IB DP Maths Topic 7.5 Confidence intervals for the mean of a normal population HL Paper 3

 

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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.

[9]
a.

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 \(\mu \) .

[10]
b.
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]

a.

(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]

b.

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.

a.

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.

b.

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

The random variable X is normally distributed with unknown mean \(\mu \) and unknown variance \({\sigma ^2}\). A random sample of 20 observations on X gave the following results.

\[\sum {x = 280,{\text{ }}\sum {{x^2} = 3977.57} } \]

Find unbiased estimates of \(\mu \) and \({\sigma ^2}\).

[3]
a.

Determine a 95 % confidence interval for \(\mu \).

[3]
b.

Given the hypotheses

\[{{\text{H}}_0}:\mu  = 15;{\text{ }}{{\text{H}}_1}:\mu  \ne 15,\]

find the p-value of the above results and state your conclusion at the 1 % significance level.

[4]
c.
Answer/Explanation

Markscheme

\(\bar x = 14\)     A1

\(s_{n – 1}^2 = \frac{{3977.57}}{{19}} – \frac{{{{280}^2}}}{{380}}\)     (M1)

\( = 3.03\)     A1

[3 marks]

Note: Accept any notation for these estimates including \(\mu \) and \({\sigma ^2}\).

Note: Award M0A0 for division by 20.

a.

the 95% confidence limits are

\(\bar x \pm t\sqrt {\frac{{s_{n – 1}^2}}{n}} \)     (M1)

Note: Award M0 for use of z.

 

ie, \(14 \pm 2.093\sqrt {\frac{{3.03}}{{20}}} \)     (A1)

Note:FT their mean and variance from (a).

 

giving [13.2, 14.8]     A1

Note: Accept any answers which round to 13.2 and 14.8.

 

[3 marks]

b.

Use of t-statistic \(\left( { = \frac{{14 – 15}}{{\sqrt {\frac{{3.03}}{{20}}} }}} \right)\)     (M1)

Note:FT their mean and variance from (a).

 

Note: Award M0 for use of z.

 

Note: Accept \(\frac{{15 – 14}}{{\sqrt {\frac{{3.03}}{{20}}} }}\).

 

\( = – 2.569 \ldots \)     (A1)

Note: Accept \(2.569 \ldots \)

 

\(p{\text{ – value}} = 0.009392 \ldots  \times 2 = 0.0188\)     A1

Note: Accept any answer that rounds to 0.019.

 

Note: Award (M1)(A1)A0 for any answer that rounds to 0.0094.

 

insufficient evidence to reject \({{\text{H}}_0}\) (or equivalent, eg accept \({{\text{H}}_0}\) or reject \({{\text{H}}_1}\))     R1

Note:FT on their p-value.

 

[4 marks]

c.

Examiners report

In (a), most candidates estimated the mean correctly although many candidates failed to obtain a correct unbiased estimate for the variance. The most common error was to divide \(\sum {{x^2}} \) by \(20\) instead of \(19\). For some candidates, this was not a costly error since we followed through their variance into (b) and (c).

a.

In (b) and (c), since the variance was estimated, the confidence interval and test should have been carried out using the t-distribution. It was extremely disappointing to note that many candidates found a Z-interval and used a Z-test and no marks were awarded for doing this. Candidates should be aware that having to estimate the variance is a signpost pointing towards the t-distribution.

b.

In (b) and (c), since the variance was estimated, the confidence interval and test should have been carried out using the t-distribution. It was extremely disappointing to note that many candidates found a Z-interval and used a Z-test and no marks were awarded for doing this. Candidates should be aware that having to estimate the variance is a signpost pointing towards the t-distribution.

c.

Question

A traffic radar records the speed, \(v\) kilometres per hour (\({\text{km}}\,{{\text{h}}^{-{\text{1}}}}\)), of cars on a section of a road.

The following table shows a summary of the results for a random sample of 1000 cars whose speeds were recorded on a given day.

Using the data in the table,

(i)     show that an estimate of the mean speed of the sample is 113.21 \({\text{km}}\,{{\text{h}}^{-{\text{1}}}}\);

(ii)     find an estimate of the variance of the speed of the cars on this section of the road.

[4]
a.

Find the 95% confidence interval, \(I\), for the mean speed.

[2]
b.

Let \(J\) be the 90% confidence interval for the mean speed.

Without calculating \(J\), explain why \(J \subset I\).

[2]
c.
Answer/Explanation

Markscheme

(i)     \(\bar v = \frac{1}{{1000}}(55 \times 5 + 65 \times 13 +  \ldots  + 145 \times 31)\)     A1M1

Note: A1 for mid-points, M1 for use of the formula.

\( = \frac{{113\,210}}{{1000}} = 113.21\)     AG

(ii)     \({s^2} = \frac{{{{(55 – 113.21)}^2} \times 5 + {{(65 – 113.21)}^2} \times 13 +  \ldots  + {{(145 – 113.21)}^2} \times 31}}{{999}}\)     (M1)

\( = \frac{{362\,295.9}}{{999}} = 362.6585 \ldots  = 363\)     A1

Note: Award A1 if answer rounds to 362 or 363.

Note: Condone division by 1000.

[4 marks]

a.

\(\bar v \pm \frac{{{t_{0.025}} \times s}}{{\sqrt n }}\)     (M1)

hence the confidence interval \(I = [112.028,{\text{ }}114.392]\)     A1

Note:     Accept answers which round to 112 and 114.

Note:     Condone the use of \({z_{0.025}}\) for \({t_{0.025}}\) and \(\sigma \) for \(s\).

[2 marks]

b.

less confidence implies narrower interval     R2

Note:     Accept equivalent statements or arguments having a meaningful diagram and/or relevant percentiles.

hence the confidence interval \(I\) at the 95% level contains the confidence interval \(J\) at the 90% level     AG

[2 marks]

c.

Examiners report

In (a)(i), the candidates were required to show that the estimate of the mean is 113.21 so that those who stated simply ‘Using my GDC, mean = 113.21’ were given no credit. Candidates were expected to indicate that the interval midpoints were used and to show the appropriate formula. In (a)(ii), division by either 999 or 1000 was accepted, partly because of the large sample size and partly because the question did not ask for an unbiased estimate of variance.

a.

 

b.

Solutions to (c) were often badly written, often quite difficult to understand exactly what was being stated.

c.

Question

A manufacturer of stopwatches employs a large number of people to time the winner of a \(100\) metre sprint. It is believed that if the true time of the winner is \(\mu \) seconds, the times recorded are normally distributed with mean \(\mu \) seconds and standard deviation \(0.03\) seconds.

The times, in seconds, recorded by six randomly chosen people are

\[9.765,{\text{ }}9.811,{\text{ }}9.783,{\text{ }}9.797,{\text{ }}9.804,{\text{ }}9.798.\]

Calculate a \(99\% \) confidence interval for \(\mu \). Give your answer correct to three decimal places.

[4]
a.

Interpret the result found in (a).

[2]
b.

Find the confidence level of the interval that corresponds to halving the width of the \(99\% \) confidence interval. Give your answer as a percentage to the nearest whole number.

[3]
c.
Answer/Explanation

Markscheme

the (unbiased) estimate of \(\mu \) is 9.793     (A1)

the \(99\% \) CI is \(9.793 \pm 2.576\frac{{0.03}}{{\sqrt 6 }}\)     (M1)(A1)

\( = [9.761,{\text{ }}9.825]\)     A1

Note:     Accept \(9.762\) and \(9.824\).

[4 marks]

a.

if this process is carried out a large number of times     A1

(approximately) \(99\% \) of the intervals will contain \(\mu \)     A1

Note:     Award A1A1 for a consideration of any specific large value of times \((n \ge 100)\).

[2 marks]

b.

METHOD 1

If the interval is halved, \(2.576\) becomes \(1.288\)     M1

normal tail probability corresponding to \(1.288 = 0.0988 \ldots \)     A1

confidence level \( = 80\% \)     A1

METHOD 2

half width \( = 0.5 \times 0.063\) or \(0.062\) or \(0.064 = 0.0315\) or \(0.031\) or \(0.032\)     M1

\(\frac{{2z \times 0.03}}{{\sqrt 6 }} = 0.0315\) or \(0.031\) or \(0.032\)

giving \(z = 1.285 \ldots \) or \(1.265 \ldots \) or \(1.306 \ldots \)     A1

confidence level \( = 80\% \) or \(79\% \) or \(81\% \)     A1

Note:     Follow through values from (a).

[3 marks]

Total [9 marks]

c.

Examiners report

The intention in (a) was that candidates should input the data into their calculators and use the software to give the confidence interval. However, as in Question 2, many candidates calculated the mean and variance by hand and used the appropriate formulae to determine the confidence limits. Again valuable time was used up and opportunity for error introduced.

a.

Answers to (b) were extremely disappointing with the vast majority giving an incorrect interpretation of a confidence interval. The most common answer given was along the lines of ‘There is a 99% probability that the interval [9.761, 9.825] contains \(\mu \)’. This is incorrect since the interval and \(\mu \) are both constants; the statement that the interval [9.761, 9.825] contains \(\mu \) is either true or false, there is no question of probability being involved. Another common response was ‘I am 99% confident that the interval [9.761, 9.825] contains \(\mu \)’. This is unsatisfactory partly because 99% confident is really a euphemism for 99% probability and partly because it answers the question ‘What is a 99% confidence interval for \(\mu \)by simply rearranging the words without actually going anywhere. The expected answer was that if the sampling was carried out a large number of times, then approximately 99% of the calculated confidence intervals would contain \(\mu \). A more rigorous response would be that a 99% confidence interval for \(\mu \) is an observed value of a random interval which contains \(\mu \) with probability 0.99 just as the number \(\bar x\) is an observed value of the random variable \(\bar X\). The concept of a confidence interval is a difficult one at this level but confidence intervals are part of the programme and so therefore is their interpretation. In view of the widespread misunderstanding of confidence intervals, partial credit was given on this occasion for interpretations involving 99% probability or confidence but this will not be the case in future examinations.

b.

Many candidates solved (c) correctly, mostly using Method 2 in the mark scheme.

c.

Question

It is known that the standard deviation of the heights of men in a certain country is \(15.0\) cm.

One hundred men from that country, selected at random, had their heights measured.

The mean of this sample was \(185\) cm. Calculate a \(95\% \) confidence interval for the mean height of the population.

[3]
a.

A second random sample of size \(n\) is taken from the same population. Find the minimum value of \(n\) needed for the width of a \(95\% \) confidence interval to be less than \(3\) cm.

[4]
b.
Answer/Explanation

Markscheme

valid attempt to use \(\bar x \pm z\frac{\sigma }{{\sqrt n }}\)     (M1)

\([182,{\text{ }}188]\)     A1A1

Note:     Accept answers that round to the correct \(3\) sf.

[3 marks]

a.

\(1.96 \times \frac{{15.0}}{{\sqrt n }} < 1.5\)     M1A1

\(n > {\left( {\frac{{15.0}}{{1.5}} \times 1.96} \right)^2}\)     (M1)

Note:     Award M1 for attempting to solve the inequality.

Note:     Allow the use of \( = \).

minimum value \(n = 385\)     A1

[4 marks]

Total [7 marks]

b.

Examiners report

[N/A]

a.

[N/A]

b.

Question

A smartphone’s battery life is defined as the number of hours a fully charged battery can be used before the smartphone stops working. A company claims that the battery life of a model of smartphone is, on average, 9.5 hours. To test this claim, an experiment is conducted on a random sample of 20 smartphones of this model. For each smartphone, the battery life, \(b\) hours, is measured and the sample mean, \({\bar b}\), calculated. It can be assumed the battery lives are normally distributed with standard deviation 0.4 hours.

It is then found that this model of smartphone has an average battery life of 9.8 hours.

State suitable hypotheses for a two-tailed test.

[1]
a.

Find the critical region for testing \({\bar b}\) at the 5 % significance level.

[4]
b.

Find the probability of making a Type II error.

[3]
c.

Another model of smartphone whose battery life may be assumed to be normally distributed with mean μ hours and standard deviation 1.2 hours is tested. A researcher measures the battery life of six of these smartphones and calculates a confidence interval of [10.2, 11.4] for μ.

Calculate the confidence level of this interval.

[4]
d.
Answer/Explanation

Markscheme

Note: In question 3, accept answers that round correctly to 2 significant figures.

\({{\text{H}}_0}\,{\text{:}}\,\mu  = 9.5{\text{;}}\,\,{{\text{H}}_1}\,{\text{:}}\,\mu  \ne 9.5\)     A1

[1 mark]

a.

Note: In question 3, accept answers that round correctly to 2 significant figures.

the critical values are \(9.5 \pm 1.95996 \ldots  \times \frac{{0.4}}{{\sqrt {20} }}\)     (M1)(A1)

i.e. 9.3247…, 9.6753…

the critical region is \({\bar b}\) < 9.32, \({\bar b}\) > 9.68     A1A1

Note: Award A1 for correct inequalities, A1 for correct values.

Note: Award M0 if t-distribution used, note that t(19)97.5 = 2.093 …

[4 marks]

b.

Note: In question 3, accept answers that round correctly to 2 significant figures.

\(\bar B \sim {\text{N}}\left( {9.8,\,{{\left( {\frac{{0.4}}{{\sqrt {20} }}} \right)}^2}} \right)\)     (A1)

\({\text{P}}\left( {9.3247 \ldots  < \bar B < 9.6753 \ldots } \right)\)     (M1)

=0.0816     A1

Note: FT the critical values from (b). Note that critical values of 9.32 and 9.68 give 0.0899.

[3 marks]

c.

Note: In question 3, accept answers that round correctly to 2 significant figures.

METHOD 1

\(X \sim {\text{N}}\left( {{\text{10}}{\text{.8,}}\,\frac{{{{1.2}^2}}}{6}} \right)\)     (M1)(A1)

P(10.2 < X < 11.4) = 0.7793…     (A1)

confidence level is 77.9%    A1

Note: Accept 78%.

METHOD 2

\(11.4 – 10.2 = 2z \times \frac{{1.2}}{{\sqrt 6 }}\)      (M1)

\(z = 1.224 \ldots \)     (A1)

P(−1.224… < Z < 1.224…) = 0.7793…      (A1)

confidence level is 77.9%      A1

Note: Accept 78%.

[4 marks]

d.

Examiners report

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d.

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 \({\sigma ^2}\).

Determine unbiased estimates of \(\mu \) and \({\sigma ^2}\).

[3]
a.

Determine a 99 % confidence interval for \(\mu \) .

[2]
b.

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]
c.
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]

a.

the 99 % confidence interval for \(\mu \) is [118, 126]     A1A1

[2 marks]

b.

(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]

c.

Examiners report

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a.

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b.

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c.

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