Confidence intervals

2 A die is biased. The mean and variance of a random sample of 70 scores on this die are found to be 3.61 and 2.70 respectively. Calculate a \(95 \%\) confidence interval for the population mean score.

3

1. Give a reason for using a sample rather than the whole population in carrying out a statistical investigation.
2. Tennis balls of a certain brand are known to have a mean height of bounce of 64.7 cm , when dropped from a height of 100 cm . A change is made in the manufacturing process and it is required to test whether this change has affected the mean height of bounce. 100 new tennis balls are tested and it is found that their mean height of bounce when dropped from a height of 100 cm is 65.7 cm and the unbiased estimate of the population variance is \(15 \mathrm {~cm} ^ { 2 }\).
   1. Calculate a \(95 \%\) confidence interval for the population mean.
   2. Use your answer to part (ii) (a) to explain what conclusion can be drawn about whether the change has affected the mean height of bounce.

3 The time taken in minutes for a certain daily train journey has a normal distribution with standard deviation 5.8. For a random sample of 20 days the journey times were noted and the mean journey time was found to be 81.5 minutes.

1. Calculate a \(98 \%\) confidence interval for the population mean journey time.
   A student was asked for the meaning of this confidence interval. The student replied as follows.
   'The times for \(98 \%\) of these journeys are likely to be within the confidence interval.'
2. Explain briefly whether this statement is true or not.
   Two independent 98\% confidence intervals are found.
3. Given that at least one of these intervals contains the population mean, find the probability that both intervals contain the population mean.

The heights, in metres, of a random sample of 8 trees of a particular type are as follows. 14.2 11.3 10.8 8.4 12.8 11.5 12.1 9.2 Assuming that heights of trees of this type are normally distributed, calculate a 95% confidence interval for the mean height of trees of this type. [6]

3 A consumer group, interested in the mean fat content of a particular type of sausage, takes a random sample of 20 sausages and sends them away to be analysed. The percentage of fat in each sausage is as follows. $$\begin{array} { l l l l l l l l l l l l l l l l l l l l } 26 & 27 & 28 & 28 & 28 & 29 & 29 & 30 & 30 & 31 & 32 & 32 & 32 & 33 & 33 & 34 & 34 & 34 & 35 & 35 \end{array}$$ Assume that the percentage of fat is normally distributed with mean \(\mu\), and that the standard deviation is known to be 3 .

1. Calculate a 98\% confidence interval for the population mean percentage of fat.
2. The manufacturer claims that the mean percentage of fat in sausages of this type is 30 . Use your answer to part (i) to determine whether the consumer group should accept this claim.

The weights, in grams, of apples are assumed to follow a normal distribution. The weights of apples sold by a supermarket have variance \(\sigma_1^2\). A random sample of 4 apples from the supermarket had weights 114, 100, 119, 123.

1. Find a 95\% confidence interval for \(\sigma_1^2\). [7]

The weights of apples sold on a market stall have variance \(\sigma_M^2\). A second random sample of 7 apples was taken from the market stall. The sample variance \(s_M^2\) of the apples was 318.8.

1. [(b)] Stating your hypotheses clearly test, at the 1\% level of significance, whether or not there is evidence that \(\sigma_M^2 > \sigma_1^2\). [5]

1 A coin is thrown 100 times and it shows heads 60 times. Calculate an approximate \(98 \%\) confidence interval for the probability, \(p\), that the coin shows heads on any throw.

1 In a survey of 2000 randomly chosen adults, 1602 said that they owned a smartphone. Calculate an approximate \(95 \%\) confidence interval for the proportion of adults in the whole population who own a smartphone.

3 A die is biased so that the probability that it shows a six on any throw is \(p\).

1. In an experiment, the die shows a six on 22 out of 100 throws. Find an approximate \(97 \%\) confidence interval for \(p\).
2. The experiment is repeated and another \(97 \%\) confidence interval is found. Find the probability that exactly one of the two confidence intervals includes the true value of \(p\).

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]

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?

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]

A normally distributed variable, \(X\), has unknown mean \(\mu\) and unknown standard deviation \(\sigma\). A sample of 10 values of \(X\) was taken. From these 10 values, a 95% confidence interval for \(\mu\) was calculated to be $$(30.47, 32.93)$$ Use this confidence interval to find unbiased estimates for \(\mu\) and \(\sigma^2\). [4 marks]

3 Based on a random sample of 700 people living in a certain area, a confidence interval for the proportion, \(p\), of all people living in that area who had travelled abroad was found to be \(0.5672 < p < 0.6528\).

1. Find the proportion of people in the sample who had travelled abroad.
2. Find the confidence level of this confidence interval. Give your answer correct to the nearest integer.

Petra is studying a particular species of bird. She takes a random sample of 12 birds from nature reserve \(A\) and measures the wing span, \(x \mathrm {~cm}\), for each bird. She then calculates a \(95 \%\) confidence interval for the population mean wing span, \(\mu \mathrm { cm }\), for birds of this species, assuming that wing spans are normally distributed. Later, she is not able to find the summary of the results for the sample, but she knows that the \(95 \%\) confidence interval is \(25.17 \leqslant \mu \leqslant 26.83\). Find the values of \(\sum x\) and \(\sum x ^ { 2 }\) for this sample. Petra also measures the wing spans of a random sample of 7 birds from nature reserve \(B\). Their wing spans, \(y \mathrm {~cm}\), are as follows. $$\begin{array} { l l l l l l l } 23.2 & 22.4 & 27.6 & 25.3 & 28.4 & 26.5 & 23.6 \end{array}$$ She believes that the mean wing span of birds found in nature reserve \(A\) is greater than the mean wing span of birds found in nature reserve \(B\). Assuming that this second sample also comes from a normal distribution, with variance the same as the first distribution, test, at the \(10 \%\) significance level, whether there is evidence to support Petra's belief.

1 The result of a memory test is known to be normally distributed with mean \(\mu\) and standard deviation 1.9. It is required to have a \(95 \%\) confidence interval for \(\mu\) with a total width of less than 2.0 . Find the least possible number of tests needed to achieve this.

A film-buff is interested in how long it takes for the credits to roll at the end of a movie. She takes a random sample of 20 movies from those that she has bought on DVD and finds that the credits on these films last for a total of 46 minutes and 15 seconds

1. Assuming that the time for the credits to roll follows a Normal distribution with a standard deviation of 23 seconds, use her data to calculate a 90% confidence interval for the mean time taken for the credits to roll. [5]
2. Find the minimum number of movies she would need to have included in her sample for her confidence interval to have a width of less than 10 seconds. [5]
3. Explain why her sample might not be representative of all movies. [1]

4 A certain train journey takes place every day throughout the year. The time taken, in minutes, for the journey is normally distributed with variance 11.2.

1. The mean time for a random sample of \(n\) of these journeys was found. A \(94 \%\) confidence interval for the population mean time was calculated and was found to have a width of 1.4076 minutes, correct to 4 decimal places. Find the value of \(n\).
2. A passenger noted the times for 50 randomly chosen journeys in January, February and March. Give a reason why this sample is unsuitable for use in finding a confidence interval for the population mean time.
3. A researcher took 4 random samples and a \(94 \%\) confidence interval for the population mean was found from each sample. Find the probability that exactly 3 of these confidence intervals contain the true value of the population mean.

The heights of a random sample of 10 imported orchids are measured. The mean height of the sample is found to be 20.1 cm. The heights of the orchids are normally distributed. Given that the population standard deviation is 0.5 cm,

1. estimate limits between which 95\% of the heights of the orchids lie, [3]
2. find a 98\% confidence interval for the mean height of the orchids. [4]

A grower claims that the mean height of this type of orchid is 19.5 cm.

1. Comment on the grower's claim. Give a reason for your answer. [2]

The heights of a random sample of 10 imported orchids are measured. The mean height of the sample is found to be 20.1 cm. The heights of the orchids are normally distributed. Given that the population standard deviation is 0.5 cm,

1. estimate limits between which 95\% of the heights of the orchids lie, [3]
2. find a 98\% confidence interval for the mean height of the orchids. [4]

A grower claims that the mean height of this type of orchid is 19.5 cm.

1. Comment on the grower's claim. Give a reason for your answer. [2]

6 An anthropologist was studying the inhabitants of two islands, Raloa and Tangi. Part of the study involved the incidence of blood group type A. The blood of 80 randomly chosen inhabitants of Raloa and 85 randomly chosen inhabitants of Tangi was tested. The number of inhabitants with type A blood was 28 for the Raloa sample and 46 for the Tangi sample. The anthropologist calculated \(90 \%\) confidence intervals for the population proportions of inhabitants with type A blood. They were \(( 0.262,0.438 )\) for Raloa and \(( 0.452,0.630 )\) for Tangi, where each figure is correct to 3 decimal places. It is known that \(43 \%\) of the world's population have type A blood.

1. State, giving your reasons, whether there is evidence for the following assertions about the proportions of people with type A blood.
   1. The proportion in Raloa is different from the world proportion.
   2. The proportion in Tangi is different from the world proportion.
   3. Carry out a suitable test, at the \(2 \%\) significance level, of whether the proportions of people with type A blood differ on the two islands.

3 A researcher wishes to estimate the proportion, \(p\), of houses in London Road that have only one occupant. He takes a random sample of 64 houses in London Road and finds that 8 houses in the sample have only one occupant. Using this sample, he calculates that an approximate \(\alpha \%\) confidence interval for \(p\) has width 0.130 . Find \(\alpha\) correct to the nearest integer.

1. A random sample of the daily sales (in £s) of a small company is taken and, using tables of the normal distribution, a 99\% confidence interval for the mean daily sales is found to be
   (123.5, 154.7)

Find a \(95 \%\) confidence interval for the mean daily sales of the company.
(6)

1 The result of a fitness trial is a random variable \(X\) which is normally distributed with mean \(\mu\) and standard deviation 2.4. A researcher uses the results from a random sample of 90 trials to calculate a \(98 \%\) confidence interval for \(\mu\). What is the width of this interval?

1 An investigation was carried out of the lengths of commuters' journeys. For a random sample of 500 commuters, the mean journey time was 75 minutes, and the standard deviation was 40 minutes.

1. Calculate a 95\% confidence interval for the mean journey time.
2. Explain whether you need to assume that journey times are normally distributed.

1. The volume of water in a bottle has a normal distribution with unknown mean, \(\mu\) millilitres, and known standard deviation, \(\sigma\) millilitres.

A random sample of 150 of the bottles of water gave a 95\% confidence interval for \(\mu\) of
(327.84, 329.76)

1. Using the confidence interval given, test whether or not \(\mu = 328\) State your hypotheses clearly and write down the significance level you have used. A second random sample, of 200 of these bottles of water, had a mean volume of 328 millilitres.
2. Calculate a 98\% confidence interval for \(\mu\) based on this second sample. You must show all steps in your working.
   (Solutions relying entirely on calculator technology are not acceptable.) Using five different random samples of 200 of these bottles of water, five \(98 \%\) confidence intervals for \(\mu\) are to be found.
3. Calculate the probability that more than 3 of these intervals will contain \(\mu\)