Mathematics 9709/61 — October/November 2024

Walkthrough

We are given a sample of 150 deer with mean height $1.42\text{ m}$ and known population standard deviation $0.35\text{ m}$. Because the sample size is large and $\sigma$ is known, the sample mean is approximately normally distributed. A 96% confidence interval leaves 2% in each tail of the normal distribution, so the correct $z$-value is about 2.054. The standard error of the sample mean is $\sigma/\sqrt{n}$, and the confidence interval is the sample mean plus or minus $z$ times this standard error. This gives $1.36 < \mu < 1.48$.

Key Takeaways

• A confidence interval for a population mean uses the sample mean and the standard error.
• The $z$-value is determined by the confidence level and the two tails.
• The interval is a statement about the population mean, not about individual data values.

Common Mistakes

• Using $z = 1.96$ (the 95% value) instead of $z = 2.054$ for 96%.
• Forgetting to divide by $\sqrt{n}$ when computing the standard error.
• Giving only the margin of error instead of the full interval.

Things to Be Careful About

• The tail probability is $0.02$, not $0.04$, because the 4% error is split between the two tails.
• The final interval should be rounded to 3 significant figures: $1.36 < \mu < 1.48$.
• Include the units (m) where appropriate.

Walkthrough

Bubay claims that 96% of deer heights are inside the confidence interval. However, the interval was constructed for the population mean, not for individual observations. A 96% confidence interval means that if we repeatedly took samples and calculated intervals, about 96% of those intervals would contain the true population mean. Individual deer heights have a standard deviation of $0.35\text{ m}$, so they vary much more than the narrow interval $1.36$ to $1.48$ m. Therefore Bubay is not correct.

Key Takeaways

• A confidence interval estimates a population parameter, such as the mean.
• It does not describe the distribution of individual data values.

Common Mistakes

• Saying that 96% of deer heights lie in the interval.
• Not mentioning that the interval is about the population mean.

Things to Be Careful About

• The explanation must state both that Bubay is wrong and that the interval is for the population mean.
• Do not confuse the confidence level with a probability for individual observations.

Walkthrough

We are given only summary statistics from a sample of 75 students: the total of all times, $\sum t = 230$, and the total of their squares, $\sum t^2 = 930$. The best single estimate of the population mean is simply the sample mean, so we divide the total by the sample size.

For the population variance, we must use the unbiased estimator. Because the sample mean is itself estimated from the data, dividing by $n$ would tend to underestimate the true variance. The correct divisor is $n-1$, giving

This formula is equivalent to the more familiar $s^2 = \dfrac{1}{n-1}\sum (t-\bar{t})^2$, but it is written in summary form so that only $\sum t$ and $\sum t^2$ are needed.

Key Takeaways

• The sample mean $\bar{t}$ is an unbiased estimate of the population mean $\mu$.
• The unbiased estimate of the population variance uses divisor $n-1$, not $n$.
• Given summary statistics, use $$s^2 = \frac{1}{n-1}\left(\sum t^2 - \frac{(\sum t)^2}{n}\right).$$

Common Mistakes

• Using $n$ instead of $n-1$ in the variance formula.
• Forgetting to subtract $\frac{(\sum t)^2}{n}$ before dividing by $n-1$.
• Rounding intermediate values too early, which can change the final 3-significant-figure answer.

Things to Be Careful About

• Keep the exact value $\frac{230}{75}$ and $\frac{337}{111}$ for later calculations, even though the rounded answers are $3.07$ and $3.04$.
• The variance estimate is $s^2$, not $s$; the units are minutes squared.

Walkthrough

We now assume that the estimates from part (a) are the true population mean and variance. Since the sample size is 75, which is large, the Central Limit Theorem tells us that the sample mean $\bar{T}$ is approximately normally distributed with mean $\mu$ and variance $\sigma^2/n$.

We want the value $a$ such that the probability that the sample mean exceeds $a$ is $0.234$. This is an upper-tail probability. We first convert $\bar{T}$ to a standard normal variable:

The upper-tail probability $0.234$ corresponds to a $z$-value. Since $1 - 0.234 = 0.766$, we look up $\Phi^{-1}(0.766)$, which is $0.726$. This is the number of standard errors above the mean that $a$ must be.

Then we set

and solve for $a$. We use the variance from part (a) divided by $n$ before taking the square root, because the standard error of the sample mean is $\sigma/\sqrt{n}$.

Key Takeaways

• The Central Limit Theorem allows us to treat the sample mean as approximately normal when $n$ is large.
• The sample mean has mean $\mu$ and variance $\sigma^2/n$, so its standard deviation is $\sigma/\sqrt{n}$.
• To find a threshold from a tail probability, use the inverse normal distribution to get the corresponding $z$-value.

Common Mistakes

• Using the probability $0.766$ directly as the $z$-value. The mark scheme states that using $0.766$ as the $z$-value is incorrect; you must use $\Phi^{-1}(0.766) = 0.726$.
• Forgetting to divide the variance by $75$ before taking the square root.
• Using $\sigma^2$ instead of $\sigma$ in the standardisation formula.
• Making a sign error: since the tail is an upper tail, $a$ is above the mean, so the $z$-value is positive.

Things to Be Careful About

• Use the exact or sufficiently accurate values from part (a) to avoid rounding errors.
• The standard error of the sample mean is $\sqrt{\sigma^2/n}$, not $\sigma^2/n$.
• The final answer should be given to 3 significant figures, as requested by the mark scheme.

Walkthrough

The problem asks us to find the constant $a$ in the probability density function. The key property of any PDF is that the total area under the curve equals 1. Since $f(x)$ is zero outside $2 \le x \le 3$, we integrate over this interval and set the result equal to 1.

We integrate term by term. The antiderivative of $\frac{a}{x^2}$ is $-\frac{a}{x}$, because the derivative of $-\frac{a}{x}$ is $\frac{a}{x^2}$. The antiderivative of $-\frac{18}{x^3}$ is $\frac{9}{x^2}$, because the derivative of $\frac{9}{x^2}$ is $-\frac{18}{x^3}$.

Evaluating at the limits: at $x = 3$, the antiderivative equals $-\frac{a}{3} + 1$; at $x = 2$, it equals $-\frac{a}{2} + \frac{9}{4}$. Subtracting the lower value from the upper value gives $\frac{a}{6} - \frac{5}{4}$. Setting this equal to 1 and solving gives $a = \frac{27}{2}$.

Key Takeaways

The defining property of a probability density function is that the total area under it over its entire support equals 1. This is the standard method for finding an unknown constant in a PDF. The problem also reinforces integration of negative powers of $x$.

Common Mistakes

• Forgetting to set the integral equal to 1.
• Sign errors when subtracting the lower-limit value. Since $\left(-\frac{a}{2} + \frac{9}{4}\right)$ is subtracted, it becomes $+\frac{a}{2} - \frac{9}{4}$.
• Incorrectly integrating $-\frac{18}{x^3}$ — the antiderivative is $\frac{9}{x^2}$, not $-\frac{9}{x^2}$.

Things to Be Careful About

The mark scheme requires showing the correct substitution of both limits. An unsupported answer for $a$ would not receive full marks.

Walkthrough

The expected value of a continuous random variable is $\text{E}(X) = \int x f(x)\,dx$ over the support. We first multiply $f(x)$ by $x$. With $a = \frac{27}{2}$, we have $x f(x) = \frac{27}{2x} - \frac{18}{x^2}$.

We integrate from 2 to 3. The antiderivative of $\frac{27}{2x}$ is $\frac{27}{2}\ln x$, using $\int \frac{1}{x}\,dx = \ln x$. The antiderivative of $-\frac{18}{x^2}$ is $\frac{18}{x}$, since the derivative of $\frac{18}{x}$ is $-\frac{18}{x^2}$.

Evaluating at the limits: at $x = 3$, we get $\frac{27}{2}\ln 3 + 6$; at $x = 2$, we get $\frac{27}{2}\ln 2 + 9$. Subtracting gives $\frac{27}{2}(\ln 3 - \ln 2) - 3$. Using the log law $\ln 3 - \ln 2 = \ln\frac{3}{2}$, we obtain $\frac{27}{2}\ln\frac{3}{2} - 3$.

Key Takeaways

The formula for the expected value of a continuous variable is $\text{E}(X) = \int x f(x)\,dx$. This problem also reinforces the integral $\int \frac{1}{x}\,dx = \ln x$ and the log subtraction law $\ln a - \ln b = \ln\frac{a}{b}$.

Common Mistakes

• Forgetting to multiply $f(x)$ by $x$ before integrating.
• Misapplying the log law: $\ln 3 - \ln 2 = \ln\frac{3}{2}$, not $\ln(3-2)$.
• Arithmetic errors when substituting the limits, particularly with the $\frac{18}{x}$ term.

Things to Be Careful About

The coefficient $\frac{27}{2}$ multiplies both logarithms, so it stays attached throughout. The mark scheme requires showing the correct substitution of both limits for full marks.

Walkthrough

The question asks whether the true mean is different from the magazine's stated value of $12.7$ cm. A one-tailed test would be used if the scientist expected the mean to be specifically greater than $12.7$ or specifically less than $12.7$. Since the claim is simply that the mean is different, the test must detect deviations in both directions, so it should be two-tailed.

Key Takeaways

A two-tailed test is used when the alternative hypothesis is of the form $\mu \neq \mu_0$. The choice is made from the wording of the problem, not from the observed sample mean.

Common Mistakes

Choosing a one-tailed test just because the sample mean, $11.942$, is below $12.7$. The direction of the test must be based on the hypothesis being tested, not on the sample result.

Things to Be Careful About

The phrase "different from" means "not equal to", so the test must be two-tailed. If the problem said "greater than" or "less than", then a one-tailed test would be appropriate.

Walkthrough

The scientist is testing whether the true mean worm length is different from $12.7$ cm, so the hypotheses are:

The sample mean is found by dividing the sum of the lengths by the sample size:

Since the population standard deviation is known, $\sigma = 2.3$, and the worm lengths are normally distributed, the sample mean is normally distributed with standard error

Standardising the sample mean under the null hypothesis gives the test statistic

At the 1% significance level, a two-tailed test has $0.005$ in each tail, so the critical values are $\pm 2.576$. Since $|z| = 2.330 < 2.576$, the test statistic is not in the critical region. Equivalently, the one-tailed probability is $0.00989$, which is greater than $0.005$, so the result is not significant. Therefore $H_0$ is not rejected.

In context, there is insufficient evidence to conclude that the mean worm length is different from $12.7$ cm.

Key Takeaways

A hypothesis test for a population mean when the population is normal and the standard deviation is known uses the $z$-statistic

The word "different" means the test is two-tailed, so the significance level must be split equally between the two tails. A conclusion in a hypothesis test should be worded as "insufficient evidence" rather than as a definite statement that $H_0$ is true.

Common Mistakes

• Using a one-tailed test simply because the sample mean is below $12.7$. The direction must be decided from the hypothesis before looking at the data.
• Comparing $z = -2.330$ with the one-tailed critical value $2.326$ instead of the two-tailed critical value $2.576$.
• Forgetting to divide $\sigma$ by $\sqrt{n}$ when computing the standard error.
• Using the sample standard deviation instead of the given population standard deviation.
• Stating that "the mean is 12.7" as a conclusion; this is too definite and loses the final mark.

Things to Be Careful About

• Both hypotheses must be stated explicitly. A correct numerical comparison without hypotheses would lose the first $B1$ mark.
• At the 1% level, a two-tailed test has $0.005$ in each tail. If using a one-tailed $p$-value, compare it with $0.005$, or double it and compare with $0.01$. The mark scheme accepts $0.00989 > 0.005$.
• The critical value $z^* = 2.576$ may be quoted to different precision, but it is usually accepted in the range $2.574$ to $2.579$.
• The conclusion must be in context and should not contradict the fact that $H_0$ was not rejected.

Walkthrough

Desk $A$ has a Poisson rate of 1.8 customers per 10 minutes. Since the required interval is 15 minutes, the mean number of arrivals is scaled by $15/10$, giving $\lambda = 2.7$.

The event "more than 2 customers" is $X \geq 3$. It is easier to use the complement:

For a Poisson distribution, $P(X = r) = e^{-\lambda}\frac{\lambda^r}{r!}$. Therefore $P(X \leq 2)$ is the sum of the terms for $r = 0, 1, 2$. Substituting $\lambda = 2.7$ and subtracting from 1 gives 0.506.

Key Takeaways

• The mean of a Poisson distribution scales directly with the length of the interval.
• For a "more than" probability, use the complement to avoid adding infinitely many terms.
• The Poisson formula $P(X = r) = e^{-\lambda}\frac{\lambda^r}{r!}$ is the core tool.

Common Mistakes

• Using $\lambda = 1.8$ for the 15-minute period instead of scaling to 2.7.
• Writing $P(X > 2)$ as $P(X = 3)$ instead of $1 - P(X \leq 2)$.
• Giving only the unsupported answer 0.506; the mark scheme requires the expression to be shown.

Things to Be Careful About

• "More than 2" means $X \geq 3$, not $X \geq 2$.
• Use the exact expression before rounding; premature rounding can change the final 3 significant figures.
• The final answer should be given to 3 significant figures.

Walkthrough

The rates are given per 10 minutes. In a 5-minute period, each rate is halved:

Because the two arrival processes are independent, the total number of arrivals in the same fixed interval is also Poisson, with mean equal to the sum of the means:

"Less than 4" means the total can be 0, 1, 2, or 3. Using the Poisson probability formula for each value and adding them gives the required probability.

Key Takeaways

• The sum of independent Poisson random variables is Poisson.
• When changing the time interval, the mean rate is multiplied by the fraction of the base interval used.
• For a "less than" event, include all integer values from 0 up to the stated boundary minus 1.

Common Mistakes

• Using the original 10-minute rates 1.8 and 2.1 instead of scaling to 5 minutes.
• Adding the rates before scaling: the correct total mean is $0.9 + 1.05 = 1.95$, not $1.8 + 2.1 = 3.9$.
• Including $P(T = 4)$ when asked for $T < 4$.
• Giving only the final value without showing the Poisson expression; the mark scheme requires the expression to be seen.

Things to Be Careful About

• "Less than 4" excludes 4, so include only $T = 0, 1, 2, 3$.
• Use the scaled, not original, rates.
• Keep enough decimal places in the intermediate sum before multiplying by $e^{-1.95}$.
• The final answer should be given to 3 significant figures.

Walkthrough

The rate at desk $B$ is 2.1 customers per 10 minutes. If $t$ is the waiting time in minutes, the mean number of customers in that period is

The probability of seeing at least one customer is one minus the probability of seeing none:

We want this to be at least 0.90, so set up the inequality:

Rearrange to isolate the exponential:

Taking natural logarithms:

Dividing by $-0.21$ reverses the inequality:

Because the inspector must wait a whole number of minutes and 10 minutes gives a probability below 0.90, the minimum whole number of minutes is 11.

Key Takeaways

• A Poisson rate can be scaled to any time interval by multiplying by the interval length.
• "At least one" is often best handled by the complement $1 - P(0)$.
• Solving $e^{-k t} \leq p$ requires taking logs and carefully reversing the inequality when dividing by a negative number.
• The final time must be a whole number of minutes, so round up when necessary.

Common Mistakes

• Using $\lambda = 2.1$ for every waiting time without scaling by the number of minutes.
• Forgetting that the probability of at least one is $1 - e^{-\lambda}$, not $e^{-\lambda}$.
• Solving for $\lambda$ and then forgetting to convert back to minutes.
• Rounding 10.96 down to 10; 10 minutes gives a probability below 0.90, so the minimum is 11.
• Giving an unsupported answer; the mark scheme expects the inequality and log step to be shown.

Things to Be Careful About

• The rate is per 10 minutes, so the per-minute rate is $0.21$.
• When taking logs, $\ln(0.10)$ is negative, so the inequality direction must be handled correctly.
• The answer must be given as a whole number of minutes; the minimum time is 11 minutes.
• If using $x$ as the number of 10-minute intervals, then $\lambda = 2.1x$ and the final time is $10.96$ minutes, still 11 minutes.

Walkthrough

We want the probability of a Type I error, which is the probability of rejecting $H_0$ when $H_0$ is actually true. The council observes accidents over two randomly chosen years, so the total $X$ follows a Poisson distribution with parameter $2 \times 3.3 = 6.6$ under $H_0$. The test is one-tailed because the council wants to check whether $\lambda$ has decreased, so we reject $H_0$ only for small values of $X$. To find the critical region at the 5% level, we compute $P(X \leq 2)$ and $P(X \leq 3)$. Since $P(X \leq 2) = 0.0400 < 0.05$ and $P(X \leq 3) = 0.105 > 0.05$, the critical region is $X \leq 2$. The Type I error probability is exactly the probability of this critical region under $H_0$, which is $0.0400$.

Key Takeaways

A Type I error is rejecting a true null hypothesis, and its probability equals the probability of the critical region under $H_0$. To find the critical region for a Poisson test, compute cumulative probabilities until the probability exceeds the significance level.

Common Mistakes

Forgetting to double $\lambda$ to 6.6 for the two-year period. Only computing $P(X \leq 2)$ without checking $P(X \leq 3)$ to confirm the critical region. Using $\lambda = 3.3$ instead of 6.6.

Things to Be Careful About

The critical region is the largest set of small values with total probability at most 0.05. The Type I error is the probability of that region under $H_0$. Give the answer to 3 significant figures.

Walkthrough

We carry out the hypothesis test. The null hypothesis is that the mean number of accidents per two-year period is still 6.6, and the alternative is that it has decreased. From part (a) we already know $P(X \leq 2) = 0.0400$. Since $0.0400 < 0.05$, the observed value $X = 2$ falls in the critical region, so we reject $H_0$. The conclusion must be in context: there is evidence that the mean number of accidents has decreased.

Key Takeaways

A hypothesis test for a Poisson mean involves comparing the probability of the observed outcome (or more extreme) with the significance level. If this probability is less than the significance level, reject the null hypothesis.

Common Mistakes

Not stating the hypotheses. Concluding that the mean "has definitely decreased" instead of "there is evidence that it has decreased". Using a two-tailed test when the question specifies a decrease.

Things to Be Careful About

Use $P(X \leq 2)$, not $P(X = 2)$. The conclusion must be in context and not overstate certainty.

Walkthrough

A Type II error is failing to reject $H_0$ when $H_1$ is true. The critical region from part (a) is $X \leq 2$, so we fail to reject $H_0$ when $X > 2$. Under the true value $\lambda = 0.6$, the two-year total is $X \sim \text{Po}(1.2)$. The Type II error probability is $P(X > 2) = 1 - P(X \leq 2)$. Compute $P(X \leq 2) = e^{-1.2}(1 + 1.2 + 1.2^2/2) = 0.8795$, so $P(X > 2) = 1 - 0.8795 = 0.1205 \approx 0.121$.