Unbiased estimates then CI

1 The diameters, \(x\) millimetres, of a random sample of 200 discs made by a certain machine were recorded. The results are summarised below. $$n = 200 \quad \Sigma x = 2520 \quad \Sigma x ^ { 2 } = 31852$$

1. Calculate a 95\% confidence interval for the population mean diameter.
2. Jean chose 40 random samples and used each sample to calculate a 95\% confidence interval for the population mean diameter. How many of these 40 confidence intervals would be expected to include the true value of the population mean diameter?

4 The masses, in grams, of a certain type of plum are normally distributed with mean \(\mu\) and variance \(\sigma ^ { 2 }\). The masses, \(m\) grams, of a random sample of 150 plums of this type were found and the results are summarised by \(\Sigma m = 9750\) and \(\Sigma m ^ { 2 } = 647500\).

1. Calculate unbiased estimates of \(\mu\) and \(\sigma ^ { 2 }\).
2. Calculate a 98\% confidence interval for \(\mu\). Two more random samples of plums of this type are taken and a \(98 \%\) confidence interval for \(\mu\) is calculated from each sample.
3. Find the probability that neither of these two intervals contains \(\mu\).

5 The masses, \(m\) grams, of a random sample of 80 strawberries of a certain type were measured and summarised as follows. $$n = 80 \quad \Sigma m = 4200 \quad \Sigma m ^ { 2 } = 229000$$

1. Find unbiased estimates of the population mean and variance.
2. Calculate a 98\% confidence interval for the population mean. 50 random samples of size 80 were taken and a \(98 \%\) confidence interval for the population mean, \(\mu\), was found from each sample.
3. Find the number of these 50 confidence intervals that would be expected to include the true value of \(\mu\).

5 The volumes, \(v\) millilitres, of juice in a random sample of 50 bottles of Cooljoos are measured and summarised as follows. $$n = 50 \quad \Sigma v = 14800 \quad \Sigma v ^ { 2 } = 4390000$$

1. Find unbiased estimates of the population mean and variance.
2. An \(\alpha \%\) confidence interval for the population mean, based on this sample, is found to have a width of 5.45 millilitres. Find \(\alpha\). Four random samples of size 10 are taken and a \(96 \%\) confidence interval for the population mean is found from each sample.
3. Find the probability that these 4 confidence intervals all include the true value of the population mean.

5 The 150 oranges in a random sample from a certain supplier were weighed and the masses, \(X\) grams, were recorded. The results are summarised below. $$n = 150 \quad \Sigma x = 14910 \quad \Sigma x ^ { 2 } = 1525000$$

1. Calculate a \(99 \%\) confidence interval for the population mean of \(X\).
2. The supplier claims that the mean mass of his oranges is 100 grams. Use your answer to part (i) to explain whether this claim should be accepted.
3. State briefly why the sample should be random.

3 The masses, in grams, of bags of flour are normally distributed with mean \(\mu\). The masses, \(m\) grams, of a random sample of 50 bags are summarised by \(\Sigma m = 25110\) and \(\Sigma m ^ { 2 } = 12610300\).

1. Calculate a \(96 \%\) confidence interval for \(\mu\), giving the end-points correct to 1 decimal place.
   Another random sample of 50 bags of flour is taken and a \(99 \%\) confidence interval for \(\mu\) is calculated.
2. Without calculation, state whether this confidence interval will be wider or narrower than the confidence interval found in part (i). Give a reason for your answer.

4 Diameters of golf balls are known to be normally distributed with mean \(\mu \mathrm { cm }\) and standard deviation \(\sigma \mathrm { cm }\). A random sample of 130 golf balls was taken and the diameters, \(x \mathrm {~cm}\), were measured. The results are summarised by \(\Sigma x = 555.1\) and \(\Sigma x ^ { 2 } = 2371.30\).

1. Calculate unbiased estimates of \(\mu\) and \(\sigma ^ { 2 }\).
2. Calculate a \(97 \%\) confidence interval for \(\mu\).
3. 300 random samples of 130 balls are taken and a \(97 \%\) confidence interval is calculated for each sample. How many of these intervals would you expect not to contain \(\mu\) ?

4 The volumes of juice in bottles of Apricola are normally distributed. In a random sample of 8 bottles, the volumes of juice, in millilitres, were found to be as follows. $$\begin{array} { l l l l l l l l } 332 & 334 & 330 & 328 & 331 & 332 & 329 & 333 \end{array}$$

1. Find unbiased estimates of the population mean and variance. A random sample of 50 bottles of Apricola gave unbiased estimates of 331 millilitres and 4.20 millilitres \({ } ^ { 2 }\) for the population mean and variance respectively.
2. Use this sample of size 50 to calculate a \(98 \%\) confidence interval for the population mean.
3. The manufacturer claims that the mean volume of juice in all bottles is 333 millilitres. State, with a reason, whether your answer to part (ii) supports this claim.

6 A variable \(X\) takes values \(1,2,3,4,5\), and these values are generated at random by a machine. Each value is supposed to be equally likely, but it is suspected that the machine is not working properly. A random sample of 100 values of \(X\), generated by the machine, gives the following results. $$n = 100 \quad \Sigma x = 340 \quad \Sigma x ^ { 2 } = 1356$$

1. Find a 95\% confidence interval for the population mean of the values generated by the machine.
2. Use your answer to part (i) to comment on whether the machine may be working properly.

3 A random sample of 80 precision-engineered cylindrical components is checked as part of a quality control process. The diameters of the cylinders should be 25.00 cm . Accurate measurements of the diameters, \(x \mathrm {~cm}\), for the sample are summarised by $$\Sigma ( x - 25 ) = 0.44 , \quad \Sigma ( x - 25 ) ^ { 2 } = 0.2287 .$$

1. Calculate a \(99 \%\) confidence interval for the population mean diameter of the components.
2. For the calculation in part (i) to be valid, is it necessary to assume that component diameters are normally distributed? Justify your answer.

1. The weights, \(x \mathrm {~kg}\), of each of 10 watermelons selected at random from Priya's shop were recorded. The results are summarised as follows

$$\sum x = 114.2 \quad \sum x ^ { 2 } = 1310.464$$

1. Calculate unbiased estimates of the mean and the variance of the weights of the watermelons in Priya's shop. Priya researches the weight of watermelons, for the variety she has in her shop, and discovers that the weights of these watermelons are normally distributed with a standard deviation of 0.8 kg
2. Calculate a \(95 \%\) confidence interval for the mean weight of watermelons in Priya's shop. Give the limits of your confidence interval to 2 decimal places. Priya claims that the confidence interval in part (b) suggests that nearly all of the watermelons in her shop weigh more than 10.5 kg
3. Use your answer to part (b) to estimate the smallest proportion of watermelons in her shop that weigh less than 10.5 kg

1 A machine fills bottles with mineral water.
The machine is checked every day to ensure that it is working correctly. On a particular day a random sample of 100 bottles is taken. The volume of water, \(x\) millilitres, for each bottle is measured and each measurement is coded using $$y = x - 1000$$ The results are summarised below $$\sum y = 847 \quad \sum y ^ { 2 } = 13510.09$$

1. 1. Show that the value of the unbiased estimate of the mean of \(x\) is 1008.47
   2. Calculate the unbiased estimate of the variance of \(x\) The machine was initially set so that the volume of water in a bottle had a mean value of 1010 millilitres. Later, a test at the \(5 \%\) significance level is used to determine whether or not the mean volume of water in a bottle has changed. If it has changed then the machine is stopped and reset.
2. Write down suitable null and alternative hypotheses for a 2-tailed test.
3. Find the critical region for \(\bar { X }\) in the above test.
4. Using your answer to part (a) and your critical region found in part (c), comment on whether or not the machine needs to be stopped and reset.
   Give a reason for your answer.
5. Explain why the use of \(\sigma ^ { 2 } = s ^ { 2 }\) is reasonable in this situation.

5. Paul takes the company bus to work. According to the bus timetable he should arrive at work at 0831. Paul believes the bus is not reliable and often arrives late. Paul decides to test the arrival time of the bus and carries out a survey. He records the values of the random variable $$X = \text { number of minutes after } 0831 \text { when the bus arrives. }$$ His results are summarised below. $$n = 15 \quad \sum x = 60 \quad \sum x ^ { 2 } = 1946$$

1. Calculate unbiased estimates of the mean, \(\mu\), and the variance of \(X\). Using the mean of Paul's sample and given \(X \sim \mathrm {~N} \left( \mu , 10 ^ { 2 } \right)\)
2. 1. calculate a 95\% confidence interval for the mean arrival time at work for this company bus.
   2. State an assumption you made about the values in the sample obtained by Paul.
3. Comment on Paul's belief. Justify your answer.

7 A factory produces 3-litre bottles of mineral water. The volume of water in a bottle has previously had a mean value of 3020 millilitres. Following a stoppage for maintenance, the volume of water, \(x\) millilitres, in each of a random sample of 100 bottles is measured and the following data obtained, where \(y = x - 3000\). $$\sum y = 1847.0 \quad \sum ( y - \bar { y } ) ^ { 2 } = 6336.00$$

1. Carry out a hypothesis test, at the \(5 \%\) significance level, to investigate whether the mean volume of water in a bottle has changed.
   (8 marks)
2. Subsequent measurements establish that the mean volume of water in a bottle produced by the factory after the stoppage is 3020 millilitres. State whether a Type I error, a Type II error or no error was made when carrying out the test in part (a).
   (l mark)

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

   Mean	19.68
   s	1.38
   N	40

   Result
   Z Estimate of a Mean

   Mean	19.68
   s	1.38
   SE	0.2182
   N	40
   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.

2 The pH value, \(X\), which is a measure of acidity, was measured for soil taken from a random sample of 20 villages in which rhododendrons grow well. The results are summarised below, where \(\bar { x }\) denotes the sample mean. You may assume that the sample is selected from a normal population. $$\Sigma x = 114 \quad \Sigma ( x - \bar { x } ) ^ { 2 } = 2.382$$

1. Calculate a \(98 \%\) confidence interval for the mean pH value in villages where rhododendrons grow well, giving 3 decimal places in your answer.
2. Comment, justifying your answer, on a suggestion that the average pH value in villages where rhododendrons grow well is 5.5.

An examination involved writing an essay. In order to compare the time taken to write the essay by students from 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 \(n_A = 12\), \(\Sigma t_A = 257\), \(\Sigma t_A^2 = 5629\), \(n_B = 8\), \(\Sigma t_B = 206\), \(\Sigma t_B^2 = 5359\). Stating any required assumptions, calculate a \(95\%\) confidence interval for the difference in the population means. [8] State, giving a reason, whether your confidence interval supports the statement that the population means, for the two colleges, are equal. [1]

The number, \(x\), of beech trees was counted in each of \(50\) randomly chosen regions of equal size in beech forests in country \(A\). The number, \(y\), of beech trees was counted in each of \(40\) randomly chosen regions of the same equal size in beech forests in country \(B\). The results are summarised as follows. $$\Sigma x = 1416 \quad \Sigma x^2 = 41100 \quad \Sigma y = 888 \quad \Sigma y^2 = 20140$$ Find a \(95\%\) confidence interval for the difference between the mean number of beech trees in regions of this size in country \(A\) and in country \(B\). [9]

A random sample of 10 observations of a normal variable \(X\) gave the following summarised data, where \(\bar{x}\) is the sample mean. $$\Sigma x = 222.8 \qquad \Sigma(x - \bar{x})^2 = 4.12$$ Find a 95% confidence interval for the population mean. [5]

100 randomly chosen adults each throw a ball once. The length, \(l\) metres, of each throw is recorded. The results are summarised below. $$n = 100 \qquad \sum l = 3820 \qquad \sum l^2 = 182200$$ Calculate a 94% confidence interval for the population mean length of throws by adults. [6]

The heights, \(x\) m, of a random sample of 50 adult males from country A were recorded. The heights, \(y\) m, of a random sample of 40 adult males from country B were also recorded. The results are summarised as follows. $$\sum x = 89.0 \qquad \sum x^2 = 159.4 \qquad \sum y = 67.2 \qquad \sum y^2 = 113.1$$ Find a 95% confidence interval for the difference between the mean heights of adult males from country A and adult males from country B. [8]

The error, \(X\) °C, made in measuring a patient's temperature at a local doctors' surgery may be modelled by a normal distribution with mean \(\mu\) and standard deviation \(\sigma\). The errors, \(x\) °C, made in measuring the temperature of each of a random sample of \(10\) patients are summarised below. $$\sum x = 0.35 \quad \text{and} \quad \sum(x - \bar{x})^2 = 0.12705$$ Construct a \(99\%\) confidence interval for \(\mu\), giving the limits to three decimal places. [5 marks]