5.05d Confidence intervals: using normal distribution

5 A random sample of 12 seventeen-year-old boys and a random sample of 14 seventeen-year-old girls were given a certain task. The times, \(t\) minutes, taken to complete the task by the members of the two samples are summarised as follows.

1. Stating any necessary assumption(s), find a \(95 \%\) symmetric confidence interval for the difference in the average times taken to complete the task by seventeen-year-old boys and seventeen-year-old girls.
2. State with a reason whether the confidence interval calculated in part (i) suggests that there may in fact be no difference in the average times taken by seventeen-year-old boys and by seventeen-year-old girls.

3 Small amounts of a potentially hazardous chemical are discharged into a river from a nearby industrial site. A random sample of size 6 was taken from the river and the concentration of the chemical present in each item was measured in grams per litre. The results are shown below. $$\begin{array} { l l l l l l } 1.64 & 1.53 & 1.78 & 1.60 & 1.73 & 1.77 \end{array}$$

1. Assuming that the sample was taken from a normal distribution with known variance 0.01 , construct a \(99 \%\) confidence interval for the mean concentration of the chemical present in the river.
2. If instead the sample was taken from a normal distribution, but with unknown variance, construct a revised \(99 \%\) confidence interval for the mean concentration of the chemical present in the river.
3. If the mean concentration of the chemical in the river exceeds 1.8 grams per litre, then remedial action needs to be taken. Comment briefly on the need for remedial action in the light of the results in parts (a) and (b).

8 Specimens of rain were collected at random from the north and south sides of an island and analysed for sulphur content. The results (in suitable units) are given below. Assume that the sulphur contents have normal distributions with population means \(\mu _ { N }\) and \(\mu _ { S }\) and a common, but unknown, variance.

1. Calculate a symmetric \(95 \%\) confidence interval for the difference in population mean sulphur contents of the rain on the north and south sides of the island, \(\mu _ { S } - \mu _ { N }\).
2. Comment on a claim that the mean sulphur content is the same on both sides of the island.

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 random sample of 8 observations of a normal random variable \(X\) gave the following summarised data, where \(\overline{x}\) denotes the sample mean. $$\Sigma x = 42.5 \quad \Sigma(x - \overline{x})^2 = 15.519$$ Test, at the 5% significance level, whether the population mean of \(X\) is greater than 4.5. [7] Calculate a 95% confidence interval for the population mean of \(X\). [3]

A random sample of 8 observations of a normal random variable \(X\) gave the following summarised data, where \(\bar{x}\) denotes the sample mean. $$\Sigma x = 42.5 \quad \Sigma(x - \bar{x})^2 = 15.519$$ Test, at the 5\% significance level, whether the population mean of \(X\) is greater than 4.5. [7] Calculate a 95\% confidence interval for the population mean of \(X\). [3]

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 machine produces metal discs whose diameters have a normal distribution. The mean of this distribution is intended to be \(10\) cm. Accuracy is checked by measuring the diameters of a random sample of six discs. The diameters, in cm, are as follows. 10.03 \quad 10.02 \quad 9.98 \quad 10.06 \quad 10.08 \quad 10.01 Calculate a 99\% confidence interval for the mean diameter of all discs produced by the machine. [5] Deduce a 99\% confidence interval for the mean circumference of all discs produced by the machine. [1]

150 sheep, chosen from a large flock of sheep, were divided into two groups of 75. Over a fixed period, one group had their grazing controlled and the other group grazed freely. The gains in weight, in kg, were recorded for each animal and the table below shows the sample means and the unbiased estimates of the population variances for the two samples.

	Sample mean	Unbiased estimate of population variance
Controlled grazing	19.14	20.54
Free grazing	15.36	9.84

It is required to test whether the population mean for sheep having their grazing controlled exceeds the population mean for sheep grazing freely by less than 5 kg. State, giving a reason, if it is necessary for the validity of the test to assume that the two population variances are equal. [1] Stating any other assumption, carry out the test at the 5\% significance level. [8]

The mean Intelligence Quotient (IQ) of a random sample of 15 pupils at School A is 109. The mean IQ of a random sample of 20 pupils at School B is 112. You may assume that the IQs for the populations from which these samples are taken are normally distributed, and that both distributions have standard deviation 15. Find a 90% confidence interval for \(\mu_B - \mu_A\), where \(\mu_A\) and \(\mu_B\) are the population mean IQs. [6]

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]

There are a large number of students at a particular college. The heights, in metres, of a random sample of 8 students are as follows. $$1.75 \quad 1.72 \quad 1.62 \quad 1.70 \quad 1.82 \quad 1.75 \quad 1.68 \quad 1.84$$ You may assume that heights of students are normally distributed.

1. Test, at the 5\% significance level, whether the population mean height of students at this college is greater than 1.70 metres. [7]
2. Find a 95\% confidence interval for the population mean height of students at this college. [3]

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]

Each of a random sample of 80 adults gave an estimate, \(h\) metres, of the height of a particular building. The results were summarised as follows. $$n = 80 \quad \sum h = 2048 \quad \sum h^2 = 52760$$

1. Calculate unbiased estimates of the population mean and variance. [3]
2. Using this sample, the upper boundary of an \(\alpha\%\) confidence interval for the population mean is 26.0. Find the value of \(\alpha\). [4]

In a survey of 300 randomly chosen adults in Rickton, 134 said that they exercised regularly. This information was used to calculate an \(\alpha\)% confidence interval for the proportion of adults in Rickton who exercise regularly. The upper bound of the confidence interval was found to be 0.487, correct to 3 significant figures. Find the value of \(\alpha\) correct to the nearest integer. [4]

The heights of a certain species of deer are known to have standard deviation \(0.35\) m. A zoologist takes a random sample of \(150\) of these deer and finds that the mean height of the deer in the sample is \(1.42\) m.

1. Calculate a \(96\%\) confidence interval for the population mean height. [3]
2. Bubay says that \(96\%\) of deer of this species are likely to have heights that are within this confidence interval. Explain briefly whether Bubay is correct. [1]

A doctor wishes to investigate the mean fat content in low-fat burgers. He takes a random sample of 15 burgers and sends them to a laboratory where the mass, in grams, of fat in each burger is determined. The results are as follows. \(9 \quad 7 \quad 8 \quad 9 \quad 6 \quad 11 \quad 7 \quad 9 \quad 8 \quad 9 \quad 8 \quad 10 \quad 7 \quad 9 \quad 9\) Assume that the mass, in grams, of fat in low-fat burgers is normally distributed with mean \(\mu\) and that the population standard deviation is 1.3.

1. Calculate a 99\% confidence interval for \(\mu\). [4]
2. Explain whether it was necessary to use the Central Limit theorem in the calculation in part (i). [2]
3. The manufacturer claims that the mean mass of fat in burgers of this type is 8 g. Use your answer to part (i) to comment on this claim. [2]

The time taken for a particular type of paint to dry was measured for a sample of 150 randomly chosen points on a wall. The sample mean was 192.4 minutes and an unbiased estimate of the population variance was 43.6 minutes\(^2\). Find a 98\% confidence interval for the mean drying time. [3]

The time taken, \(T\) minutes, for a special anti-rust paint to dry was measured for a random sample of 120 painted pieces of metal. The sample mean was 51.2 minutes and an unbiased estimate of the population variance was 37.4 minutes\(^2\). Determine a 99% confidence interval for the mean drying time. [3]

35% of a random sample of \(n\) students walk to college. This result is used to construct an approximate 98% confidence interval for the population proportion of students who walk to college. Given that the width of this confidence interval is 0.157, correct to 3 significant figures, find \(n\). [5]

Leaves from a certain type of tree have lengths that are distributed with standard deviation 3 cm. A random sample of 6 of these leaves is taken and the mean length of this sample is found to be 8 cm.

1. Calculate a 95\% confidence interval for the population mean length. [3]
2. Write down the probability that the whole 95\% confidence interval will lie below the population mean. [1]

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]

A factory produces steel sheets whose weights \(X\) kg, are such that \(X \sim \text{N}(\mu, \sigma^2)\) A random sample of these sheets is taken and a 95\% confidence interval for \(\mu\) is found to be (29.74, 31.86)

1. Find, to 2 decimal places, the standard error of the mean. [3]
2. Hence, or otherwise, find a 90\% confidence interval for \(\mu\) based on the same sample of sheets. [3]

Using four different random samples, four 90\% confidence intervals for \(\mu\) are to be found.

1. Calculate the probability that at least 3 of these intervals will contain \(\mu\). [3]

The weights of tubs of margarine are known to be normally distributed. A random sample of 10 tubs of margarine were weighed, to the nearest gram, and the results were as follows. $$498 \quad 502 \quad 500 \quad 496 \quad 509 \quad 504 \quad 511 \quad 497 \quad 506 \quad 499$$

1. Find unbiased estimates of the mean and the variance of the population from which this sample was taken. [5]

Given that the population standard deviation is 5.0 g,

1. estimate limits, to 2 decimal places, between which 90\% of the weights of the tubs lie, [2]
2. find a 95\% confidence interval for the mean weight of the tubs. [5]

A second random sample of 15 tubs was found to have a mean weight of 501.9 g.

1. Stating your hypotheses clearly and using a 1\% level of significance, test whether or not the mean weight of these tubs is greater than 500 g. [5]

The weights of tubs of margarine are known to be normally distributed. A random sample of 10 tubs of margarine were weighed, to the nearest gram, and the results were as follows. 498 502 500 496 509 504 511 497 506 499

1. Find unbiased estimates of the mean and the variance of the population from which this sample was taken. [5]

Given that the population standard deviation is 5.0 g,

1. estimate limits, to 2 decimal places, between which 90\% of the weights of the tubs lie, [2]
2. find a 95\% confidence interval for the mean weight of the tubs. [5]

A second random sample of 15 tubs was found to have a mean weight of 501.9 g.

1. Stating your hypotheses clearly and using a 1\% level of significance, test whether or not the mean weight of these tubs is greater than 500 g. [5]