Question
The random variable X is normally distributed with unknown mean \(\mu \) and unknown variance \({\sigma ^2}\) . A random sample of 10 observations on X was taken and the following 95 % confidence interval for \(\mu \) was correctly calculated as [4.35, 4.53] .
(a) Calculate an unbiased estimate for
(i) \(\mu \) ,
(ii) \({\sigma ^2}\) .
(b) The value of \(\mu \) is thought to be 4.5, so the following hypotheses are defined.\[{{\text{H}}_0}:\mu = 4.5;{\text{ }}{{\text{H}}_1}:\mu < 4.5\]
(i) Find the p-value of the observed sample mean.
(ii) State your conclusion if the significance level is
(a) 1 %,
(b) 10 %.
▶️Answer/Explanation
Markscheme
(a) (i) \(\bar x = \frac{{4.35 + 4.53}}{2} = 4.44\) (estimate of \(\mu \)) A2
(ii) Degrees of freedom = 9 (A1)
Critical value of t = 2.262 (A1)
\(2.262 \times \frac{s}{{\sqrt {10} }} = 0.09\) M1A1
\(s = 0.12582…\) (A1)
\({s^2} = 0.0158\) (estimate of \({\sigma ^2}\)) A1
[8 marks]
(b) (i) Using t test (M1)
\(t = \frac{{4.44 – 4.5}}{{\sqrt {\frac{{0.0158}}{{10}}} }} = – 1.50800\) (Accept \( – 1.50946\)) (A1)
p-value = 0.0829 (Accept 0.0827) A2
(ii) (a) Accept \({{\text{H}}_0}\) / Reject \({{\text{H}}_1}\) . R1
(b) Reject \({{\text{H}}_0}\) / Accept \({{\text{H}}_1}\) . R1
[6 marks]
Total [14 marks]
Examiners report
Most candidates realised that the unbiased estimate of the mean was simply the central point of the confidence interval. Many candidates, however, failed to realise that, because the variance was unknown, the t-distribution was used to determine the confidence limits. In (b), although the p-value was asked for specifically, some candidates solved the problem correctly by comparing the value of their statistic with the appropriate critical values. This method was given full credit but, of course, marks were lost by their failure to give the p-value.
Question
(a) After a chemical spillage at sea, a scientist measures the amount, x units, of the chemical in the water at 15 randomly chosen sites. The results are summarised in the form \(\sum {x = 18} \) and \(\sum {{x^2} = 28.94} \). Before the spillage occurred the mean level of the chemical in the water was 1.1. Test at the 5 % significance level the hypothesis that there has been an increase in the amount of the chemical in the water.
(b) Six months later the scientist returns and finds that the mean amount of the chemical in the water at the 15 randomly chosen sites is 1.18. Assuming that this sample came from a normal population with variance 0.0256, find a 90 % confidence interval for the mean level of the chemical.
▶️Answer/Explanation
Markscheme
(a) \(\bar x = \frac{{\sum x }}{n} = 1.2\) (A1)
\(s_{n – 1}^2 = 0.524 \ldots \) (A1)
it is a one tailed test
\({{\text{H}}_0}:\mu = 1.1,{\text{ }}{{\text{H}}_1}:\mu > 1.1\) A1
EITHER
\(t = \frac{{1.2 – 1.1}}{{\sqrt {\frac{{0.524 \ldots }}{{15}}} }} = 0.535\) (M1) A1
\(v = 14\) (A1)
\({t_{crit}} = 1.761\) A1
since \(0.535 < {t_{crit}}\) we accept \({{\text{H}}_0}\) that there is no increase in the amount of the chemical R1
OR
\(p = 0.301\) A4
since \(p > 0.05\) we accept \({{\text{H}}_0}\) that there is no increase in the amount of the chemical R1
[8 marks]
(b) 90 % confidence interval \( = 1.18 \pm 1.645\sqrt {\frac{{0.0256}}{{15}}} \) (M1)A1A1A1
\( = [1.11,{\text{ }}1.25]\) A1 N5
[5 marks]
Total [13 marks]
Examiners report
This question also proved accessible to a majority of candidates with many wholly correct or nearly wholly correct answers seen. A few candidates did not recognise that part (a) was a t-distribution and part (b) was a Normal distribution, but most recognised the difference. Many candidates received an accuracy penalty on this question for not giving the final answer to part (b) to 3 significant figures.
Question
Alan and Brian are athletes specializing in the long jump. When Alan jumps, the length of his jump is a normally distributed random variable with mean 5.2 metres and standard deviation 0.1 metres. When Brian jumps, the length of his jump is a normally distributed random variable with mean 5.1 metres and standard deviation 0.12 metres. For both athletes, the length of a jump is independent of the lengths of all other jumps. During a training session, Alan makes four jumps and Brian makes three jumps. Calculate the probability that the mean length of Alan’s four jumps is less than the mean length of Brian’s three jumps.
Colin joins the squad and the coach wants to know the mean length, \(\mu \) metres, of his jumps. Colin makes six jumps resulting in the following lengths in metres.
5.21, 5.30, 5.22, 5.19, 5.28, 5.18
(i) Calculate an unbiased estimate of both the mean \(\mu \) and the variance of the lengths of his jumps.
(ii) Assuming that the lengths of these jumps are independent and normally distributed, calculate a 90 % confidence interval for \(\mu \) .
▶️Answer/Explanation
Markscheme
let \(\bar A,{\text{ }}\bar B\) denote the means of Alan’s and Brian’s jumps
attempting to find the distributions of \(\bar A,{\text{ }}\bar B\) (M1)
\(\bar A{\text{ is N}}\left( {5.2,\frac{{{{0.1}^2}}}{4}} \right)\) A1
\(\bar B{\text{ is N}}\left( {5.1,\frac{{{{0.12}^2}}}{3}} \right)\) A1
attempting to find the distribution of \(\bar A – \bar B\) (M1)
\(\bar A – \bar B{\text{ is N}}\left( {5.2 – 5.1,\frac{{{{0.1}^2}}}{4} + \frac{{{{0.12}^2}}}{3}} \right)\) (A1)(A1)
i.e. \({\text{N}}(0.1,{\text{ }}0.0073)\) A1
\({\text{P}}(\bar A < \bar B) = {\text{P}}(\bar A – \bar B < 0)\) M1
\( = 0.121\) A1
[9 marks]
(i) \(\sum {x = 31.38,{\text{ }}\sum {{x^2} = 164.1294} } \)
\(\bar x = \frac{{31.38}}{6} = 5.23\) (M1)A1
EITHER
\(s_{n – 1}^2 = \frac{{164.1294}}{5} – \frac{{{{31.38}^2}}}{{5 \times 6}} = 0.00240\) (M1)(A1)A1
OR
\({s_{n – 1}} = 0.04899 \Rightarrow s_{n – 1}^2 = 0.00240\) (M1)(A1)A1
Note: Accept the exact answer 0.0024 without an arithmetic penalty.
(ii) using the t-distribution with DF = 5 (A1)
critical value of t = 2.015 A1
90 % confidence limits are \(5.23 \pm 2.015\sqrt {\frac{{0.0024}}{6}} \) M1A1
giving [5.19, 5.27] A1 N5
[10 marks]
Examiners report
In (a), it was disappointing to note that many candidates failed to realise that the question was concerned with the mean lengths of the jumps and worked instead with the sums of the lengths.
Most candidates obtained correct estimates in (b)(i), usually directly from the GDC. In (b)(ii), however, some candidates found a z-interval instead of a t-interval.
Question
Anna cycles to her new school. She records the times taken for the first ten days with the following results (in minutes).
12.4 13.7 12.5 13.4 13.8 12.3 14.0 12.8 12.6 13.5
Assume that these times are a random sample from the \({\text{N}}(\mu ,{\text{ }}{\sigma ^2})\) distribution.
(a) Determine unbiased estimates for \(\mu \) and \({\sigma ^2}\).
(b) Calculate a 95 % confidence interval for \(\mu \).
(c) Before Anna calculated the confidence interval she thought that the value of \(\mu \) would be 12.5. In order to check this, she sets up the null hypothesis \({{\text{H}}_0}:\mu = 12.5\).
(i) Use the above data to calculate the value of an appropriate test statistic. Find the corresponding p-value using a two-tailed test.
(ii) Interpret your p-value at the 1 % level of significance, justifying your conclusion.
▶️Answer/Explanation
Markscheme
(a) estimate of \(\mu = 13.1\) A1
estimate of \({\sigma ^2} = 0.416\) A1
[2 marks]
(b) using a GDC (or otherwise), the 95% confidence interval is (M1)
[12.6, 13.6] A1A1
Note: Accept open or closed intervals.
[3 marks]
(c) (i) \(t = \frac{{13.1 – 12.5}}{{0.6446 \ldots /\sqrt {10} }} = 2.94\) (M1)A1
\(v = 9\) (A1)
p-value \( = 2 \times {\text{P}}(T > 2.9433 \ldots )\) (M1)
\( = 0.0164\,\,\,\,\,\)(accept 0.0165) A1
(ii) we accept the null hypothesis (the mean travel time is 12.5 minutes) A1
because 0.0164 (or 0.0165) > 0.01 R1
Note: Allow follow through on their p-value.
[7 marks]
Total [12 marks]
Examiners report
This was well answered by many candidates. In (a), some candidates chose the wrong standard deviation from their calculator and often failed to square their result to obtain the unbiased variance estimate. Candidates should realise that it is the smaller of the two values (ie the one obtained by dividing by (n – 1)) that is required. The most common error was to use the normal distribution instead of the t-distribution. The signpost towards the t-distribution is the fact that the variance had to be estimated in (a). Accuracy penalties were often given for failure to round the confidence limits, the t-statistic or the p-value to three significant figures.
Question
The length of time, T, in months, that a football manager stays in his job before he is removed can be approximately modelled by a normal distribution with population mean \(\mu \) and population variance \({\sigma ^2}\). An independent sample of five values of T is given below.
6.5, 12.4, 18.2, 3.7, 5.4
(a) Given that \({\sigma ^2} = 9\),
(i) use the above sample to find the 95 % confidence interval for \(\mu \), giving the bounds of the interval to two decimal places;
(ii) find the smallest number of values of T that would be required in a sample for the total width of the 90 % confidence interval for \(\mu \) to be less than 2 months.
(b) If the value of \({\sigma ^2}\) is unknown, use the above sample to find the 95 % confidence interval for \(\mu \), giving the bounds of the interval to two decimal places.
▶️Answer/Explanation
Markscheme
(a) (i) as \({\sigma ^2}\) is known \({\bar x}\) is \({\text{N}}\left( {\mu ,\frac{{{\sigma ^2}}}{n}} \right)\) (M1)
CI is \(\bar x – {z^ * }\frac{\sigma }{{\sqrt n }} < \mu < \bar x + {z^ * }\frac{\sigma }{{\sqrt n }}\) (M1)
\(\bar x = 9.24,{\text{ }}{z^ * } = 1.960\) for 95 % CI (A1)
CI is \(6.61 < \mu < 11.87\) by GDC A1A1
(ii) CI is \(\bar x – {z^ * }\frac{\sigma }{{\sqrt n }} < \mu < \bar x + {z^ * }\frac{\sigma }{{\sqrt n }}\)
require \(2 \times 1.645\frac{3}{{\sqrt n }} < 2\) R1A1
\(4.935 < \sqrt n \) (A1)
\(24.35 < n\) A1
so smallest value for n = 25 A1
Note: Accept use of table.
[10 marks]
(b) as \({\sigma ^2}\) is not known \({\bar x}\) has the t distribution with v = 4 (M1)(A1)
CI is \(\bar x – {t^ * }\frac{{{s_{n – 1}}}}{{\sqrt n }} < \mu < \bar x + {t^ * }\frac{{{s_{n – 1}}}}{{\sqrt n }}\)
\(\bar x = 9.24,{\text{ }}{s_{n – 1}} = 5.984,{\text{ }}{t^ * } = 2.776\) for 95 % CI (A1)
CI is \(1.81 < \mu < 16.67\) by GDC A1A1
[5 marks]
Total [15 marks]
Examiners report
The 2 confidence intervals were generally done well by using a calculator. Some marks were dropped by not giving the answers to 2 decimal places as required. Weak candidates did not realise that (b) was a t interval. Part (a) (ii) was not as well answered and often it was the first step that was the problem.
Question
A shopper buys 12 apples from a market stall and weighs them with the following results (in grams).
117, 124, 129, 118, 124, 116, 121, 126, 118, 121, 122, 129
You may assume that this is a random sample from a normal distribution with mean \(\mu \) and variance \({\sigma ^2}\).
a.Determine unbiased estimates of \(\mu \) and \({\sigma ^2}\).[3]
b.Determine a 99 % confidence interval for \(\mu \) .[2]
c.The stallholder claims that the mean weight of apples is 125 grams but the shopper claims that the mean is less than this.
(i) State suitable hypotheses for testing these claims.
(ii) Calculate the p-value of the above sample.
(iii) Giving a reason, state which claim is supported by your p-value using a 5 % significance level.[5]
▶️Answer/Explanation
Markscheme
unbiased estimate of \(\mu = 122\) A1
unbiased estimate of \({\sigma ^2} = 4.4406{ \ldots ^2} = 19.7\) (M1)A1
Note: Award (M1)A0 for 4.44.
[3 marks]
the 99 % confidence interval for \(\mu \) is [118, 126] A1A1
[2 marks]
(i) \({{\text{H}}_0}:\mu = 125;{\text{ }}{{\text{H}}_1}:\mu < 125\) A1
(ii) p-value = 0.0220 A2
(iii) the shopper’s claim is supported because \(0.0220 < 0.05\) A1R1
[5 marks]
Examiners report
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