# Hypothesis tests two means

For a random person selected from a population, let $$X$$ be their Instagram followers and $$Y$$ be their Facebook followers. We randomly sample the followers of $$n$$ people, $$(X_1, Y_1), ..., (X_n, Y_n)$$. Let $$D = X - Y$$ and $$D_i = X_i - Y_i$$ for $$i = 1, ..., n$$. Suppose we can accurately model $$D$$ as a normal random variable, but whose mean $$\mu_D$$ and variance $$\sigma^2_D$$ are unknown.

(a) We take $$n = 7$$ samples:

$$X_i: 118, 136, 125, 121, 111, 142, 114$$

$$Y_i: 81, 91, 63, 78, 59, 93, 83$$

Determine a $$95$$% confidence interval for $$\mu_D$$.

(b) Recent research suggests that $$\mu_D = 40$$. We claim that the average difference between Instagram and Facebook followers has increased, and we take $$n = 17$$ random samples. Use the test statistic $$\frac{\bar{D} - 40}{\frac{S_D}{\sqrt{17}}}$$ to form a test at the $$0.05$$ significance level.

I tried to do this

(a) I did a few calculations and found the sample statistics as:

$$\bar{x} = 123.86, s_x = 11.42, \bar{y} = 78.29, s_y = 13.00$$

By the Welch statistic, we find the degrees of freedom as

$$\Delta = \frac{(\frac{s_x^2}{n_x} + \frac{s_y^2}{n_y})^2}{\frac{1}{n_x - 1}(\frac{s_x^2}{n_x})^2 + \frac{1}{n_y - 1}(\frac{s_y^2}{n_y})^2}$$

Plugging in the values gives $$\Delta = 11$$.

The value of $$t_{\alpha /2}$$ at $$0.025$$ with $$11$$ degrees of freedom, from the tables, is $$2.01$$.

The interval is therefore

$$((\bar{x} - \bar{y}) - t_{\alpha / 2}\cdot \sqrt{s_x^2/n_x + s_y^2/n_y}, (\bar{x} - \bar{y}) + t_{\alpha / 2}\cdot \sqrt{s_x^2/n_x + s_y^2/n_y})$$

$$= (34.6, 56.6)$$

Another way:

$$\mu_D = 45.57, s_D = 10.13$$.

$$t_{\alpha/2}$$ with $$\alpha = 0.05$$ and $$6$$ degrees of freedom is $$2.447$$.

The interval is:

$$(\bar{d} - t_{\alpha/2} \cdot \frac{s_D}{\sqrt{n}}, \bar{d} - t_{\alpha/2} \cdot \frac{s_D}{\sqrt{n}})$$

$$= (36.20, 54.94)$$

Both methods give similar values but I am not sure which is correct.

(b) $$H_0: \mu_D = 40$$ and $$H_1: \mu_D > 40$$.

$$t_{\alpha}$$ with $$\alpha = 0.05$$ and $$16$$ degrees of freedom is about $$1.75$$.

The critical region is

$$\frac{\bar{d} - 40}{\frac{s_d}{\sqrt{17}}} \geq 1.75$$

We reject the null hypothesis if the test statistic is greater than or equal to $$1.75$$.

Is what I have done correct? For (a), which method is supposed to be the right one?

b looks good to me. For a, the second method is correct. We want to treat each observation as one entity, i.e. $$d_i=x_i-y_i$$ and then conduct the confidence interval treating d as a single normal random variable with unknown mean and unknown variance. Also called a paired analysis. Option 1 would be comparing the means of two different populations.