5.05d Confidence intervals: using normal distribution

4 A new computer was bought by a local council to search council records and was tested by an employee. She searched a random sample of 500 records and the sample mean search time was found to be 2.18 milliseconds and an unbiased estimate of variance was \(1.58 ^ { 2 }\) milliseconds \({ } ^ { 2 }\).

1. Calculate a \(98 \%\) confidence interval for the population mean search time \(\mu\) milliseconds.
2. It is required to obtain a sample mean time that differs from \(\mu\) by less than 0.05 milliseconds with probability 0.95 . Estimate the sample size required.
3. State why it is unnecessary for the validity of your calculations that search time has a normal distribution.

3

1. A company packages butter. Of 50 randomly selected packs, 8 were found to have damaged wrappers. Find an approximate \(95 \%\) confidence interval for the proportion of packs with damaged wrappers.
2. The mass of a pack has a normal distribution with standard deviation 8.5 g . In a random sample of 10 packs the masses, in g , are as follows. $$\begin{array} { l l l l l l l l l l } 220 & 225 & 218 & 223 & 224 & 220 & 229 & 228 & 226 & 228 \end{array}$$ Find a 99\% confidence interval for the mean mass of a pack.

4 The weights of a particular variety (A) of tomato are known to be Normally distributed with mean 80 grams and standard deviation 11 grams.

1. Find the probability that a randomly chosen tomato of variety A weighs less than 90 grams. The weights of another variety (B) of tomato are known to be Normally distributed with mean 70 grams. These tomatoes are packed in sixes using packaging that weighs 15 grams.
2. The probability that a randomly chosen pack of 6 tomatoes of variety B , including packaging, weighs less than 450 grams is 0.8463 . Show that the standard deviation of the weight of single tomatoes of variety B is 6 grams, to the nearest gram.
3. Tomatoes of variety A are packed in fives using packaging that weighs 25 grams. Find the probability that the total weight of a randomly chosen pack of variety A is greater than the total weight of a randomly chosen pack of variety B .
4. A new variety (C) of tomato is introduced. The weights, \(c\) grams, of a random sample of 60 of these tomatoes are measured giving the following results. $$\Sigma c = 3126.0 \quad \Sigma c ^ { 2 } = 164223.96$$ Find a \(95 \%\) confidence interval for the true mean weight of these tomatoes.

1 Each month the amount of electricity, measured in kilowatt-hours ( kWh ), used by a particular household is Normally distributed with mean 406 and standard deviation 12.

1. Find the probability that, in a randomly chosen month, less than 420 kWh is used. The charge for electricity used is 14.6 pence per kWh .
2. Write down the distribution of the total charge for the amount of electricity used in any one month. Hence find the probability that, in a randomly chosen month, the total charge is more than \(\pounds 60\).
3. The household receives a bill every three months. Assume that successive months may be regarded as independent of each other. Find the value of \(b\) such that the probability that a randomly chosen bill is less than \(\pounds b\) is 0.99 . In a different household, the amount of electricity used per month was Normally distributed with mean 432 kWh . This household buys a new washing machine that is claimed to be cheaper to run than the old one. Over the next six months the amounts of electricity used, in kWh , are as follows. $$\begin{array} { l l l l l l } 404 & 433 & 420 & 423 & 413 & 440 \end{array}$$
4. Treating this as a random sample, carry out an appropriate test, with a \(5 \%\) significance level, to see if there is any evidence to suggest that the amount of electricity used per month by this household has decreased on average.

4 A timber supplier cuts wooden fence posts from felled trees. The posts are of length \(( k + X ) \mathrm { cm }\) where \(k\) is a constant and \(X\) is a random variable which has probability density function $$f ( x ) = \begin{cases} 1 + x & - 1 \leqslant x < 0 \\ 1 - x & 0 \leqslant x \leqslant 1 \\ 0 & \text { elsewhere } \end{cases}$$

1. Sketch \(\mathrm { f } ( x )\).
2. Write down the value of \(\mathrm { E } ( X )\) and find \(\operatorname { Var } ( X )\).
3. Write down, in terms of \(k\), the approximate distribution of \(\bar { L }\), the mean length of a random sample of 50 fence posts. Justify your choice of distribution.
4. In a particular sample of 50 posts, the mean length is 90.06 cm . Find a \(95 \%\) confidence interval for the true mean length of the fence posts.
5. Explain whether it is reasonable to suppose that \(k = 90\).

1

1. Define simple random sampling. Describe briefly one difficulty associated with simple random sampling.
2. Freeze-drying is an economically important process used in the production of coffee. It improves the retention of the volatile aroma compounds. In order to maintain the quality of the coffee, technologists need to monitor the drying rate, measured in suitable units, at regular intervals. It is known that, for best results, the mean drying rate should be 70.3 units and anything substantially less than this would be detrimental to the coffee. Recently, a random sample of 12 observations of the drying rate was as follows. $$\begin{array} { l l l l l l l l l l l l } 66.0 & 66.1 & 59.8 & 64.0 & 70.9 & 71.4 & 66.9 & 76.2 & 65.2 & 67.9 & 69.2 & 68.5 \end{array}$$
   1. Carry out a test to investigate at the \(5 \%\) level of significance whether the mean drying rate appears to be less than 70.3. State the distributional assumption that is required for this test.
   2. Find a 95\% confidence interval for the true mean drying rate.

2 In a particular chain of supermarkets, one brand of pasta shapes is sold in small packets and large packets. Small packets have a mean weight of 505 g and a standard deviation of 11 g . Large packets have a mean weight of 1005 g and a standard deviation of 17 g . It is assumed that the weights of packets are Normally distributed and are independent of each other.

1. Find the probability that a randomly chosen large packet weighs between 995 g and 1020 g .
2. Find the probability that the weights of two randomly chosen small packets differ by less than 25 g .
3. Find the probability that the total weight of two randomly chosen small packets exceeds the weight of a randomly chosen large packet.
4. Find the probability that the weight of one randomly chosen small packet exceeds half the weight of a randomly chosen large packet by at least 5 g .
5. A different brand of pasta shapes is sold in packets of which the weights are assumed to be Normally distributed with standard deviation 14 g . A random sample of 20 packets of this pasta is found to have a mean weight of 246 g . Find a \(95 \%\) confidence interval for the population mean weight of these packets.

1 A certain industrial process requires a supply of water. It has been found that, for best results, the mean water pressure in suitable units should be 7.8. The water pressure is monitored by taking measurements at regular intervals. On a particular day, a random sample of the measurements is as follows. $$\begin{array} { l l l l l l l l l } 7.50 & 7.64 & 7.68 & 7.51 & 7.70 & 7.85 & 7.34 & 7.72 & 7.74 \end{array}$$ These data are to be used to carry out a hypothesis test concerning the mean water pressure.

1. Why is a test based on the Normal distribution not appropriate in this case?
2. What distributional assumption is needed for a test based on the \(t\) distribution?
3. Carry out a \(t\) test, with a \(2 \%\) level of significance, to see whether it is reasonable to assume that the mean pressure is 7.8 .
4. Explain what is meant by a \(95 \%\) confidence interval.
5. Find a \(95 \%\) confidence interval for the actual mean water pressure.

3 In the manufacture of child car seats, a resin made up of three ingredients is used. The ingredients are two polymers and an impact modifier. The resin is prepared in batches. Each ingredient is supplied by a separate feeder and the amount supplied to each batch, in kg, is assumed to be Normally distributed with mean and standard deviation as shown in the table below. The three feeders are also assumed to operate independently of each other.

1. Find the probability that, in a randomly chosen batch of resin, there is no more than 2100 kg of polymer 1.
2. Find the probability that, in a randomly chosen batch of resin, the amount of polymer 1 exceeds the amount of polymer 2 by at least 400 kg .
3. Find the value of \(b\) such that the total amount of the ingredients in a randomly chosen batch exceeds \(b \mathrm {~kg} 95 \%\) of the time.
4. Polymer 1 costs \(\pounds 1.20\) per kg, polymer 2 costs \(\pounds 1.30\) per kg and the impact modifier costs \(\pounds 0.80\) per kg. Find the mean and variance of the total cost of a batch of resin.
5. Each batch of resin is used to make a large number of car seats from which a random sample of 50 seats is selected in order that the tensile strength (in suitable units) of the resin can be measured. From one such sample, the \(99 \%\) confidence interval for the true mean tensile strength of the resin in that batch was calculated as \(( 123.72,127.38 )\). Find the mean and standard deviation of the sample.

2 Pat makes and sells fruit cakes at a local market. On her stall a sign states that the average weight of the cakes is 1 kg . A trading standards officer carries out a routine check of a random sample of 8 of Pat's cakes to ensure that they are not underweight, on average. The weights, in kg , that he records are as follows. $$\begin{array} { l l l l l l l l } 0.957 & 1.055 & 0.983 & 0.917 & 1.015 & 0.865 & 1.013 & 0.854 \end{array}$$

1. On behalf of the trading standards officer, carry out a suitable test at a \(5 \%\) level of significance, stating your hypotheses clearly. Assume that the weights of Pat's fruit cakes are Normally distributed.
2. Find a 95\% confidence interval for the true mean weight of Pat's fruit cakes. Pat's husband, Tony, is the owner of a factory which makes and supplies fruit cakes to a large supermarket chain. A large random sample of \(n\) of these cakes has mean weight \(\bar { x } \mathrm {~kg}\) and variance \(0.006 \mathrm {~kg} ^ { 2 }\).
3. Write down, in terms of \(n\) and \(\bar { x }\), a \(95 \%\) confidence interval for the true mean weight of cakes produced in Tony's factory.
4. What is the size of the smallest sample that should be taken if the width of the confidence interval in part (iii) is to be 0.025 kg at most?

1 Gerry runs 5000 -metre races for his local athletics club. His coach has been monitoring his practice times for several months and he believes that they can be modelled using a Normal distribution with mean 15.3 minutes. The coach suggests that Gerry should try running with a pacemaker in order to see if this can improve his times. Subsequently a random sample of 10 of Gerry's times with the pacemaker is collected to see if any reduction has been achieved. The sample of times (in minutes) is as follows. $$\begin{array} { l l l l l l l l l l } 14.86 & 15.00 & 15.62 & 14.44 & 15.27 & 15.64 & 14.58 & 14.30 & 15.08 & 15.08 \end{array}$$

1. Why might a \(t\) test be used for these data?
2. Using a \(5 \%\) significance level, carry out the test to see whether, on average, Gerry's times have been reduced.
3. What is meant by 'a \(5 \%\) significance level'? What would be the consequence of decreasing the significance level?
4. Find a \(95 \%\) confidence interval for the true mean of Gerry's times using a pacemaker.

4 The weights of Avonley Blue cheeses made by a small producer are found to be Normally distributed with mean 10 kg and standard deviation 0.4 kg .

1. Find the probability that a randomly chosen cheese weighs less than 9.5 kg . One particular shop orders four Avonley Blue cheeses each week from the producer. From experience, the shopkeeper knows that the weekly demand from customers for Avonley Blue cheese is Normally distributed with mean 35 kg and standard deviation 3.5 kg . In the interests of food hygiene, no cheese is kept by the shopkeeper from one week to the next.
2. Find the probability that, in a randomly chosen week, demand from customers for Avonley Blue will exceed the supply. Following a campaign to promote Avonley Blue cheese, the shopkeeper finds that the weekly demand for it has increased by \(30 \%\) (i.e. the mean and standard deviation are both increased by \(30 \%\) ). Therefore the shopkeeper increases his weekly order by one cheese.
3. Find the probability that, in a randomly chosen week, demand will now exceed supply.
4. Following complaints, the cheese producer decides to check the mean weight of the Avonley Blue cheeses. For a random sample of 12 cheeses, she finds that the mean weight is 9.73 kg . Assuming that the population standard deviation of the weights is still 0.4 kg , find a \(95 \%\) confidence interval for the true mean weight of the cheeses and comment on the result. Explain what is meant by a 95\% confidence interval. RECOGNISING ACHIEVEMENT

1 Technologists at a company that manufactures paint are trying to develop a new type of gloss paint with a shorter drying time than the current product. In order to test whether the drying time has been reduced, the technologists paint a square metre of each of the new and old paints on each of 10 different surfaces. The lengths of time, in hours, that each square metre takes to dry are as follows.

1. Explain why a paired sample is used in this context.
2. The mean reduction in drying time is to be investigated. Why might a \(t\) test be appropriate in this context and what assumption needs to be made?
3. Using a significance level of \(5 \%\), carry out a test to see if there appears to be any reduction in mean drying time.
4. Find a 95\% confidence interval for the true mean reduction in drying time.

3 The triathlon is a sports event in which competitors take part in three stages, swimming, cycling and running, one straight after the other. The winner is the competitor with the shortest overall time. In this question the times for the separate stages are assumed to be Normally distributed and independent of each other. For a particular triathlon event in which there was a very large number of competitors, the mean and standard deviation of the times, measured in minutes, for each stage were as follows.

1. For a randomly chosen competitor, find the probability that the swimming time is between 10 and 13 minutes.
2. For a randomly chosen competitor, find the probability that the running time exceeds the swimming time by more than 10 minutes.
3. For a randomly chosen competitor, find the probability that the swimming and running times combined exceed \(\frac { 2 } { 3 }\) of the cycling time.
4. In a different triathlon event the total times, in minutes, for a random sample of 12 competitors were as follows. $$\begin{array} { l l l l l l l l l l l l } 103.59 & 99.04 & 85.03 & 81.34 & 106.79 & 89.14 & 98.55 & 98.22 & 108.87 & 116.29 & 102.51 & 92.44 \end{array}$$ Find a 95\% confidence interval for the mean time of all competitors in this event.
5. Discuss briefly whether the assumptions of Normality and independence for the stages of triathlon events are reasonable.

1 In the past, the times for workers in a factory to complete a particular task had a known median of 7.4 minutes. Following a review, managers at the factory wish to know if the median time to complete the task has been reduced.

1. A random sample of 12 times, in minutes, gives the following results. $$\begin{array} { l l l l l l l l l l l l } 6.90 & 7.23 & 6.54 & 7.62 & 7.04 & 7.33 & 6.74 & 6.45 & 7.81 & 7.71 & 7.50 & 6.32 \end{array}$$ Carry out an appropriate test using a \(5 \%\) level of significance.
2. Some time later, a much larger random sample of times gives the following results. $$n = 80 \quad \sum x = 555.20 \quad \sum x ^ { 2 } = 3863.9031$$ Find a \(95 \%\) confidence interval for the true mean time for the task. Justify your choice of which distribution to use.
3. Describe briefly one advantage and one disadvantage of having a \(99 \%\) confidence interval instead of a \(95 \%\) confidence interval.

2 A company supplying cattle feed to dairy farmers claims that its new brand of feed will increase average milk yields by 10 litres per cow per week. A farmer thinks the increase will be less than this and decides to carry out a statistical investigation using a paired \(t\) test. A random sample of 10 dairy cows are given the new feed and then their milk yields are compared with their yields when on the old feed. The yields, in litres per week, for the 10 cows are as follows.

1. Why is it sensible to use a paired test?
2. State the condition necessary for a paired \(t\) test.
3. Assuming the condition stated in part (ii) is met, carry out the test, using a significance level of \(5 \%\), to see whether it appears that the company's claim is justified.
4. Find a 95\% confidence interval for the mean increase in the milk yield using the new feed.

4 The probability density function of a random variable \(X\) is given by $$\mathrm { f } ( x ) = \begin{cases} k x & 0 \leqslant x \leqslant a \\ k ( 2 a - x ) & a < x \leqslant 2 a \\ 0 & \text { otherwise } \end{cases}$$ where \(a\) and \(k\) are positive constants.

1. Sketch \(\mathrm { f } ( x )\). Hence explain why \(\mathrm { E } ( X ) = a\).
2. Show that \(k = \frac { 1 } { a ^ { 2 } }\).
3. Find \(\operatorname { Var } ( X )\) in terms of \(a\). In order to estimate the value of \(a\), a random sample of size 50 is taken from the distribution. It is found that the sample mean and standard deviation are \(\bar { x } = 1.92\) and \(s = 0.8352\).
4. Construct a symmetrical \(95 \%\) confidence interval for \(a\). Give one reason why the answer is only approximate.
5. A non-statistician states that the probability that \(a\) lies in the interval found in part (iv) is 0.95 . Comment on this statement. \section*{END OF QUESTION PAPER} \section*{OCR \(^ { \text {® } }\)}

1 An industrial process produces components. Some of the components contain faults. The number of faults in a component is modelled by the random variable \(X\) with probability function $$\mathrm { P } ( X = x ) = \theta ( 1 - \theta ) ^ { x } \quad \text { for } x = 0,1,2 , \ldots$$ where \(\theta\) is a parameter with \(0 < \theta < 1\). The numbers of faults in different components are independent.
A random sample of \(n\) components is inspected. \(n _ { 0 }\) are found to have no faults, \(n _ { 1 }\) to have one fault and the remainder \(\left( n - n _ { 0 } - n _ { 1 } \right)\) to have two or more faults.

1. Find \(\mathrm { P } ( X \geqslant 2 )\) and hence show that the likelihood is $$\mathrm { L } ( \theta ) = \theta ^ { n _ { 0 } + n _ { 1 } } ( 1 - \theta ) ^ { 2 n - 2 n _ { 0 } - n _ { 1 } }$$
2. Find the maximum likelihood estimator \(\hat { \theta }\) of \(\theta\). You are not required to verify that any turning point you locate is a maximum.
3. Show that \(\mathrm { E } ( X ) = \frac { 1 - \theta } { \theta }\). Deduce that another plausible estimator of \(\theta\) is \(\tilde { \theta } = \frac { 1 } { 1 + \bar { X } }\) where \(\bar { X }\) is the sample mean. What additional information is needed in order to calculate the value of this estimator?
4. You are given that, in large samples, \(\tilde { \theta }\) may be taken as Normally distributed with mean \(\theta\) and variance \(\theta ^ { 2 } ( 1 - \theta ) / n\). Use this to obtain a \(95 \%\) confidence interval for \(\theta\) for the case when 100 components are inspected and it is found that 92 have no faults, 6 have one fault and the remaining 2 have exactly four faults each.

1 The random variable \(X\) has the Normal distribution with mean 0 and variance \(\theta\), so that its probability density function is $$\mathrm { f } ( x ) = \frac { 1 } { \sqrt { 2 \pi \theta } } \mathrm { e } ^ { - x ^ { 2 } / 2 \theta } , \quad - \infty < x < \infty$$ where \(\theta ( \theta > 0 )\) is unknown. A random sample of \(n\) observations from \(X\) is denoted by \(X _ { 1 } , X _ { 2 } , \ldots , X _ { n }\).

1. Find \(\hat { \theta }\), the maximum likelihood estimator of \(\theta\).
2. Show that \(\hat { \theta }\) is an unbiased estimator of \(\theta\).
3. In large samples, the variance of \(\hat { \theta }\) may be estimated by \(\frac { 2 \hat { \theta } ^ { 2 } } { n }\). Use this and the results of parts (i) and (ii) to find an approximate \(95 \%\) confidence interval for \(\theta\) in the case when \(n = 100\) and \(\Sigma X _ { i } ^ { 2 } = 1000\).

3

1. Explain the meaning of the following terms in the context of hypothesis testing: Type I error, Type II error, operating characteristic, power.
2. A market research organisation is designing a sample survey to investigate whether expenditure on everyday food items has increased in 2011 compared with 2010. For one of the populations being studied, the random variable \(X\) is used to model weekly expenditure, in \(\pounds\), on these items in 2011, where \(X\) is Normally distributed with mean \(\mu\) and variance \(\sigma ^ { 2 }\). As the corresponding mean value in 2010 was 94 , the hypotheses to be examined are $$\begin{aligned} & \mathrm { H } _ { 0 } : \mu = 94 \\ & \mathrm { H } _ { 1 } : \mu > 94 \end{aligned}$$ By comparison with the corresponding 2010 value, \(\sigma ^ { 2 }\) is assumed to be 25 .
   The following criteria for the survey are laid down.
   A random sample of size \(n\) is to be taken and the usual Normal test based on \(\bar { X }\) is to be used, with a critical value of \(c\) such that \(\mathrm { H } _ { 0 }\) is rejected if the value of \(\bar { X }\) exceeds \(c\). Find \(c\) and the smallest value of \(n\) that is required.
3. Sketch the power function of an ideal test for examining the hypotheses in part (ii).

6 The times taken by employees in a factory to complete a certain task have a normal distribution with mean \(\mu\) seconds and standard deviation \(\sigma\) seconds, both of which are unknown. Based on a random sample of 20 employees, the symmetric \(95 \%\) confidence interval for \(\mu\) is \(( 481,509 )\). Calculate a symmetric \(90 \%\) confidence interval for \(\mu\).
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8 Part of a research study of identical twins who had been separated at birth involved a random sample of 9 pairs, in which one twin had been raised by the natural parents and the other by adoptive parents. The IQ scores of these twins were measured, with the following results. It may be assumed that the difference in IQ scores has a normal distribution. The mean IQ scores of separated twins raised by natural parents and by adoptive parents are denoted by \(\mu _ { N }\) and \(\mu _ { A }\) respectively. Obtain a \(90 \%\) confidence interval for \(\mu _ { N } - \mu _ { A }\). One of the researchers claimed that there was no evidence of a difference between the two population means. State, giving a reason, whether the confidence interval supports this claim.

A study was made of the acidity levels in farmland on opposite sides of an island. The levels were measured at six randomly chosen points on the eastern side and at five randomly chosen points on the western side. The values obtained, in suitable units, are denoted by \(x _ { E }\) and \(x _ { W }\) respectively. The sample means \(\bar { x } _ { E }\) and \(\bar { x } _ { W }\), and unbiased estimates of the two population variances, \(s _ { E } ^ { 2 }\) and \(s _ { W } ^ { 2 }\), are as follows. $$\bar { x } _ { E } = 5.035 , s _ { E } ^ { 2 } = 0.0231 , \bar { x } _ { W } = 4.782 , s _ { W } ^ { 2 } = 0.0195 .$$ The population means on the eastern and western sides are denoted by \(\mu _ { E }\) and \(\mu _ { W }\) respectively. State suitable hypotheses for a test for a difference between the mean acidity levels on the two sides of the island. Stating any required assumptions, obtain the rejection region for a test at the \(5 \%\) significance level of whether the mean acidity levels differ on the two sides of the island. Give the conclusion of the test. Find the largest value of \(a\) for which the samples above provide evidence at the \(5 \%\) significance level that \(\mu _ { E } - \mu _ { W } > a\).

8 An examination involved writing an essay. In order to compare the time taken to write the essay by students in two large colleges, a sample of 12 students from college \(A\) and a sample of 8 students from college \(B\) were randomly selected. The times, \(t _ { A }\) and \(t _ { B }\), taken for these students to write the essay were measured, correct to the nearest minute, and are summarised by $$n _ { A } = 12 , \quad \Sigma t _ { A } = 257 , \quad \Sigma t _ { A } ^ { 2 } = 5629 , \quad n _ { B } = 8 , \quad \Sigma t _ { B } = 206 , \quad \Sigma t _ { B } ^ { 2 } = 5359$$ Stating any required assumptions, calculate a \(95 \%\) confidence interval for the difference in the population means. State, giving a reason, whether your confidence interval supports the statement that the population means, for the two colleges, are equal.

8 A company decides that its employees should follow an exercise programme for 30 minutes each day, with the aim that they lose weight and increase productivity. The weights, in kg , of a random sample of 8 employees at the start of the programme and after following the programme for 6 weeks are shown in the table. Assuming that loss in weight is normally distributed, find a 95\% confidence interval for the mean loss in weight of the company's employees. Test at the \(5 \%\) significance level whether, after the exercise programme, there is a reduction of more than 2.5 kg in the population mean weight.