Sampling and estimation — practice questions

We are given $n=8$, $\sum x = 1240$, and $\sum x^2 = 192600$.

1. Unbiased estimate of the population mean ($\mu$): The unbiased estimate for $\mu$ is the sample mean, $\bar{x}$. $\bar{x} = \frac{\sum x}{n} = \frac{1240}{8} = 155$ So, the unbiased estimate of the population mean mass is 155 g.

2. Unbiased estimate of the population variance ($\sigma^2$): The unbiased estimate for $\sigma^2$ is $s^2$. $s^2 = \frac{1}{n-1} \left( \sum x^2 - \frac{(\sum x)^2}{n} \right)$ $s^2 = \frac{1}{8-1} \left( 192600 - \frac{(1240)^2}{8} \right)$ $s^2 = \frac{1}{7} \left( 192600 - \frac{1537600}{8} \right)$ $s^2 = \frac{1}{7} (192600 - 192200)$ $s^2 = \frac{1}{7} (400) \approx 57.1428...$ So, the unbiased estimate of the population variance is 57.1 (to 3 s.f.).

We are given: Sample size, $n = 50$ Sample mean, $\bar{x} = 245$ minutes Population standard deviation, $\sigma = 12$ minutes Confidence level = 95%

1. Find the z-value: For a 95% confidence interval, the z-value is 1.96. This corresponds to the value that leaves 2.5% in each tail of the standard normal distribution.

2. Apply the confidence interval formula: The formula is $\bar{x} \pm z \times \frac{\sigma}{\sqrt{n}}$. CI = $245 \pm 1.96 \times \frac{12}{\sqrt{50}}$

3. Calculate the margin of error: Margin of error = $1.96 \times \frac{12}{\sqrt{50}} \approx 1.96 \times \frac{12}{7.071} \approx 1.96 \times 1.697 \approx 3.326$

4. Calculate the interval: Lower bound = $245 - 3.326 = 241.674$ Upper bound = $245 + 3.326 = 248.326$

5. State the final answer: The 95% confidence interval for the mean charging time is (241.7, 248.3) minutes, to 1 decimal place. This means we are 95% confident that the true mean charging time for all cars of this model lies between 241.7 and 248.3 minutes.