5.05d Confidence intervals: using normal distribution

1 The masses of a certain variety of plums are known to have standard deviation 13.2 g . A random sample of 200 of these plums is taken and the mean mass of the plums in the sample is found to be 62.3 g .

1. Calculate a \(98 \%\) confidence interval for the population mean mass.
2. State with a reason whether it was necessary to use the Central Limit theorem in the calculation in part (i).

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.

3 A random sample of 150 students attending a college is taken, and their travel times, \(t\) minutes, are measured. The data are summarised by \(\Sigma t = 4080\) and \(\Sigma t ^ { 2 } = 159252\).

1. Calculate unbiased estimates of the population mean and variance.
2. Calculate a \(94 \%\) confidence interval for the population mean travel time.

4

1. Give a reason why, in carrying out a statistical investigation, a sample rather than a complete population may be used.
2. Rose wishes to investigate whether men in her town have a different life-span from the national average of 71.2 years. She looks at government records for her town and takes a random sample of the ages of 110 men who have died recently. Their mean age in years was 69.3 and the unbiased estimate of the population variance was 65.61.
   1. Calculate a \(90 \%\) confidence interval for the population mean and explain what you understand by this confidence interval.
   2. State with a reason what conclusion about the life-span of men in her town Rose could draw from this confidence interval.

3 A survey was conducted to find the proportion of people owning DVD players. It was found that 203 out of a random sample of 278 people owned a DVD player.

1. Calculate a \(97 \%\) confidence interval for the true proportion of people who own a DVD player. A second survey to find the proportion of people owning DVD players was conducted at 10 o'clock on a Thursday morning in a shopping centre.
2. Give one reason why this is not a satisfactory sample.

3

1. Explain what is meant by the term 'random sample'. In a random sample of 350 food shops it was found that 130 of them had Special Offers.
2. Calculate an approximate \(95 \%\) confidence interval for the proportion of all food shops with Special Offers.
3. Estimate the size of a random sample required for an approximate \(95 \%\) confidence interval for this proportion to have a width of 0.04 .

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 \(\mu\) ?

2 The lengths of sewing needles in travel sewing kits are distributed normally with mean \(\mu \mathrm { mm }\) and standard deviation 1.5 mm . A random sample of \(n\) needles is taken. Find the smallest value of \(n\) such that the width of a \(95 \%\) confidence interval for the population mean is at most 1 mm .

1 There are 18 people in Millie's class. To choose a person at random she numbers the people in the class from 1 to 18 and presses the random number button on her calculator to obtain a 3-digit decimal. Millie then multiplies the first digit in this decimal by two and chooses the person corresponding to this new number. Decimals in which the first digit is zero are ignored.

1. Give a reason why this is not a satisfactory method of choosing a person. Millie obtained a random sample of 5 people of her own age by a satisfactory sampling method and found that their heights in metres were \(1.66,1.68,1.54,1.65\) and 1.57 . Heights are known to be normally distributed with variance \(0.0052 \mathrm {~m} ^ { 2 }\).
2. Find a \(98 \%\) confidence interval for the mean height of people of Millie's age.

1 In a survey of 1000 randomly chosen adults, 605 said that they used email. Calculate a \(90 \%\) confidence interval for the proportion of adults in the whole population who use email.

3 The masses of sweets produced by a machine are normally distributed with mean \(\mu\) grams and standard deviation 1.0 grams. A random sample of 65 sweets produced by the machine has a mean mass of 29.6 grams.

1. Find a \(99 \%\) confidence interval for \(\mu\). The manufacturer claims that the machine produces sweets with a mean mass of 30 grams.
2. Use the confidence interval found in part (i) to draw a conclusion about this claim.
3. Another random sample of 65 sweets produced by the machine is taken. This sample gives a \(99 \%\) confidence interval that leads to a different conclusion from that found in part (ii). Assuming that the value of \(\mu\) has not changed, explain how this can be possible.

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 In order to obtain a random sample of people who live in her town, Jane chooses people at random from the telephone directory for her town.

1. Give a reason why Jane's method will not give a random sample of people who live in the town. Jane now uses a valid method to choose a random sample of 200 people from her town and finds that 38 live in apartments.
2. Calculate an approximate \(99 \%\) confidence interval for the proportion of all people in Jane's town who live in apartments.
3. Jane uses the same sample to give a confidence interval of width 0.1 for this proportion. This interval is an \(x \%\) confidence interval. Find the value of \(x\).

3 Joshi suspects that a certain die is biased so that the probability of showing a six is less than \(\frac { 1 } { 6 }\). He plans to throw the die 25 times and if it shows a six on fewer than 2 throws, he will conclude that the die is biased in this way.

1. Find the probability of a Type I error and state the significance level of the test. Joshi now decides to throw the die 100 times. It shows a six on 9 of these throws.
2. Calculate an approximate \(95 \%\) confidence interval for the probability of showing a six on one throw of this die.

2 Heights of a certain species of animal are known to be normally distributed with standard deviation 0.17 m . A conservationist wishes to obtain a \(99 \%\) confidence interval for the population mean, with total width less than 0.2 m . Find the smallest sample size required.

1 A random sample of 80 values of a variable \(X\) is taken and these values are summarised below. $$n = 80 \quad \Sigma x = 150.2 \quad \Sigma x ^ { 2 } = 820.24$$ Calculate unbiased estimates of the population mean and variance of \(X\) and hence find a \(95 \%\) confidence interval for the population mean of \(X\).

4 In a survey a random sample of 150 households in Nantville were asked to fill in a questionnaire about household budgeting.

1. The results showed that 33 households owned more than one car. Find an approximate \(99 \%\) confidence interval for the proportion of all households in Nantville with more than one car. [4]
2. The results also included the weekly expenditure on food, \(x\) dollars, of the households. These were summarised as follows. $$n = 150 \quad \Sigma x = 19035 \quad \Sigma x ^ { 2 } = 4054716$$ Find unbiased estimates of the mean and variance of the weekly expenditure on food of all households in Nantville.
3. The government has a list of all the households in Nantville numbered from 1 to 9526. Describe briefly how to use random numbers to select a sample of 150 households from this list.

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.

5 Raman is researching the heights of male giraffes in a particular region. Raman assumes that the heights of male giraffes in this region are normally distributed. He takes a random sample of 8 male giraffes from the region and measures the height, in metres, of each giraffe. These heights are as follows. $$\begin{array} { c c c c c c c c } 5.2 & 5.8 & 4.9 & 6.1 & 5.5 & 5.9 & 5.4 & 5.6 \end{array}$$

1. Find a \(90 \%\) confidence interval for the population mean height of male giraffes in this region. [5]
   Raman claims that the population mean height of male giraffes in the region is less than 5.9 metres.
2. Test at the \(2.5 \%\) significance level whether this sample provides sufficient evidence to support Raman's claim.

1 The times taken by members of a large quiz club to complete a challenge have a normal distribution with mean \(\mu\) minutes. The times, \(x\) minutes, are recorded for a random sample of 8 members of the club. The results are summarised as follows, where \(\bar { x }\) is the sample mean. $$\bar { x } = 33.8 \quad \sum ( x - \bar { x } ) ^ { 2 } = 94.5$$ Find a 95\% confidence interval for \(\mu\).

1 The lengths of the leaves of a particular type of tree are normally distributed with mean \(\mu \mathrm { cm }\). The lengths, \(x \mathrm {~cm}\), of a random sample of 12 leaves of this type are recorded. The results are summarised as follows. $$\sum x = 91.2 \quad \sum x ^ { 2 } = 695.8$$ Find a 95\% confidence interval for \(\mu\).

2 Shane is studying the lengths of the tails of male red kangaroos. He takes a random sample of 14 male red kangaroos and measures the length of the tail, \(x \mathrm {~m}\), for each kangaroo. He then calculates a \(90 \%\) confidence interval for the population mean tail length, \(\mu \mathrm { m }\), of male red kangaroos. He assumes that the tail lengths are normally distributed and finds that \(1.11 \leqslant \mu \leqslant 1.14\). Find the values of \(\sum x\) and \(\sum x ^ { 2 }\) for this sample.

1 The times taken by members of a large cycling club to complete a cross-country circuit have a normal distribution with mean \(\mu\) minutes. The times taken, \(x\) minutes, are recorded for a random sample of 14 members of the club. The results are summarised as follows, where \(\bar { x }\) is the sample mean. $$\bar { x } = 42.8 \quad \sum ( x - \bar { x } ) ^ { 2 } = 941.5$$ Find a 95\% confidence interval for \(\mu\).

2 A rowing club has a large number of members.A random sample of 12 of these members is taken and the pulse rate,\(x\) beats per minute(bpm),of each is measured after a 30 -minute training session.A \(98 \%\) confidence interval for the population mean pulse rate,\(\mu \mathrm { bpm }\) ,is calculated from the sample as \(64.22 < \mu < 68.66\) .

1. Find the values of \(\sum x\) and \(\sum x ^ { 2 }\) .
2. State an assumption that is necessary for the confidence interval to be valid. \includegraphics[max width=\textwidth, alt={}, center]{b5ff998a-fcb6-4a1b-ae86-ec66b0dccc3c-04_2718_38_141_2009}

6 Seva is investigating the lengths of the tails of adult wallabies in two regions of Australia, \(X\) and \(Y\). He chooses a random sample of 50 adult wallabies from region \(X\) and records the lengths, \(x \mathrm {~cm}\), of their tails. He also chooses a random sample of 40 adult wallabies from region \(Y\) and records the lengths, \(y \mathrm {~cm}\), of their tails. His results are summarised as follows. $$\sum x = 1080 \quad \sum x ^ { 2 } = 23480 \quad \sum y = 940 \quad \sum y ^ { 2 } = 22220$$ It cannot be assumed that the population variances of the two distributions are the same.

1. Find a \(90 \%\) confidence interval for the difference between the population mean lengths of the tails of adult wallabies in regions \(X\) and \(Y\). \includegraphics[max width=\textwidth, alt={}, center]{b5ff998a-fcb6-4a1b-ae86-ec66b0dccc3c-10_2718_38_141_2010} The population mean lengths of the tails of adult wallabies in regions \(X\) and \(Y\) are \(\mu _ { X } \mathrm {~cm}\) and \(\mu _ { Y } \mathrm {~cm}\) respectively.
2. Test, at the \(10 \%\) significance level, the null hypothesis \(\mu _ { Y } - \mu _ { X } = 1.1\) against the alternative hypothesis \(\mu _ { Y } - \mu _ { X } > 1.1\). State your conclusion in the context of the question.
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