Question
An automatic machine is used to fill bottles of water. The amount delivered, \(Y\) ml , may be assumed to be normally distributed with mean \(\mu \) ml and standard deviation \(8\) ml . Initially, the machine is adjusted so that the value of \(\mu \) is \(500\). In order to check that the value of \(\mu \) remains equal to \(500\), a random sample of 10 bottles is selected at regular intervals, and the mean amount of water, \(\overline y \) , in these bottles is calculated. The following hypotheses are set up.
\({{\rm{H}}_0}:\mu = 500\) ; \({{\rm{H}}_1}:\mu \ne 500\)
The critical region is defined to be \(\left( {\overline y < 495} \right) \cup \left( {\overline y > 505} \right)\) .
(i) Find the significance level of this procedure.
(ii) Some time later, the actual value of \(\mu \) is \(503\). Find the probability of a Type II error.
▶️Answer/Explanation
Markscheme
(i) Under \({{\rm{H}}_0}\) , the distribution of \({\overline y }\) is N(500, 6.4) . (A1)
Significance level \( = {\rm{P}}\overline y < 495\) or \( > 505|{{\rm{H}}_0}\) M2
\( = 2 \times 0.02405\) (A1)
\( = 0.0481\) A1 N5
Note: Using tables, answer is \(0.0478\).
(ii) The distribution of \(\overline y \) is now N(\(503\), \(6.4\)) . (A1)
P(Type ΙΙ error) \( = {\rm{P}}(495 < \overline y < 505)\) (M1)
\( = 0.785\) A1 N3
Note: Using tables, answer is \(0.784\).
[8 marks]
Question
A machine fills containers with grass seed. Each container is supposed to weigh \(28\) kg. However the weights vary with a standard deviation of \(0.54\) kg. A random sample of \(24\) bags is taken to check that the mean weight is \(28\) kg.
A.a.Assuming the series for \({{\rm{e}}^x}\) , find the first five terms of the Maclaurin series for\[\frac{1}{{\sqrt {2\pi } }}{{\rm{e}}^{\frac{{ – {x^2}}}{2}}} {\rm{ .}}\][3]
A.b.(i) Use your answer to (a) to find an approximate expression for the cumulative distributive function of \({\rm{N}}(0,1)\) .
(ii) Hence find an approximate value for \({\rm{P}}( – 0.5 \le Z \le 0.5)\) , where \(Z \sim {\rm{N}}(0,1)\) .[6]
B.a.State and justify an appropriate test procedure giving the null and alternate hypotheses.[5]
B.b.What is the critical region for the sample mean if the probability of a Type I error is to be \(3.5\%\)?[7]
B.c.If the mean weight of the bags is actually \(28\).1 kg, what would be the probability of a Type II error?[2]
▶️Answer/Explanation
Markscheme
\({{\rm{e}}^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} + \ldots \)
\({{\rm{e}}^{\frac{{ – {x^2}}}{2}}} = 1 + \left( { – \frac{{{x^2}}}{2}} \right) + \frac{{{{\left( { – \frac{{{x^2}}}{2}} \right)}^2}}}{{2!}} + \frac{{{{\left( { – \frac{{{x^2}}}{2}} \right)}^3}}}{{3!}} + \frac{{{{\left( { – \frac{{{x^2}}}{2}} \right)}^4}}}{{4!}} + \ldots \) M1A1
\(\frac{1}{{\sqrt {2\pi } }}{{\rm{e}}^{\frac{{ – {x^2}}}{2}}} = \frac{1}{{\sqrt {2\pi } }}\left( {1 – \frac{{{x^2}}}{2} + \frac{{{x^4}}}{8} – \frac{{{x^6}}}{{48}} + \frac{{{x^8}}}{{384}}} \right)\) A1
[3 marks]
(i) \(\frac{1}{{\sqrt {2\pi } }}\int_0^x {1 – \frac{{{t^2}}}{2}} + \frac{{{t^4}}}{8} – \frac{{{t^6}}}{{48}} + \frac{{{t^8}}}{{384}}{\rm{d}}t\) M1
\( = \frac{1}{{\sqrt {2\pi } }}\left( {x – \frac{{{x^3}}}{6} + \frac{{{x^5}}}{{40}} – \frac{{{x^7}}}{{336}} + \frac{{{x^9}}}{{3456}}} \right)\) A1
\({\rm{P}}(Z \le x) = 0.5 + \frac{1}{{\sqrt {2\pi } }}\left( {x – \frac{{{x^3}}}{6} + \frac{{{x^5}}}{{40}} – \frac{{{x^7}}}{{336}} + \frac{{{x^9}}}{{3456}} – \ldots } \right)\) R1A1
(ii) \({\rm{P}}( – 0.5 \le Z \le 0.5) = \frac{2}{{\sqrt {2\pi } }}\left( {0.5 – \frac{{{{0.5}^3}}}{6} + \frac{{{{0.5}^5}}}{{40}} – \frac{{{{0.5}^7}}}{{336}} + \frac{{{{0.5}^9}}}{{3456}} – \ldots } \right)\) M1
\( = 0.38292 = 0.383\) A1
[6 marks]
this is a two tailed test of the sample mean \(\overline x \)
we use the central limit theorem to justify assuming that R1
\(\overline X \sim {\rm{N}}\left( {28,\frac{{{{0.54}^2}}}{{24}}} \right)\) R1A1
\({{\rm{H}}_0}:\mu = 28\) A1
\({{\rm{H}}_1}:\mu \ne 28\) A1
[5 marks]
since \({\text{P(Type I error)}} = 0.035\) , critical value \(2.108\) (M1)A1
and (\(\overline x \le 28 – 2.108\sqrt {\frac{{{{0.54}^2}}}{{24}}} \) or \(\overline x \ge 28 + 2.108\sqrt {\frac{{{{0.54}^2}}}{{24}}} \) ) (M1)(A1)(A1)
\(\overline x \le 27.7676\) or \(\overline x \ge 28.2324\)
so \(\overline x \le 27.8\) or \(\overline x \ge 28.2\) A1A1
[7 marks]
if \(\mu = 28.1\)
\(\overline X \sim {\rm{N}}\left( {28.1,\frac{{{{0.54}^2}}}{{24}}} \right)\) R1
\({\text{P(Type II error)}} = {\rm{P}}(27.7676 < \overline x < 28.2324)\)
\( = 0.884\) A1
Note: Depending on the degree of accuracy used for the critical region the answer for part (c) can be anywhere from \(0.8146\) to \(0.879\).
[2 marks]
Examiners report
The derivation of a series from a given one by substitution seems not to be well known. This made finding the required series from \(({{\rm{e}}^x})\) in part (a) to be much more difficult than it need have been. The fact that this part was worth only 3 marks was a clear hint that an easy derivation was possible.
In part (b)(i) the \(0.5\) was usually missing which meant that this part came out incorrectly.
The conditions required in part (a) were rarely stated correctly and some candidates were unable to state the hypotheses precisely. There was some confusion with “less than” and “less than or equal to”.
There was some confusion with “less than” and “less than or equal to”.
Levels of accuracy in the body of the question varied wildly leading to a wide range of answers to part (c).
Question
In a large population of sheep, their weights are normally distributed with mean \(\mu \) kg and standard deviation \(\sigma \) kg. A random sample of \(100\) sheep is taken from the population.
The mean weight of the sample is \(\bar X\) kg.
a.State the distribution of \(\bar X\) , giving its mean and standard deviation.[2]
b.The sample values are summarized as \(\sum {x = 3782} \) and \(\sum {{x^2} = 155341} \) where \(x\) kg is the weight of a sheep.
(i) Find unbiased estimates for \(\mu \) and \({\sigma ^2}\).
(ii) Find a \(95\%\) confidence interval for \(\mu \).[6]
c.Test, at the \(1\%\) level of significance, the null hypothesis \(\mu = 35\) against the alternative hypothesis that \(\mu > 35\).[5]
▶️Answer/Explanation
Markscheme
\(\bar X \sim N\left( {\mu ,{\text{ }}\frac{{{\sigma ^2}}}{{100}}} \right)\) A1A1
Note: Award A1 for \(N\), A1 for the parameters.
(i) \(\bar x = \frac{{\sum x }}{n} = \frac{{3782}}{{100}} = 37.8\) A1
\(s_{n – 1}^2 = \frac{{155341}}{{99}} – \frac{{{{3782}^2}}}{{9900}} = 124\) M1A1
(ii) \(95\% CI = 37.82 \pm 1.98\sqrt {\frac{{124.3006}}{{100}}} \) (M1)(A1)
\( = (35.6,{\text{ }}40.0)\) A1
METHOD 1
one tailed t-test A1
testing \(37.82\) A1
\(99\) degrees of freedom
reject if \(t > 2.36\) A1
t-value being tested is \(2.5294\) A1
since \(2.5294 > 2.36\) we reject the null hypothesis and accept the alternative hypothesis R1
METHOD 2
one tailed t-test (A1)
\(p = 0.00650\) A3
since \(p{\text{ – value}} < 0.01\) we reject the null hypothesis and accept the alternative hypothesis R1
Examiners report
Almost all candidates recognised the sample distribution as normal but were not always successful in stating the mean and the standard deviation. Similarly almost all candidates knew how to find an unbiased estimator for \(\mu \), but a number failed to find the correct answer for the unbiased estimator for \({\sigma ^2}\). Most candidates were successful in finding the 95% confidence interval for \(\mu \). In part c) many fully correct answers were seen but a significant number of candidates did not recognise they were working with a t-distribution.
Almost all candidates recognised the sample distribution as normal but were not always successful in stating the mean and the standard deviation. Similarly almost all candidates knew how to find an unbiased estimator for \(\mu \), but a number failed to find the correct answer for the unbiased estimator for \({\sigma ^2}\). Most candidates were successful in finding the 95% confidence interval for \(\mu \). In part c) many fully correct answers were seen but a significant number of candidates did not recognise they were working with a t-distribution.
Almost all candidates recognised the sample distribution as normal but were not always successful in stating the mean and the standard deviation. Similarly almost all candidates knew how to find an unbiased estimator for \(\mu \), but a number failed to find the correct answer for the unbiased estimator for \({\sigma ^2}\). Most candidates were successful in finding the 95% confidence interval for \(\mu \). In part c) many fully correct answers were seen but a significant number of candidates did not recognise they were working with a t-distribution.
Question
The weights of apples, in grams, produced on a farm may be assumed to be normally distributed with mean \(\mu \) and variance \({\sigma ^2}\) .
a.The farm manager selects a random sample of \(10\) apples and weighs them with the following results, given in grams.\[82, 98, 102, 96, 111, 95, 90, 89, 99, 101\]
(i) Determine unbiased estimates for \(\mu \) and \({\sigma ^2}\) .
(ii) Determine a \(95\%\) confidence interval for \(\mu \) .[5]
b.The farm manager claims that the mean weight of apples is \(100\) grams but the buyer from the local supermarket claims that the mean is less than this. To test these claims, they select a random sample of \(100\) apples and weigh them. Their results are summarized as follows, where \(x\) is the weight of an apple in grams.\[\sum {x = 9831;\sum {{x^2} = 972578} } \]
(i) State suitable hypotheses for testing these claims.
(ii) Determine the \(p\)-value for this test.
(iii) At the \(1\%\) significance level, state which claim you accept and justify your answer.[5]
▶️Answer/Explanation
Markscheme
(i) from the GDC,
unbiased estimate for \(\mu = 96.3\) A1
unbiased estimate for \({\sigma ^2} = 8.028{ \ldots ^2} = 64.5\) (M1)A1
(ii) \(95\%\) confidence interval is [\(90.6\), \(102\)] A1A1
Note: Accept \(102.0\) as the upper limit.
[5 marks]
(i) \({H_0}:\mu = 100;{H_1}:\mu < 100\) A1
(ii) \(\overline x = 98.31,{S_{n – 1}} = 7.8446 \ldots \) (A1)
\(p\)-value \( = 0.0168\) A1
(iii) the farm manager’s claim is accepted because \(0.0168 > 0.01\) A1R1
[5 marks]
Examiners report
[N/A]
[N/A]