Question
(a) A random sample of 8 boxes of cereal from a certain supplier was taken. Each box was weighed and the masses in grams were as follows.
261 249 259 252 255 256 258 254
Find unbiased estimates of the population mean and variance.
(b) The supplier claims that the mean mass of boxes of cereal is 253 g. A quality control officer suspects that the mean mass is actually more than 253 g. In order to test this claim, he weighs a random sample of 100 boxes of cereal and finds that the total mass is 25 360 g.
(i) Given that the population standard deviation of the masses is 3.5 g, test at the 5% significance level whether the population mean mass is more than 253 g.
An employee says, “This test is invalid because it uses the normal distribution, but we do not know whether the masses of the boxes are normally distributed.”
(ii) Explain briefly whether this statement is true or not.
▶️Answer/Explanation
Solution: –
(a) Unbiased Estimates of Mean and Variance
Given data:
- Estimate of Population Mean:
- Estimate of Population Variance (unbiased):
Final Answers:
(b) Hypothesis Test
(i) Test at 5% significance level
Given Data:
- Sample size:
- Sample mean:
- Population standard deviation:
- Claimed mean:
- Significance level:
Hypotheses:
Test Statistic:
Critical Value at
for one-tailed test:
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