Question
The function \(f\) is defined by \(f(x) = \ln (1 + \sin x)\) .
When a scientist measures the concentration \(\mu \) of a solution, the measurement obtained may be assumed to be a normally distributed random variable with mean \(\mu \) and standard deviation \(1.6\).
A.a.Show that \(f”(x) = \frac{{ – 1}}{{1 + \sin x}}\) .[4]
A.b.Determine the Maclaurin series for \(f(x)\) as far as the term in \({x^4}\) .[6]
A.c.Deduce the Maclaurin series for \(\ln (1 – \sin x)\) as far as the term in \({x^4}\) .[2]
A.d.By combining your two series, show that \(\ln \sec x = \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{12}} + \ldots \) .[4]
A.e.Hence, or otherwise, find \(\mathop {\lim }\limits_{x \to 0} \frac{{\ln \sec x}}{{x\sqrt x }}\) .[2]
B.a.He makes 5 independent measurements of the concentration of a particular solution and correctly calculates the following confidence interval for \(\mu \) .
[\(22.7\) , \(26.1\)]
Determine the confidence level of this interval.[5]
B.b.He is now given a different solution and is asked to determine a \(95\%\) confidence interval for its concentration. The confidence interval is required to have a width less than \(2\). Find the minimum number of independent measurements required.[5]
▶️Answer/Explanation
Markscheme
\(f'(x) = \frac{{\cos x}}{{1 + \sin x}}\) M1A1
\(f”(x) = \frac{{ – \sin x(1 + \sin x) – {{\cos }^2}x}}{{{{(1 + \sin x)}^2}}}\) M1
\( = \frac{{ – \sin x – 1}}{{{{(1 + \sin x)}^2}}}\) A1
\( = \frac{{ – 1}}{{1 + \sin x}}\) AG
[4 marks]
\(f”'(x) = \frac{{\cos x}}{{{{(1 + \sin x)}^2}}}\) A1
\({f^{iv}}(x) = \frac{{ – \sin x{{(1 + \sin x)}^2} – 2(1 + \sin x){{\cos }^2}x}}{{{{(1 + \sin x)}^4}}}\) A1
\(f(0) = 0\) , \(f'(0) = 1\) , \(f”(0) = – 1\) , \(f”'(0) = 1\) , \({f^{iv}}(0) = – 2\) (A2)
Note: Award A1 for 2 errors and A0 for more than 2 errors.
\(\ln (1 + \sin x) = x – \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} – \frac{{{x^4}}}{{12}} + \ldots \) M1A1
[6 marks]
\(\ln (1 – \sin x) = \ln (1 + \sin ( – x)) = – x – \frac{{{x^2}}}{2} – \frac{{{x^3}}}{6} – \frac{{{x^4}}}{{12}} + \ldots \) M1A1
[2 marks]
Adding, M1
\(\ln (1 – {\sin ^2}x) = \ln {\cos ^2}x\) A1
\( = – {x^2} – \frac{{{x^4}}}{6} + \ldots \) A1
\(\ln \cos x = – \frac{{{x^2}}}{2} – \frac{{{x^4}}}{{12}} + \ldots \) A1
\(\ln \sec x = \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{12}} + \ldots \) AG
[4 marks]
\(\frac{{\ln \sec x}}{{x\sqrt x }} = \frac{{\sqrt x }}{2} + \frac{{{x^2}\sqrt x }}{{12}} + \ldots \) M1
Limit \( = 0\) A1
[2 marks]
Interval width \( = 26.1 – 22.7 = 3.4\)
So \(3.4 = 2z \times \frac{{1.6}}{{\sqrt 5 }}\) M1A1
\(z = 2.375 \ldots \) A1
Probability \( = 0.9912\) A1
Confidence level \( = 2 \times 0.4912 = 98.2\% \) A1
[5 marks]
\(z\)-value \( = 1.96\) A1
We require
\(2 \times \frac{{1.96 \times 1.6}}{{\sqrt n }} < 2\) M1A1
Whence \(n > 9.83\) A1
So we need \(n = 10\) A1
Note: Accept \( = \) signs throughout.
[5 marks]
Question
a.The weights, \(X\) grams, of tomatoes may be assumed to be normally distributed with mean \(\mu \) grams and standard deviation \(\sigma \) grams. Barry weighs \(21\) tomatoes selected at random and calculates the following statistics.\[\sum {x = 1071} \) ; \(\sum {{x^2} = 54705} \]
(i) Determine unbiased estimates of \(\mu \) and \({\sigma ^2}\) .
(ii) Determine a \(95\%\) confidence interval for \(\mu \) .[8]
b.The random variable \(Y\) has variance \({\sigma ^2}\) , where \({\sigma ^2} > 0\) . A random sample of \(n\) observations of \(Y\) is taken and \(S_{n – 1}^2\) denotes the unbiased estimator for \({\sigma ^2}\) .
By considering the expression
\({\rm{Var}}({S_{n – 1}}) = {\rm{E}}(S_{n – 1}^2) – {\left\{ {E\left. {({S_{n – 1}})} \right\}} \right.^2}\) ,
show that \(S_{n – 1}^{}\) is not an unbiased estimator for \(\sigma \) .[5]
▶️Answer/Explanation
Markscheme
(i) \(\overline x = \frac{{1071}}{{21}} = 51\) A1
\(S_{n – 1}^2 = \frac{{54705}}{{20}} – \frac{{{{1071}^2}}}{{20 \times 21}} = 4.2\) M1A1
(ii) degrees of freedom \( = 20\) ; \(t\)-value \( = 2.086\) (A1)(A1)
\(95\%\) confidence limits are
\(51 \pm 2.086\sqrt {\frac{{4.2}}{{21}}} \) (M1)(A1)
leading to \(\left[ {50.1,51.9} \right]\) A1
[8 marks]
\({\rm{Var}}({S_{n – 1}}) > 0\) A1
\(E(S_{n – 1}^2) = {\sigma ^2}\) (A1)
substituting in the given equation,
\({\sigma ^2} – E(S_{n – 1}^{}) > 0\) M1
it follows that
\(E(S_{n – 1}^{}) < \sigma \) A1
this shows that \({S_{n – 1}}\) is not an unbiased estimator for \(\sigma \) since that would require = instead of \( < \) R1
[5 marks]
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]
(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]