\sqrt{\frac{s_0^2}{n_0} + \frac{s_y^2}{n_y}}} \overset{\cdot}{\sim} t_\nu$$ is approximately Student's $t$-distributed with $$\nu = \frac{\left(\frac{s_o^2}{n_o} + \frac{s_y^2}{n_y}\right)^2}{\frac{s_o^4}{n_o^2 (n_o-1)} + \frac{s_y^4}{n_y^2 (n_y - 1)}}$$ degrees of freedom, and where $n_o = n_y = 10$ are the sample sizes of each group. The denominator of $T$ is indeed the (sample) standard deviation of the mean difference $\bar x_o - \bar x_y$ and so is a true standard error of the mean difference; but it is not a pooled variance because we are not assuming that the samples from $X_o$ and $X_y$ are drawn from distributions with equal variance. In such a case, assuming independence of observations within a sample and between samples, we easily calculate $$\begin{align*}\operatorname{Var}[\bar X_o - \bar X_y] &\overset{\text{ind}}{=} \operatorname{Var}[\bar X_o] + \operatorname{Var}[\bar X_y] \\ &= \operatorname{Var}[(X_{o1} + X_{o2} + \cdots + X_{on_o})/n_o] + \operatorname{Var}[(X_{y1} + X_{y2} + \cdots + X_{yn_y})/n_y] \\ &\overset{\text{ind}}{=} \frac{n_o\operatorname{Var}[X_o]}{n_o^2} + \frac{n_y \operatorname{Var}[X_y]}{n_y^2} \\ &= \frac{\sigma_o^2}{n_o} + \frac{\sigma_y^2}{n_y}, \end{align*}$$ and substitution of the sample variances results in the claimed relationship.

It is important to note that the distribution of the aforementioned test statistic is only approximately $t$-distributed. However, in practice, the Satterthwaite approximation of $\nu$ works very well, and with it, one can reliably compute $p$-values and make the desired inferences regarding $\mu_o - \mu_y$.

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