Mathematics 9709/61 — May/June 2020

Walkthrough

First add all seven lengths to get the total $\sum x = 34.7$. Dividing by $n=7$ gives the sample mean $\bar{x}=4.9571$, which is an unbiased estimate of the population mean.

For the variance, we cannot simply average squared deviations about the sample mean using divisor $n$, because that would underestimate the population variance. The unbiased estimate uses divisor $n-1$. Using the computational form:

$$s^2 = \frac{\sum x^2 - \frac{(\sum x)^2}{n}}{n-1}$$

We first find $\sum x^2 = 175.15$, then substitute. The result is $0.522857$, which rounds to $0.523$ to 3 significant figures.

Key Takeaways

An unbiased estimate of the population mean is the sample mean. An unbiased estimate of the population variance is obtained by dividing by $n-1$, not $n$. The formula $\sum x^2 - (\sum x)^2/n$ is a convenient way to compute the corrected sum of squares.

Common Mistakes

Dividing by $n$ instead of $n-1$ gives $0.448$ and loses the unbiasedness mark. Forgetting to square each data value when computing $\sum x^2$. Rounding intermediate values too early can change the final 3 sf answer.

Things to Be Careful About

Use unrounded values in the variance calculation. The final answers should be given to 3 significant figures: $4.96$ and $0.523$. The divisor is $6$, not $7$.

Walkthrough

For a normal population, the sample mean $\bar{X}$ is normally distributed with mean $\mu$ and standard error $\sigma/\sqrt{n}$. The mark scheme follows through the sample variance estimate $s^2=0.523$ from part (a), so the standard error is $\sqrt{0.523/7} = 0.2733$.

A 95% confidence interval uses the two-tailed critical value $z=1.96$. The interval is $\bar{x} \pm 1.96 \times \text{SE}$. Substituting $\bar{x}=4.9571$ gives lower limit $4.4214$ and upper limit $5.4928$, so the interval is $4.42$ to $5.49$ to 3 sf.

Key Takeaways

For a normal population, the sample mean is normally distributed. A 95% confidence interval for the population mean is formed by adding and subtracting $1.96$ standard errors from the sample mean.

Common Mistakes

Using the wrong critical value (e.g. $1.645$ for 90% or $2.576$ for 99%). Forgetting to divide the variance by $n$ before taking the square root. Using the sample standard deviation instead of the standard error.

Things to Be Careful About

The mark scheme uses the follow-through sample variance $s^2=0.523$ from part (a) in the standard error, not the stated true variance $0.55$. If the given true variance $0.55$ were used instead, the interval would be $4.41$ to $5.51$; however the expected mark-scheme answer is $4.42$ to $5.49$. Always use unrounded values for the mean and standard error until the final rounding.