Question 10[Maximum mark: 6]
A manufacturer of chocolates produces them in individual packets, claiming to have an average of 85 chocolates per packet.
Talha bought 30 of these packets in order to check the manufacturer’s claim.
Given that the number of individual chocolates is x , Talha found that, from his 30 packets,
∑x = 2506 and ∑x2 = 209 738 .
a. Find an unbiased estimate for the mean number (μ) of chocolates per packet. [1]
b. Use the formula \(s_{n-1}^{2}= \frac{\sum x^{2}-\frac{(\sum x)^{2}}{n}}{n-1}\)
to determine an unbiased estimate for the variance of the number of chocolates per packet. [2]
c. Find a 95 % confidence interval for μ . You may assume that all conditions for a
confidence interval have been met. [2]
d. Suggest, with justification, a valid conclusion that Talha could make. [1]
▶️Answer/Explanation
(a) \(\bar{X}= \frac{\sum x}{n}= \frac{2506}{30}83.5 (83.5333.. )\)
(b) \(s^{2}_{n-1}= \frac{\sum x^{2}-\frac{(\sum x)^{2}}{n}}{n-1}= \frac{209738-\frac{2506^{2}}{30}}{29}= 13.9(13.9126..)\)
(c) (82.1,84.9)(82.1405…, 84.9261…)
(d) 85 is outside the confidence interval and therefore Talha would suggest that the manufacturer’s claim is incorrect
Question
At an early stage in analysing the marks scored by candidates in an examination paper, the examining board takes a random sample of 250 candidates and finds that the marks, \(x\) , of these candidates give \(\sum {x = 10985} \) and \(\sum {{x^2} = 598736} \).
a.Calculate a 90% confidence interval for the population mean mark μ for this paper.[4]
b.The null hypothesis μ = 46.5 is tested against the alternative hypothesis μ < 46.5 at the λ% significance level. Determine the set of values of λ for which the null hypothesis is rejected in favour of the alternative hypothesis.[4]
▶️Answer/Explanation
Markscheme
\(\bar x = 43.94\) (A1)
unbiased variance estimate = 466.0847 (A1)
Note: Accept sample variance = 464.2204.
⇒ 90% confidence interval is (41.7,46.2) A1A1
[4 marks]
Z-value is −1.87489 or −1.87866 (A1)
probability is 0.0304 or 0.0301 (A1)
⇒ λ ≥ 3.01 (M1)A1
[4 marks]
Question
A random sample \({X_1},{\text{ }}{X_2},{\text{ }} \ldots ,{\text{ }}{X_n}\) is taken from the normal distribution \({\text{N}}(\mu ,{\text{ }}{\sigma ^2})\), where the value of \(\mu \) is unknown but the value of \({\sigma ^2}\) is known. The mean of the sample is denoted by \(\bar X\).
A mathematics teacher, wishing to apply the above result, generates some artificial data, assumes a value for the variance and calculates the following 95% confidence interval for \(\mu \),
\[[3.12,{\text{ }}3.25].\]
The teacher asks Alun to interpret this result. Alun makes the following statement. “The value of \(\mu \) lies in the interval \([3.12,{\text{ }}3.25]\) with probability 0.95.”
a.i.State the distribution of \(\frac{{\bar X – \mu }}{{\frac{\sigma }{{\sqrt n }}}}\).[1]
a.ii.Hence show that, with probability 0.95,
\[\bar X – 1.96\frac{\sigma }{{\sqrt n }} \leqslant \mu \leqslant \bar X + 1.96\frac{\sigma }{{\sqrt n }}.\][4]
b.i.Explain briefly why this is an incorrect statement.[1]
b.ii.Give a correct interpretation.[1]
▶️Answer/Explanation
Markscheme
\(\frac{{\bar X – \mu }}{{\frac{\sigma }{{\sqrt n }}}}\) is \({\text{N}}(0,{\text{ }}1)\) or it has the Z-distribution A1
attempt to make a probability statement R1
therefore with probability 0.95,
\( – 1.96 \leqslant \frac{{\bar X – \mu }}{{\frac{\sigma }{{\sqrt n }}}} \leqslant 1.96\) A1
\( – 1.96\frac{\sigma }{{\sqrt n }} \leqslant \bar X – \mu \leqslant 1.96\frac{\sigma }{{\sqrt n }}\) A1
\(1.96\frac{\sigma }{{\sqrt n }} \geqslant \mu – \bar X \geqslant – 1.96\frac{\sigma }{{\sqrt n }}\) A1
\(\bar X + 1.96\frac{\sigma }{{\sqrt n }} \geqslant \mu \geqslant \bar X – 1.96\frac{\sigma }{{\sqrt n }}\)
Note: Award the final A1 for either of the above two lines.
\(\bar X – 1.96\frac{\sigma }{{\sqrt n }} \leqslant \mu \leqslant \bar X + 1.96\frac{\sigma }{{\sqrt n }}\) AG
Question
Let \({X_k}\) be independent normal random variables, where \({\rm{E}}({X_k}) = \mu \) and \(Var({X_k}) = \sqrt k \) , for \(k = 1,2, \ldots \) .
The random variable \(Y\) is defined by \(Y = \sum\limits_{k = 1}^6 {\frac{{{{( – 1)}^{k + 1}}}}{{\sqrt k }}} {X_k}\) .
a.(i) Find \({\rm{E}}(Y)\) in the form \(p\mu \) , where \(p \in \mathbb{R}\) .
(ii) Find \(k\) if \({\rm{Var}}({X_k}) < {\rm{Var}}(Y) < {\rm{Var}}({X_{k + 1}})\) .[5]
b.A random sample of \(n\) values of \(Y\) was found to have a mean of \(8.76\).
(i) Given that \(n = 10\) , determine a \(95\%\) confidence interval for \(\mu \) .
(ii) The width of the confidence interval needs to be halved. Find the appropriate value of \(n\) .[6]
▶️Answer/Explanation
Markscheme
(i) \({\rm{E}}(Y) = \frac{1}{{\sqrt 1 }}\mu – \frac{1}{{\sqrt 2 }}\mu + \frac{1}{{\sqrt 3 }}\mu – \frac{1}{{\sqrt 4 }}\mu + \frac{1}{{\sqrt 5 }}\mu – \frac{1}{{\sqrt 6 }}\mu \) (M1)
\( = 0.409\) \((209)\mu \) A1
Note: Accept answers which round to \(0.41\).
(ii) \(Var(Y) = \frac{1}{{\sqrt 1 }} + \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 3 }} + \frac{1}{{\sqrt 4 }} + \frac{1}{{\sqrt 5 }} + \frac{1}{{\sqrt 6 }}\) (M1)
\( = 3.64\) \((3.6399 \ldots )\) A1
\(({\rm{Var}}({X_{13}}) = 3.61;{\rm{Var}}({X_{14}}) = 3.74) \Rightarrow k = 13\) A1
[5 marks]
(i) \(95\%\) CI for is \({\rm{E}}(Y)\)
\(8.76 \pm 1.96\sqrt {\frac{{3.6399 \ldots }}{{10}}} \) (M1)
\( = \left[ {7.58,9.94} \right]\) A1A1
Note: Accept \(\left[ {7.6,9.9} \right]\) . Do not penalize answers given to more than 3sf.
Since \(\mu = \frac{{{\rm{E}}(Y)}}{{0.409 \ldots }}\) , CI for \(\mu \) is \(\left[ {18.5,24.3} \right]\) A1
Note: Do not penalize answers given to more than 3sf.
(ii) width of a CI is inversely proportional to the square root of \(n\) (M1)
so \(n = 40\) A1
[6 marks]
Question
Bill buys two biased coins from a toy shop.
a.The shopkeeper claims that when one of the coins is tossed, the probability of obtaining a head is \(0.6\). Bill wishes to test this claim by tossing the coin \(250\) times and counting the number of heads obtained.
(i) State suitable hypotheses for this test.
(ii) He obtains \(140\) heads. Find the \(p\)-value of this result and determine whether or not it supports the shopkeeper’s claim at the \(5\%\) level of significance.[6]
b.Bill tosses the other coin a large number of times and counts the number of heads obtained. He correctly calculates a \(95\%\) confidence interval for the probability that when this coin is tossed, a head is obtained. This is calculated as [\(0.35199\), \(0.44801\)] where the end-points are correct to five significant figures.
Determine
(i) the number of times the coin was tossed;
(ii) the number of heads obtained.[7]
▶️Answer/Explanation
Markscheme
(i) \({{\rm{H}}_0}:p = 0.6\) ; \({{\rm{H}}_1}:p \ne 0.6\) A1A1
(ii) EITHER
using a normal approximation, \(p\)-value \( = 0.197\) A2
Note: Award A1 for \(0.0984\).
the shopkeeper’s claim is supported A1
because \(0.197 > 0.05\) R1
OR
using binomial distribution, \(p\)-value \( = 0.221\) A2
Note: Award A1 for \(0.110\).
the shopkeeper’s claim is supported A1
because \(0.221 > 0.05\) R1
Note: Follow through the candidate’s \(p\)-value for A1R1.
Note: Accept \(p\)-values correct to two significant figures.
[6 marks]
(i) \(\hat p = \frac{{0.35199 + 0.44801}}{2} = 0.4\) A1
width of CI \( = 3.92\sqrt {\frac{{0.4 \times 0.6}}{n}} \) M1
\(3.92\sqrt {\frac{{0.4 \times 0.6}}{n}} = 0.44801 – 0.35199 = 0.096(02)\) A1
solving,
\(n = {\left( {\frac{{3.92}}{{0.096(02)}}} \right)^2} \times 0.24\) (M1)
\( = 400\) A1
(ii) \(x = n\widehat p = 400 \times 0.4 = 160\) M1A1
[7 marks]