5.05d Confidence intervals: using normal distribution

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Edexcel S4 2017 June Q3
11 marks Standard +0.3
3. The lengths, \(X \mathrm {~mm}\), of the wings of adult blackbirds follow a normal distribution. A random sample of 5 adult blackbirds is taken and the lengths of the wings are measured. The results are summarised below $$\sum x = 655 \text { and } \sum x ^ { 2 } = 85845$$
  1. Test, at the \(10 \%\) level of significance, whether or not the mean length of an adult blackbird's wing is less than 135 mm . State your hypotheses clearly.
  2. Find the \(90 \%\) confidence interval for the variance of the lengths of adult blackbirds' wings. Show your working clearly.
Edexcel S4 2017 June Q5
11 marks Challenging +1.2
  1. Jamland and Goodjam are two suppliers of jars of jam. The weights of the jars of jam produced by each supplier can be assumed to be normally distributed with unknown, but equal, variances. A random sample of 20 jars of jam is taken from those supplied by Jamland.
Based on this sample, the 95\% confidence interval for the mean weight of a jar of Jamland jam, in grams, is
[0pt] [ 492, 507 ] A random sample of 10 jars of jam is selected from those supplied by Goodjam. The weight of each jar of Goodjam jam, \(y\) grams, is recorded. The results are summarised as follows $$\bar { y } = 480 \quad s _ { y } ^ { 2 } = 280$$ Find a 90\% confidence interval for the value by which the mean weight of a jar of jam supplied by Jamland exceeds the mean weight of a jar of jam supplied by Goodjam.
Edexcel S4 2018 June Q2
13 marks Standard +0.3
  1. Jeremiah currently uses a Fruity model of juicer. He agrees to trial a new model of juicer, Zesty. The amounts of juice extracted, \(x \mathrm { ml }\), from each of 9 randomly selected oranges, using the Zesty are summarised as
$$\sum x = 468 \quad \sum x ^ { 2 } = 24560$$ Given that the amounts of juice extracted follow a normal distribution,
  1. calculate a 95\% confidence interval for
    1. the mean amount of juice extracted from an orange using the Zesty,
    2. the standard deviation of the amount of juice extracted from an orange using the Zesty. Jeremiah knows that, for his Fruity, the mean amount of juice extracted from an orange is 38 ml and the standard deviation of juice extracted from an orange is 5 ml . He decides that he will replace his Fruity with a Zesty if both
      • the mean for the Zesty is more than \(20 \%\) higher than the mean for his Fruity and
  2. the standard deviation for the Zesty is less than 5.5 ml .
  3. Using your answers to part (a), explain whether or not Jeremiah should replace his Fruity with the Zesty.
Edexcel S4 Q4
14 marks Standard +0.8
4. Gill, a member of the accounts department in a large company, is studying the expenses claims of company employees. She assumes that the claims, in \(\pounds\), follow a normal distribution with mean \(\mu\) and variance \(\sigma ^ { 2 }\). As a first stage in her investigation she took the following random sample of 10 claims. $$30.85,99.75,142.73,223.16,75.43,28.57,53.90,81.43,68.62,43.45 .$$
  1. Find a 95\% confidence interval for \(\mu\). The chief accountant would like a \(95 \%\) confidence interval where the difference between the upper confidence limit and the lower confidence limit is less than 20 .
  2. Show that \(\frac { \sigma ^ { 2 } } { n } < 26.03\) (to 2 decimal places), where \(n\) is the size of the sample required to achieve this. Gill decides to use her original sample of 10 to obtain a value for \(\sigma ^ { 2 }\) so that the chance of her value being an underestimate is 0.01 .
  3. Find such a value for \(\sigma ^ { 2 }\).
  4. Use this value for \(\sigma ^ { 2 }\) to estimate the size of sample the chief accountant requires.
Edexcel S4 Q5
16 marks Standard +0.8
5. An educational researcher is testing the effectiveness of a new method of teaching a topic in mathematics. A random sample of 10 children were taught by the new method and a second random sample of 9 children, of similar age and ability, were taught by the conventional method. At the end of the teaching, the same test was given to both groups of children. The marks obtained by the two groups are summarised in the table below.
New methodConventional method
Mean \(( \bar { x } )\)82.378.2
Standard deviation \(( s )\)3.55.7
Number of students \(( n )\)109
  1. Stating your hypotheses clearly and using a \(5 \%\) level of significance, investigate whether or not
    1. the variance of the marks of children taught by the conventional method is greater than that of children taught by the new method,
    2. the mean score of children taught by the conventional method is lower than the mean score of those taught by the new method.
      [0pt] [In each case you should give full details of the calculation of the test statistics.]
  2. State any assumptions you made in order to carry out these tests.
  3. Find a 95\% confidence interval for the common variance of the marks of the two groups.
OCR MEI Further Statistics B AS 2018 June Q1
6 marks Moderate -0.8
1 The birth weights, in kilograms, of a random sample of 9 captive-bred elephants are as follows. $$\begin{array} { l l l l l l l l l } 94 & 138 & 130 & 118 & 146 & 165 & 82 & 115 & 69 \end{array}$$ A researcher uses software to produce a \(99 \%\) confidence interval for the mean birth weight of captive-bred elephants. The output from the software is shown in Fig. 1. \begin{table}[h]
Result
T Estimate of a Mean
Mean
s
SE
N
df
Lower limit
Upper limit
Interval
\captionsetup{labelformat=empty} \caption{Fig. 1}
\end{table}
  1. State an assumption about the distribution of the population from which these weights come that is necessary in order to produce this interval.
  2. State the confidence interval which the software gives, in the form \(a < \mu < b\).
  3. Explain
OCR MEI Further Statistics B AS 2018 June Q5
10 marks Moderate -0.3
5 The flight time between two airports is known to be Normally distributed with mean 3.75 hours and standard deviation 0.21 hours. A new airline starts flying the same route. The flight times for a random sample of 12 flights with the new airline are shown in the spreadsheet (Fig. 5), together with the sample mean. \begin{table}[h]
ABCDEFGHIJKL
13.5953.7233.5843.6433.6693.6973.5503.6743.9243.5633.3303.706
2
3Mean3.638
\captionsetup{labelformat=empty} \caption{Fig. 5}
\end{table} \section*{(i) In this question you must show detailed reasoning.} You should assume that:
  • the flight times for the new airline are Normally distributed,
  • the standard deviation of the flight times is still 0.21 hours.
Carry out a test at the \(5 \%\) significance level to investigate whether the mean flight time for the new airline is less than 3.75 hours.
(ii) If both of the assumptions in part (i) were false, name an alternative test that you could carry out to investigate average flight times, stating any assumption necessary for this test.
(iii) If instead the flight times were still Normally distributed but the standard deviation was not known to be 0.21 hours, name another test that you could carry out.
OCR MEI Further Statistics B AS 2018 June Q6
10 marks Standard +0.3
6 A company has a large fleet of cars. It is claimed that use of a fuel additive will reduce fuel consumption. In order to test this claim a researcher at the company randomly selects 40 of the cars. The fuel consumption of each of the cars is measured, both with and without the fuel additive. The researcher then calculates the difference \(d\) litres per kilometre between the two figures for each car, where \(d\) is the fuel consumption without the additive minus the fuel consumption with the additive. The sample mean of \(d\) is 0.29 and the sample standard deviation is 1.64 .
  1. Showing your working, find a 95\% confidence interval for the population mean difference.
  2. Explain whether the confidence interval suggests that, on average, the fuel additive does reduce fuel consumption.
  3. Explain why you can construct the interval in part (i) despite not having any information about the distribution of the population of differences.
  4. Explain why the sample used was random.
OCR MEI Further Statistics B AS 2019 June Q6
11 marks Standard +0.3
6 The label on a pack of strawberries in a large batch states that it holds 250 g of strawberries. A random sample of 40 packs from the batch is selected and software is used to produce a \(95 \%\) confidence interval for the mean weight of strawberries per pack. An extract from the software output is shown in Fig. 6. \begin{table}[h]
Sample Mean248.92
Standard Error0.61506
Sample Size40
Confidence Level0.95
Interval\(248.92 \pm 1.2055\)
\captionsetup{labelformat=empty} \caption{Fig. 6}
\end{table}
  1. Explain whether the confidence interval suggests that the mean weight of strawberries per pack in the batch is different from 250 g .
  2. A manager looking at the data says that the conclusion would have been different if a \(90 \%\) confidence interval had been used.
    Determine whether the manager is correct.
  3. Explain briefly whether or not it is appropriate for the manager to vary the confidence level before coming to any conclusions.
    [0pt]
  4. On another occasion, using the same sample size, a 95\% confidence interval for the mean weight of strawberries per pack is [248.05, 249.95].
    Find the sample variance in this case.
  5. Explain the meaning of a 95\% confidence interval.
OCR MEI Further Statistics B AS 2022 June Q1
10 marks Moderate -0.3
1 Each working day, Beth takes a bus to her place of work. She believes that the mean time that her journey takes is 30 minutes. In order to check this, Beth selects a random sample of 8 journeys. The times in minutes for these 8 journeys are as follows. \(\begin{array} { l l l l l l l l } 31.9 & 28.5 & 35.9 & 31.0 & 30.2 & 34.9 & 28.9 & 31.3 \end{array}\)
  1. What assumption does Beth need to make in order to construct a confidence interval for the mean journey time based on the \(t\) distribution?
  2. In this question you must show detailed reasoning. Given that the assumption in part (a) is valid, determine a 95\% confidence interval for the mean journey time.
  3. Explain whether the confidence interval suggests that Beth may be correct in the belief that her mean journey time is 30 minutes.
OCR MEI Further Statistics B AS 2021 November Q1
9 marks Moderate -0.8
1 Over time LED light bulbs gradually lose brightness. For a particular type of LED bulb, it is known that the mean reduction in brightness after 10000 hours is \(2.6 \%\). A manufacturer produces a new version of this bulb, which costs less to make, but is claimed to have the same reduction in brightness after 10000 hours as the previous version. In order to check this claim, a random sample of 10 bulbs is selected. For each bulb, the original brightness and the brightness after 10000 hours are measured, in suitable units. A spreadsheet is used to produce a \(95 \%\) confidence interval for the mean percentage reduction in brightness. A screenshot of the spreadsheet is shown in Fig. 1. \begin{table}[h]
ABCDEFGH1JK
1Original brightness1075112111061095110111091114112311081115
2After 10000 hours1042108410761065107010791081109110801082
3Percentage reduction3.073.302.712.742.822.712.962.852.532.96
4
5
6Sample mean (\%)2.8650
7Sample sd (\%)0.2179
8SE0.0689
9DF9
10tvalue2.262
11Lower limit2.709
12Upper limit3.021
1.3
\captionsetup{labelformat=empty} \caption{Fig. 1}
\end{table}
  1. State the confidence interval in the form \(a < \mu < b\).
  2. Explain whether the confidence interval suggests that the mean percentage reduction in brightness after 10000 hours is different from 2.6\%.
  3. Explain how the value in cell B8 was calculated.
  4. State an assumption necessary for this confidence interval to be calculated.
  5. Explain the advantage of using the same bulbs for both measurements.
OCR MEI Further Statistics B AS Specimen Q5
11 marks Standard +0.3
5 A particular alloy of bronze is specified as containing \(11.5 \%\) copper on average. A researcher takes a random sample of 14 specimens of this bronze and undertakes an analysis of each of them. The percentages of copper are found to be as follows.
11.1211.2911.4211.4311.2011.2511.65
11.3311.5611.3411.4411.2411.6011.52
The researcher uses software to draw a Normal probability plot for these data and to conduct a Kolmogorov-Smirnov test for Normality. The output is shown in Fig 5.1. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{0de8222f-7df5-4e17-ab68-0f9d84fc615d-4_428_1550_1434_299} \captionsetup{labelformat=empty} \caption{Fig 5.1}
\end{figure}
  1. Comment on what the Normal probability plot and the \(p\)-value of the test suggest about the data. The researcher uses software to produce a \(99 \%\) confidence interval for the mean percentage of copper in the alloy, based on the \(t\) distribution. The output from the software is shown in Fig 5.2. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{0de8222f-7df5-4e17-ab68-0f9d84fc615d-5_1058_615_434_726} \captionsetup{labelformat=empty} \caption{Fig 5.2}
    \end{figure}
  2. State the confidence interval which the software gives, in the form \(a < \mu < b\).
  3. (A) State an assumption necessary for the use of the \(t\) distribution in the construction of this confidence interval.
    (B) State whether the assumption in part (iii) (A) seems reasonable.
  4. Does the confidence interval suggest that the copper content is different from \(11.5 \%\), on average? Explain your answer.
  5. In the output from the software shown in Fig 5.2, SE stands for 'standard error'.
    (A) Explain what a standard error is.
    (B) Show how the standard error was calculated in this case.
  6. Suggest a way in which the researcher could produce a narrower confidence interval.
OCR MEI Further Statistics B AS Specimen Q6
8 marks Standard +0.3
6 The table below shows the mean and variance of the test scores of a random samples of 70 girls who are starting an A level Mathematics course.
Sample meanSample variance
118.8686.57
  1. Showing your working, find a \(95 \%\) confidence interval for the population mean.
  2. Explain why you can construct the interval in part (i) despite no information about the distribution of the parent population being given.
  3. The same random sample of girls repeats the test. The mean improvement in score is 0.9 . The \(95 \%\) confidence interval for the improvement is \([ - 1.5,3.3 ]\). What is the sample variance for the improvement in score?
OCR MEI Further Statistics Major 2019 June Q4
7 marks Moderate -0.3
4 Shellfish in the sea near nuclear power stations are regularly monitored for levels of radioactivity. On a particular occasion, the levels of caesium-137 (a radioactive isotope) in a random sample of 8 cockles, measured in becquerels per kilogram, were as follows. \(\begin{array} { l l l l l l l l } 2.36 & 2.97 & 2.69 & 3.00 & 2.51 & 2.45 & 2.21 & 2.63 \end{array}\) Software is used to produce a 95\% confidence interval for the level of caesium-137 in the cockles. The output from the software is shown in Fig. 4. The value for 'SE' has been deliberately omitted. T Estimate of a Mean
Confidence Level 0.95 Sample
Mean 2.6025
s 0.2793
□
0.2793 N □ 8 Result T Estimate of a Mean \begin{table}[h]
Mean2.6025
s0.2793
SE
N8
df7
Interval\(2.6025 \pm 0.2335\)
\captionsetup{labelformat=empty} \caption{Fig. 4}
\end{table}
  1. State an assumption necessary for the use of the \(t\) distribution in the construction of this confidence interval.
  2. State the confidence interval which the software gives in the form \(a < \mu < b\).
  3. In the software output shown in Fig. 4, SE stands for standard error. Find the standard error in this case.
  4. Show how the value of 0.2335 in the confidence interval was calculated.
  5. State how, using this sample, a wider confidence interval could be produced.
OCR MEI Further Statistics Major 2019 June Q7
11 marks Standard +0.3
7 A swimming coach believes that times recorded by people using stopwatches are on average 0.2 seconds faster than those recorded by an electronic timing system. In order to test this, the coach takes a random sample of 40 competitors' times recorded by both methods, and finds the differences between the times recorded by the two methods. The mean difference in the times (electronic time minus stopwatch time) is 0.1442 s and the standard deviation of the differences is 0.2580 s .
  1. Find a 95\% confidence interval for the mean difference between electronic and stopwatch times.
  2. Explain whether there is evidence to suggest that the coach's belief is correct.
  3. Explain how you can calculate the confidence interval in part (a) even though you do not know the distribution of the parent population of differences.
  4. If the coach wanted to produce a \(95 \%\) confidence interval of width no more than 0.12 s , what is the minimum sample size that would be needed, assuming that the standard deviation remains the same?
OCR MEI Further Statistics Major 2023 June Q5
13 marks Standard +0.3
5 Amari is investigating how accurately people can estimate a short time period. He asks each of a random sample of 40 people to estimate a period of 20 seconds. For each person, he starts a stopwatch and then stops it when they tell him that they think that 20 s has elapsed. The times which he records are denoted by \(x \mathrm {~s}\). You are given that \(\sum x = 765 , \quad \sum x ^ { 2 } = 15065\).
  1. Determine a 95\% confidence interval for the mean estimated time.
  2. Amari says that the confidence interval supports the suggestion that people can estimate 20 s accurately. Make two comments about Amari's statement.
  3. Discuss whether you could have constructed the confidence interval if there had only been 10 people involved in the experiment. Amari thinks that people would be able to estimate more accurately if he gave them a second attempt. He repeats the experiment with each person and again records the times. Software is used to produce a \(95 \%\) confidence interval for the mean estimated time. The output from the software is shown below. Z Estimate of a Mean Confidence level 0.95 Sample
    Mean19.68
    s1.38
    N40
    Result
    Z Estimate of a Mean
    Mean19.68
    s1.38
    SE0.2182
    N40
    Interval\(19.68 \pm 0.4277\)
  4. State the confidence interval in the form \(\mathrm { a } < \mu < \mathrm { b }\).
  5. Make two comments based on this confidence interval about Amari's opinion that second attempts result in more accurate estimates.
OCR MEI Further Statistics Major 2024 June Q5
10 marks Standard +0.3
5 A researcher is investigating whether doing yoga has any effect on quality of sleep in older people. The researcher selects a random sample of 40 older people, who then complete a yoga course. Before they start the course and again at the end, the 40 people fill in a questionnaire which measures their perceived sleep quality. The higher the score, the better is the perceived quality of sleep. The researcher uses software to produce a 90\% confidence interval for the difference in mean sleep quality (sleep quality after the course minus sleep quality before the course). The output from the software is shown below. Z Estimate of a Mean Confidence level □ 0.9 Sample
Mean0.586
\(s\)2.14
40
Result
Z Estimate of a Mean
Mean0.586
s2.14
SE0.3384
N40
Lower limit0.029
Upper limit1.143
Interval\(0.586 \pm 0.557\)
  1. Explain why the confidence interval is based on the Normal distribution even though the distribution of the population of differences is not known.
  2. Explain whether the confidence interval suggests that the mean sleep qualities before and after completing a yoga course are different.
  3. In the output from the software, SE stands for 'standard error'.
    1. Explain what standard error is.
    2. Show how the standard error was calculated in this case.
  4. A colleague of the researcher suggests that the confidence level should have been \(95 \%\) rather than \(90 \%\). Determine whether this would have made a difference to your answer to part (b).
OCR MEI Further Statistics Major 2020 November Q4
6 marks Moderate -0.3
4 An amateur meteorologist records the total rainfall at her home each day using a traditional rain gauge. This means that she has to go out each day at 9 am to read the rain gauge and then to empty it. She wants to save time by using a digital rain gauge, but she also wants to ensure that the readings from the digital gauge are similar to those of her traditional gauge. Over a period of 100 days, she uses both gauges to measure the rainfall. The meteorologist uses software to produce a 95\% confidence interval for the difference between the two readings (the traditional gauge reading minus the digital gauge reading). The output from the software is shown in Fig. 4. Although rainfall was measured over a period of 100 days, there was no rain on 40 of those days and so the sample size in the software output is 60 rather than 100. \begin{table}[h]
Z Estimate of a Mean
Confidence Level
□
0.95
Sample
Mean 0.1173
□
Result
Z Estimate of a Mean
Mean0.1173
\(\sigma\)0.5766
SE0.07444
N60
Lower Limit-0.0286
Upper Limit0.2632
Interval\(0.1173 \pm 0.1459\)
\captionsetup{labelformat=empty} \caption{Fig. 4}
\end{table}
  1. Explain why this confidence interval can be calculated even though nothing is known about the distribution of the population of differences.
  2. State the confidence interval which the software gives in the form \(a < \mu < b\).
  3. Show how the value 0.07444 (labelled SE) was calculated.
  4. Comment on whether you think that the confidence interval suggests that the two different methods of measurement are broadly in agreement.
OCR MEI Further Statistics Major 2020 November Q7
9 marks Moderate -0.8
7 The lengths in mm of a random sample of 6 one-year-old fish of a particular species are as follows. \(\begin{array} { l l l l l l } 271 & 293 & 306 & 287 & 264 & 290 \end{array}\)
  1. State an assumption required in order to find a confidence interval for the mean length of one-year-old fish of this species. Fig. 7 shows a Normal probability plot for these data. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{8d36bc92-07ac-40c3-9e75-26f2bc9d2fcc-07_599_753_646_246} \captionsetup{labelformat=empty} \caption{Fig. 7}
    \end{figure}
  2. Explain why the Normal probability plot suggests that the assumption in part (a) may be valid.
  3. In this question you must show detailed reasoning. Assuming that this assumption is true, find a 95\% confidence interval for the mean length of one-year-old fish of this species.
OCR MEI Further Statistics Major 2021 November Q1
6 marks Standard +0.3
1 When babies are born, their head circumferences are measured. A random sample of 50 newborn female babies is selected. The sample mean head circumference is 34.711 cm . The sample standard deviation head circumference is 1.530 cm .
  1. Determine a 95\% confidence interval for the population mean head circumference of newborn female babies.
  2. Explain why you can calculate this interval even though the distribution of the population of head circumferences of newborn female babies is unknown.
OCR MEI Further Statistics Major 2021 November Q7
10 marks Standard +0.3
7 A physiotherapist is investigating hand grip strength in adult women under 30 years old. She thinks that the grip strength of the dominant hand will be on average 2 kg higher than the grip strength of the non-dominant hand. The physiotherapist selects a random sample of 12 adult women under 30 years old and measures the grip strength of each of their hands. She then uses software to produce a \(95 \%\) confidence interval for the mean difference in grip strength between the two hands (dominant minus nondominant), as shown in Fig. 7. \begin{table}[h]
T Estimate of a Mean
Confidence Level0.95
Sample
\multirow{3}{*}{
}
Result
T Estimate of a Mean
Mean2.79
s3.92
SE1.13161
N12
df11
Lower Limit0.29935
Upper Limit5.28065
Interval\(2.79 \pm 2.49065\)
\captionsetup{labelformat=empty} \caption{Fig. 7} \end{table}
  1. Explain why the physiotherapist used the same people for testing their dominant and nondominant grip strengths.
  2. State any assumptions necessary in order to construct the confidence interval shown in Fig. 7.
  3. Explain whether the confidence interval supports the physiotherapist's belief.
  4. The physiotherapist then finds some data which have previously been collected on grip strength using a sample of 100 adult women. A 95\% confidence interval, based on this sample and calculated using a Normal distribution, for the mean difference in grip strength between the two hands (dominant minus non-dominant) is (1.94, 2.84).
    1. For this sample, find
WJEC Further Unit 5 2023 June Q3
11 marks Standard +0.3
3. Athletes who compete in the 400 m event have resting heart rates (RHR), measured in beats per minute, which are normally distributed with known standard deviation \(4 \cdot 7\). A random sample of 90 athletes who compete in the 400 m event is taken. Their resting heart rates are summarised by $$\sum x = 4014 \quad \text { and } \quad \sum x ^ { 2 } = 182257 .$$
  1. Find a \(99 \%\) confidence interval for the mean of the RHR of athletes who compete in the 400 m event. Give the limits of your interval correct to 1 decimal place.
  2. Without doing any further calculation, explain how the width of a \(95 \%\) confidence interval would compare to the width of your interval in part (a). Athletes who compete in the discus event have RHR which are normally distributed with known standard deviation \(\sigma\). A random sample of 100 athletes who compete in the discus event is taken. A 95\% confidence interval for the mean of the RHR is calculated as [49•4, 52•6].
  3. Determine the value of \(\sigma\) that was used to calculate this confidence interval.
  4. Referring to the confidence intervals, state, with a reason, what can be said about the RHR of athletes who compete in the 400 m event compared to the RHR of athletes who compete in the discus event.
WJEC Further Unit 5 2023 June Q5
13 marks Standard +0.3
5. The masses, \(X\), in kg, of men who work for a large company are normally distributed with mean 75 and standard deviation 10.
  1. Find the probability that the mean mass of a random sample of 5 men is less than 70 kg .
  2. The mean mass, in kg , of a random sample of \(n\) men drawn from this distribution is \(\bar { X }\). Given that \(\mathrm { P } ( \bar { X } > 80 )\) is approximately \(0 \cdot 007\), find \(n\). The masses, in kg, of women who work for the company are normally distributed with mean 68 and standard deviation 6 . A lift in the company building will not move if the total mass in the lift is more than 500 kg .
  3. A random sample of 3 men and 4 women get in the lift. Find the probability that the lift will not move.
  4. State a modelling assumption you have made in calculating your answer for part (c).
WJEC Further Unit 5 2023 June Q7
7 marks Challenging +1.2
7. Branwen intends to buy a new bike, either a Cannotrek or a Bianchondale. If there is evidence that the difference in the mean times on the two bikes over a 10 km time trial is more than 1.25 minutes, she will buy the faster bike. Otherwise, she will base her decision on other factors. She negotiates a test period to try both bikes. The times, in minutes, taken by Branwen to complete a 10 km time trial on the Cannotrek may be modelled by a normal distribution with mean \(\mu _ { C }\) and standard deviation \(0 \cdot 75\). The times, in minutes, taken by Branwen to complete a 10 km time trial on the Bianchondale may be modelled by a normal distribution with mean \(\mu _ { B }\) and standard deviation \(0 \cdot 6\). During the test period, she completes 6 time trials with a mean time of 19.5 minutes on the Cannotrek, and 5 time trials with a mean time of 17.3 minutes on the Bianchondale. She calculates a \(p \%\) confidence interval for \(\mu _ { C } - \mu _ { B }\).
  1. What would be the largest value of \(p\) that would lead Branwen to base her purchasing decision on the time trials, without considering other factors?
  2. State an assumption you have made in part (a).
AQA Further Paper 3 Statistics Specimen Q4
6 marks Standard +0.3
4 David, a zoologist, is investigating a particular species of monitor lizard. He measures the lengths, in centimetres, of a random sample of this particular species of lizard. His measured lengths are $$\begin{array} { l l l l l l l l l l } 53.2 & 57.8 & 55.3 & 58.9 & 59.0 & 60.2 & 61.8 & 62.3 & 65.4 & 66.5 \end{array}$$ The lengths may be assumed to be normally distributed.
David correctly constructed a 90\% confidence interval for the mean length of lizard using the measured lengths given and the formula \(\bar { x } \pm \left( b \times \frac { s } { \sqrt { n } } \right)\) This interval had limits of 57.63 and 62.45, correct to two decimal places.
4
  1. State the value for \(b\) used in David's formula. 4
  2. David interprets his interval and states,
    "My confidence interval indicates that exactly 90\% of the population of lizard lengths for this particular species lies between 57.63 cm and \(62.45 \mathrm {~cm} ^ { \prime \prime }\). Do you think David's statement is true? Explain your reasoning. 4
  3. David's assistant, Amina, correctly constructs a \(\beta \%\) confidence interval from David's random sample of measured lengths. Amina informs David that the width of her confidence interval is 8.54 .
    Find the value of \(\beta\).
    [0pt] [3 marks]
    Turn over for the next question