Inference using normal and t-distributions — practice questions

Let $\mu$ be the population mean volume of coffee dispensed.
1. State Hypotheses:
$H_0: \mu = 200$
$H_1: \mu < 200$
This is a one-tailed test.\

2. Choose Distribution and Significance Level:
Population variance $\sigma^2$ is unknown. Sample size $n=8$ is small. The population is stated to be normal. Therefore, we use a t-distribution.
Significance level, $\alpha = 0.05$.
Degrees of freedom, $v = n - 1 = 8 - 1 = 7$.\

3. Calculate the Test Statistic:
Sample mean, $\bar{x} = 197$.
Unbiased estimate of population variance, $s^2 = 36$, so $s = \sqrt{36} = 6$.
Test statistic: $T = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{197 - 200}{6/\sqrt{8}}$
$T = \frac{-3}{6/2.828...} = \frac{-3}{2.121...} = -1.414$\

4. Find the Critical Value:
We need the critical value for a one-tailed test at the 5% level with 7 degrees of freedom. From t-distribution tables, $t_{7}(0.05) = 1.895$. Since our test is for $\mu < 200$, the critical region is $T < -1.895$.\

5. Make a Decision and Conclude:
Our calculated test statistic is $T = -1.414$.
Since $-1.414 > -1.895$, the test statistic does not fall in the critical region.
Therefore, we do not reject $H_0$.
There is insufficient evidence at the 5% significance level to suggest that the mean volume of coffee dispensed is less than 200 ml.

Let $\mu$ be the population mean reaction time.
1. Identify Information and Choose Distribution:
Sample mean, $\bar{x} = 0.83$.
Population standard deviation, $\sigma = 0.15$.
Sample size, $n = 50$.
Confidence level = 99%.
Since the population standard deviation $\sigma$ is known, we use the normal distribution.\

2. Find the Critical Value ($z^*$):
For a 99% confidence interval, we have $1 - \alpha = 0.99$, so $\alpha = 0.01$.
We need the z-value that leaves $\alpha/2 = 0.005$ in each tail.
Looking up the value for a cumulative probability of $1 - 0.005 = 0.995$ in the normal distribution table gives $z^* = 2.576$.\

3. Calculate the Margin of Error:
Standard Error (SE) = $\frac{\sigma}{\sqrt{n}} = \frac{0.15}{\sqrt{50}} \approx 0.021213$.
Margin of Error = $z^* \times \text{SE} = 2.576 \times 0.021213 \approx 0.05466$.\

4. Construct the Interval:
CI = $\bar{x} \pm$ Margin of Error
CI = $0.83 \pm 0.05466$
CI = $(0.83 - 0.05466, 0.83 + 0.05466)$
CI = $(0.77534, 0.88466)$\

5. Conclude:
A 99% confidence interval for the mean reaction time is (0.775, 0.885) seconds (to 3 d.p.). We are 99% confident that the true mean reaction time for all drivers lies within this range.