Homework 6

Solution.

The test statistic for the scenario is

since $X$ is normal and the variance is known. The sample test statistic (under $H_0$) is then

1. The $p$-value is the probability of observing $\mean{x} = 35$ or something more extreme (i.e., numbers that give more evidence that $\mu > 32$). In this problem, this is

   $$\boxed{\P\p{Z \geq 9} \approx 0}.$$

2. We reject $H_0$ at $\alpha = 0.05$.

Solution.

1. Since the sample size is small and the variances are unknown but equal, the test statistic is

   $$T = \frac{\mean{X} - \mean{Y} - \p{\mu_X - \mu_Y}}{S_p\sqrt{\frac{1}{n_X} + \frac{1}{n_Y}}} \sim t\p{n_X + n_Y - 2},$$

   where $S_p$ is the pooled sample standard deviation. We have a one-sided test, so we reject if

   $$\boxed{t < -t_\alpha\p{27} = -1.703}.$$

2. Under $H_0$, $\mu_X - \mu_Y = 0$, so the sample test statistic is

   $$t = \frac{72.9 - 81.7}{s_p\sqrt{\frac{1}{13} + \frac{1}{16}}} = -0.8686.$$

   This is not in the critical region, so we do not reject $H_0$ at $\alpha = 0.05$.

Solution.

For parts (1)-(3), since the variances are assumed to be equal, the test statistic is

1. A $95\%$ confidence interval is

   $$\mean{x} - \mean{y} \pm t_{\frac{\alpha}{2}}\p{26} s_p\sqrt{\frac{1}{n_X} + \frac{1}{n_Y}} = \boxed{\br{-30.7909, 2.7909}}$$

2. Since this is a two-sided test, the critical region is

   $$t < -t_{\frac{\alpha}{2}}\p{26} = -2.056 \quad\text{or}\quad t > t_{\frac{\alpha}{2}}\p{26} = 2.056.$$

   In our case, the sample test statistic is

   $$t = -1.7143,$$

   which is not in the critical region. Thus, we do not reject $H_0$ at $\alpha = 0.05$.

3. Since this is a two-sided test, the $p$-value is

   $$\begin{aligned} \P\p{T \leq -\abs{t} \text{ or } T \geq \abs{t}} &= 2\P\p{T \geq \abs{t}} \\ &= 2\p{1 - \P\p{T \leq \abs{t}}} \\ &= 2\p{1 - \P\p{T \leq 1.7143}}. \end{aligned}$$

   From the $t$-table in the book,

   $$0.95 < \P\p{T \leq 1.7143} < 0.975,$$

   so the $p$-value satisfies

   $$\boxed{0.05 < p\text{-value} < 0.10}.$$

4. What the problem asks you to do is to treat the samples as differences, e.g., $d_1 = x_1 - y_1 = 49 - 133$, etc. This gives us the following sample:

   $$\begin{array}{r|rrrrrrrrrrrrrr} d & -84 & 0 & 17 & -37 & -26 & 16 & 28 & -17 & 9 -101 & -15 & -15 & 23 & 6 \end{array}$$

   We need to test the hypotheses

   $$\begin{aligned} H_0 &\colon \mu_D = 0 \\ H_1 &\colon \mu_D \neq 0. \end{aligned}$$

   In this case, we have $n = 14$ differences and the test statistic is now

   $$T = \frac{\mean{D} - \p{\mu_X - \mu_Y}}{S_D/\sqrt{n}} \sim t\p{13}.$$

   A $95\%$ confidence interval is

   $$\mean{d} \pm t_{\frac{\alpha}{2}}\p{13} \frac{s_D}{\sqrt{n}} = \boxed{\br{-36.2270, 8.2270}}.$$

   The critical region is

   $$t < -t_{\frac{\alpha}{2}}\p{13} = -2.160 \quad\text{or}\quad t > t_{\frac{\alpha}{2}}\p{13} = 2.160.$$

   The test statistic is

   $$t = -1.3605,$$

   which does not fall into the critical region, so we do not reject $H_0$ at $\alpha = 0.05$. Lastly, the $p$-value is computed like before:

   $$p\text{-value} = 2\p{1 - \P\p{T \leq 1.3605}}.$$

   From the textbook's $t$-table, $0.90 < \P\p{T \leq 1.3605} < 0.95$, so the $p$-value satisfies

   $$\boxed{0.10 < p\text{-value} < 0.20}.$$

Solution.

Since the sample size is small and the variances are assumed to be equal, our test statistic is

Our observed test statistic is

so the $p$-value is

From the $t$-table, $\P\p{T \leq 0.1970} < 0.60$, so

Thus, we do not reject $H_0$ at $\alpha = 0.01$.

Solution.

The significance level $\alpha$ is the probability that we reject $H_0$ given that $H_0$ is true. In other words,

We can use the normal approximation to a binomial distribution since $n = 100$ is large:

Then (with a continuity correction) we get

Alternatively, you can just calculate the binomial sum directly if your calculator is able to compute the binomial coefficients: