CIE A level Math -Probability & Statistics 2: 6.4 Sampling and estimation : a sample and a population: Exam Style Questions Paper 6

Solution: –

(a) Unbiased Estimates of Mean and Variance

Given data:

$$261, 249, 259, 252, 255, 256, 258, 254$$

1. Estimate of Population Mean:

$$\hat{\mu} = \frac{\sum x_i}{n} = \frac{261 + 249 + 259 + 252 + 255 + 256 + 258 + 254}{8}$$ $$= \frac{2044}{8} = 255.5$$

2. Estimate of Population Variance (unbiased):

$$s^2 = \frac{1}{n-1} \sum (x_i – \hat{\mu})^2$$ $$= \frac{1}{7} [(261 – 255.5)^2 + (249 – 255.5)^2 + (259 – 255.5)^2 + (252 – 255.5)^2 + (255 – 255.5)^2 + (256 – 255.5)^2 + (258 – 255.5)^2 + (254 – 255.5)^2]$$ $$= \frac{1}{7} [30.25 + 42.25 + 12.25 + 12.25 + 0.25 + 0.25 + 6.25 + 2.25]$$ $$= \frac{106}{7} = 15.14$$

Final Answers:

$$\hat{\mu} = \mathbf{255.5}, \quad s^2 = \mathbf{15.14}$$

(b) Hypothesis Test

(i) Test at 5% significance level

Given Data:

• Sample size:
  $n = 100$
• Sample mean:
  $$\bar{x} = \frac{25,360}{100} = 253.6$$
• Population standard deviation:
  $\sigma = 3.5$
• Claimed mean:
  $\mu_0 = 253$
• Significance level:
  $\alpha = 0.05$

Hypotheses:

$$H_0: \mu = 253 \quad \text{(Supplier’s claim)}$$ $$H_1: \mu > 253 \quad \text{(Quality officer’s suspicion)}$$

Test Statistic:

$$Z = \frac{\bar{x} – \mu_0}{\sigma / \sqrt{n}} = \frac{253.6 – 253}{3.5 / \sqrt{100}}$$ $$= \frac{0.6}{3.5 / 10} = \frac{0.6}{0.35} = 1.71$$

Critical Value at

for one-tailed test:

$$Z_{0.05} = 1.645$$

Since

, we reject

at the 5% level.

Conclusion: There is sufficient evidence to suggest that the mean mass is more than 253 g.

(ii) Validity of Normal Approximation

The employee’s concern is incorrect because of the Central Limit Theorem (CLT). Even if the population distribution is unknown or not normal, the sample mean

is approximately normal when the sample size is large (

). Since

is large, the normal approximation is valid.

Final Answer: The statement is false because the sample size is large, so the normal distribution can be used due to the Central Limit Theorem.

——–Markscheme————–

(a) $Est(\mu) = \frac{2044}{8} \qquad [=255.5]$

$Est(\sigma^2) = \frac{8}{7} \times (522348 – “255.5”^2) \quad or \quad \frac{1}{7} \times (522348 – \frac{2044^2}{8})$

= 15.1 (3 sf) or $\frac{106}{7}$

(b) (i) $H_0: \mu = 253$

$H_1: \mu > 253$

$\frac{\frac{25360}{100} – 253}{3.5 \div \sqrt{100}}$

= 1.714

1.714 > 1.645 or 0.0432 < 0.05

[Reject $H_0$] There is sufficient evidence (at 5% level) to suggest

[mean] mass is greater than 253

(b) (ii) Not true