5.05d Confidence intervals: using normal distribution

10 Engineers are investigating the speed of the internet connection received by households in two towns \(P\) and \(Q\). The speeds, in suitable units, in \(P\) and \(Q\) are denoted by \(x\) and \(y\) respectively. For a random sample of 50 houses in town \(P\) and a random sample of 40 houses in town \(Q\) the results are summarised as follows. $$\Sigma x = 240 \quad \Sigma x ^ { 2 } = 1224 \quad \Sigma y = 168 \quad \Sigma y ^ { 2 } = 754$$ Calculate a \(95 \%\) confidence interval for \(\mu _ { P } - \mu _ { Q }\), where \(\mu _ { P }\) and \(\mu _ { Q }\) are the population mean speeds for \(P\) and \(Q\). Test, at the \(1 \%\) significance level, whether \(\mu _ { P }\) is greater than \(\mu _ { Q }\).

9 A gardener \(P\) claims that a new type of fruit tree produces a higher annual mass of fruit than the type that he has previously grown. The old type of tree produced 5.2 kg of fruit per tree, on average. A random sample of 10 trees of the new type is chosen. The masses, \(x \mathrm {~kg}\), of fruit produced are summarised as follows. $$\Sigma x = 61.0 \quad \Sigma x ^ { 2 } = 384.0$$ Test, at the \(5 \%\) significance level, whether gardener \(P\) 's claim is justified, assuming a normal distribution. Another gardener \(Q\) has his own type of fruit tree. The masses, \(y \mathrm {~kg}\), of fruit produced by a random sample of 10 trees grown by gardener \(Q\) are summarised as follows. $$\Sigma y = 70.0 \quad \Sigma y ^ { 2 } = 500.6$$ Test, at the \(5 \%\) significance level, whether the mean mass of fruit produced by gardener \(Q\) 's trees is greater than the mean mass of fruit produced by gardener \(P\) 's trees. You may assume that both distributions are normal and you should state any additional assumption.

8 The number, \(x\), of a certain type of sea shell was counted at 60 randomly chosen sites, each one metre square, along the coastline in country \(A\). The number, \(y\), of the same type of shell was counted at 50 randomly chosen sites, each one metre square, along the coastline in country \(B\). The results are summarised as follows. $$\Sigma x = 1752 \quad \Sigma x ^ { 2 } = 55500 \quad \Sigma y = 1220 \quad \Sigma y ^ { 2 } = 33500$$ Find a 95\% confidence interval for the difference between the mean number of sea shells, per square metre, on the coastlines in country \(A\) and in country \(B\).

8 Weekly expenses claimed by employees at two different branches, \(A\) and \(B\), of a large company are being compared. Expenses claimed by an employee at branch \(A\) and by an employee at branch \(B\) are denoted by \(\\) x\( and \)\\( y\) respectively. A random sample of 60 employees from branch \(A\) and a random sample of 50 employees from branch \(B\) give the following summarised data. $$\Sigma x = 6060 \quad \Sigma x ^ { 2 } = 626220 \quad \Sigma y = 4750 \quad \Sigma y ^ { 2 } = 464500$$ Using a \(2 \%\) significance level, test whether, on average, employees from branch \(A\) claim the same as employees from branch \(B\).

The time taken for a randomly chosen student at College \(P\) to complete a particular puzzle has a normal distribution with mean \(\mu\) minutes. The times, \(x\) minutes, are recorded for a random sample of 8 students chosen from the college. The results are summarised as follows. $$\Sigma x = 42.8 \quad \Sigma x ^ { 2 } = 236.0$$ Find a 95\% confidence interval for \(\mu\). A test is carried out on this sample data, at the \(10 \%\) significance level. The test supports the claim that \(\mu > k\). Find the greatest possible value of \(k\). A random sample, of size 12, is taken from the students at College \(Q\). Their times to complete the puzzle give a sample mean of 4.60 minutes and an unbiased variance estimate of 1.962 minutes \({ } ^ { 2 }\). Use a 2 -sample test at the \(10 \%\) significance level to test whether the mean time for students at College \(Q\) to complete the puzzle is less than the mean time for students at College \(P\) to complete the puzzle. You should state any assumptions necessary for the test to be valid.

7 A random sample of 8 sunflower plants is taken from the large number grown by a gardener, and the heights of the plants are measured. A 95\% confidence interval for the population mean, \(\mu\) metres, is calculated from the sample data as \(1.17 < \mu < 2.03\). Given that the height of a sunflower plant is denoted by \(x\) metres, find the values of \(\Sigma x\) and \(\Sigma x ^ { 2 }\) for this sample of 8 plants.

8 A large number of long jumpers are competing in a national long jump competition. The distances, in metres, jumped by a random sample of 7 competitors are as follows. $$\begin{array} { l l l l l l l } 6.25 & 7.01 & 5.74 & 6.89 & 7.24 & 5.64 & 6.52 \end{array}$$ Assuming that distances are normally distributed, test, at the \(5 \%\) significance level, whether the mean distance jumped by long jumpers in this competition is greater than 6.2 metres. The distances jumped by another random sample of 8 long jumpers in this competition are recorded. Using the data from this sample of 8 long jumpers, a \(95 \%\) confidence interval for the population mean, \(\mu\) metres, is calculated as \(5.89 < \mu < 6.75\). Find the unbiased estimates for the population mean and population variance used in this calculation.

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.

A farmer grows two different types of cherries, Type \(A\) and Type \(B\). He assumes that the masses of each type are normally distributed. He chooses a random sample of 8 cherries of Type \(A\). He finds that the sample mean mass is 15.1 g and that a \(95 \%\) confidence interval for the population mean mass, \(\mu \mathrm { g }\), is \(13.5 \leqslant \mu \leqslant 16.7\).

1. Find an unbiased estimate for the population variance of the masses of cherries of Type \(A\).
   The farmer now chooses a random sample of 6 cherries of Type \(B\) and records their masses as follows.
   12.2
   13.3
   13.9
   14.0
   15.4
   16.4
2. Test at the \(5 \%\) significance level whether the mean mass of cherries of Type \(B\) is less than the mean mass of cherries of Type \(A\). You should assume that the population variances for the two types of cherry are equal.
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

7 The pulse rate of each member of a random sample of 25 adult UK males who exercise for a given period each week is measured in beats per minute. A \(98 \%\) confidence interval for the mean pulse rate, \(\mu\) beats per minute, for all such UK males was calculated as \(61.21 < \mu < 64.39\), based on a \(t\)-distribution.

1. Calculate the sample mean pulse rate and the standard deviation used in the calculation.
2. State an assumption necessary for the validity of the confidence interval.
3. The mean pulse rate for all UK males is 72 beats per minute. State, giving a reason, if it can be concluded that, on average, UK males who exercise have a reduced pulse rate.

A perfume manufacturer had one bottle-filling machine, but because of increased sales a second machine was obtained. In order to compare the performance of the two machines, a random sample of 50 bottles filled by the first machine and a random sample of 60 bottles filled by the second machine were checked. The volumes of the contents from the first machine, \(x _ { 1 } \mathrm { ml }\), and from the second machine, \(x _ { 2 } \mathrm { ml }\), are summarised by $$\Sigma x _ { 1 } = 1492.0 , \quad \Sigma x _ { 1 } ^ { 2 } = 44529.52 , \quad \Sigma x _ { 2 } = 1803.6 , \quad \Sigma x _ { 2 } ^ { 2 } = 54220.84 .$$ The volumes have distributions with means \(\mu _ { 1 } \mathrm { ml }\) and \(\mu _ { 2 } \mathrm { ml }\) for the first and second machines respectively. Test, at the \(2 \%\) significance level, whether \(\mu _ { 2 }\) is greater than \(\mu _ { 1 }\). Find the set of values of \(\alpha\) for which there would be evidence at the \(\alpha \%\) significance level that \(\mu _ { 2 } - \mu _ { 1 } > 0.1\).

9 A random sample of five metal rods produced by a machine is taken. Each rod is tested for hardness. The results, in suitable units, are as follows. $$\begin{array} { l l l l l } 524 & 526 & 520 & 523 & 530 \end{array}$$ Assuming a normal distribution, calculate a \(95 \%\) confidence interval for the population mean. Some adjustments are made to the machine. Assume that a normal distribution is still appropriate and that the population variance remains unchanged. A second random sample, this time of ten metal rods, is now taken. The results for hardness are as follows. $$\begin{array} { l l l l l l l l l l } 525 & 520 & 522 & 524 & 518 & 520 & 519 & 525 & 527 & 516 \end{array}$$ Stating suitable hypotheses, test at the \(10 \%\) significance level whether there is any difference between the population means before and after the adjustments.

9 The leaves from oak trees growing in two different areas \(A\) and \(B\) are being measured. The lengths, in cm , of a random sample of 7 oak leaves from area \(A\) are $$6.2 , \quad 8.3 , \quad 7.8 , \quad 9.3 , \quad 10.2 , \quad 8.4 , \quad 7.2$$ Assuming that the distribution is normal, find a 95\% confidence interval for the mean length of oak leaves from area \(A\). The lengths, in cm, of a random sample of 5 oak leaves from area \(B\) are $$5.9 , \quad 7.4 , \quad 6.8 , \quad 8.2 , \quad 8.7$$ Making suitable assumptions, which should be stated, test, at the \(5 \%\) significance level, whether the mean length of oak leaves from area \(A\) is greater than the mean length of oak leaves from area \(B\). [9]

7 The speed \(v\) at which a javelin is thrown by an athlete is measured in \(\mathrm { km } \mathrm { h } ^ { - 1 }\). The results for 10 randomly chosen throws are summarised by $$\Sigma v = 1110.8 , \quad \Sigma ( v - \bar { v } ) ^ { 2 } = 333.9$$ where \(\bar { v }\) is the sample mean.

1. Stating any necessary assumption, calculate a \(99 \%\) confidence interval for the mean speed of a throw. The results for a further 5 randomly chosen throws are now combined with the above results. It is found that the sample variance is smaller than that used in part (i).
2. State, with reasons, whether a \(95 \%\) confidence interval calculated from the combined 15 results will be wider or less wide than that found in part (i).

9 A random sample of 9 observations of a normally distributed random variable \(X\) gave the following summarised data. $$\Sigma x = 94.5 \quad \Sigma x ^ { 2 } = 993.6$$ Test, at the \(5 \%\) significance level, whether the population mean of \(X\) is 10.2 . Calculate a \(90 \%\) confidence interval for the population mean of \(X\).

A uniform plane object consists of three identical circular rings, \(X , Y\) and \(Z\), enclosed in a larger circular ring \(W\). Each of the inner rings has mass \(m\) and radius \(r\). The outer ring has mass \(3 m\) and radius \(R\). The centres of the inner rings lie at the vertices of an equilateral triangle of side \(2 r\). The outer ring touches each of the inner rings and the rings are rigidly joined together. The fixed axis \(A B\) is the diameter of \(W\) that passes through the centre of \(X\) and the point of contact of \(Y\) and \(Z\) (see diagram). It is given that \(R = \left( 1 + \frac { 2 } { 3 } \sqrt { } 3 \right) r\).

1. Show that the moment of inertia of the object about \(A B\) is \(( 7 + 2 \sqrt { } 3 ) m r ^ { 2 }\). The line \(C D\) is the diameter of \(W\) that is perpendicular to \(A B\). A particle of mass \(9 m\) is attached to \(D\). The object is now held with its plane horizontal. It is released from rest and rotates freely about the fixed horizontal axis \(A B\).
2. Find, in terms of \(g\) and \(r\), the angular speed of the object when it has rotated through \(60 ^ { \circ }\).

Fish of a certain species live in two separate lakes, \(A\) and \(B\). A zoologist claims that the mean length of fish in \(A\) is greater than the mean length of fish in \(B\). To test his claim, he catches a random sample of 8 fish from \(A\) and a random sample of 6 fish from \(B\). The lengths of the 8 fish from \(A\), in appropriate units, are as follows. $$\begin{array} { l l l l l l l l } 15.3 & 12.0 & 15.1 & 11.2 & 14.4 & 13.8 & 12.4 & 11.8 \end{array}$$ Assuming a normal distribution, find a \(95 \%\) confidence interval for the mean length of fish in \(A\). The lengths of the 6 fish from \(B\), in the same units, are as follows. $$\begin{array} { l l l l l l } 15.0 & 10.7 & 13.6 & 12.4 & 11.6 & 12.6 \end{array}$$ Stating any assumptions that you make, test at the \(5 \%\) significance level whether the mean length of fish in \(A\) is greater than the mean length of fish in \(B\). Calculate a 95\% confidence interval for the difference in the mean lengths of fish from \(A\) and from \(B\).

8 The amounts spent on the weekly food shopping by families in the big city \(P\) and the small town \(Q\) are to be compared. The amounts spent, in dollars, in \(P\) and \(Q\) are denoted by \(x\) and \(y\) respectively. For a random sample of 60 families in \(P\) and a random sample of 50 families in \(Q\), the amounts are summarised as follows. $$\Sigma x = 9600 \quad \Sigma x ^ { 2 } = 1560000 \quad \Sigma y = 7200 \quad \Sigma y ^ { 2 } = 1052500$$ Assuming a common population variance, find

1. a pooled estimate for the population variance,
2. a \(95 \%\) confidence interval for the difference in the population means in \(P\) and \(Q\).

10 A factory produces bottles of an energy juice. Two different machines are used to fill empty bottles with the juice. The manager chooses a random sample of 50 bottles filled by machine \(X\) and a random sample of 60 bottles filled by machine \(Y\). The volumes of juice, \(x\) and \(y\) respectively, measured in appropriate units, are summarised by $$\Sigma x = 45.5 , \quad \Sigma ( x - \bar { x } ) ^ { 2 } = 19.56 , \quad \Sigma y = 72.3 , \quad \Sigma ( y - \bar { y } ) ^ { 2 } = 30.25$$ where \(\bar { x }\) and \(\bar { y }\) are the sample means of the volume of juice in the bottles filled by \(X\) and \(Y\) respectively.

1. Find a 90\% confidence interval for the difference between the mean volume of juice in bottles filled by machine \(X\) and the mean volume of juice in bottles filled by machine \(Y\).
   A test at the \(\alpha \%\) significance level does not provide evidence that there is any difference in the means of the volume of juice in bottles filled by machine \(X\) and the volume of juice in bottles filled by machine \(Y\).
2. Find the set of possible values of \(\alpha\).

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

A farmer \(A\) grows two types of potato plants, Royal and Majestic. A random sample of 10 Royal plants is taken and the potatoes from each plant are weighed. The total mass of potatoes on a plant is \(x \mathrm {~kg}\). The data are summarised as follows. $$\Sigma x = 42.0 \quad \Sigma x ^ { 2 } = 180.0$$ A random sample of 12 Majestic plants is taken. The total mass of potatoes on a plant is \(y \mathrm {~kg}\). The data are summarised as follows. $$\Sigma y = 57.6 \quad \Sigma y ^ { 2 } = 281.5$$

1. Test, at the \(5 \%\) significance level, whether the population mean mass of potatoes from Royal plants is the same as the population mean mass of potatoes from Majestic plants. You may assume that both distributions are normal and you should state any additional assumption that you make.
   A neighbouring farmer \(B\) grows Crown potato plants. His plants produce 3.8 kg of potatoes per plant, on average. Farmer \(A\) claims that her Royal plants produce a higher mean mass of potatoes than Farmer \(B\) 's Crown plants.
2. Test, at the \(5 \%\) significance level, whether Farmer \(A\) 's claim is justified.

2 In the vegetable section of a local supermarket, leeks are on sale either loose (and unprepared) or prepared in packs of 4 . The weights of unprepared leeks are modelled by the random variable \(X\) which has the Normal distribution with mean 260 grams and standard deviation 24 grams. The prepared leeks have had \(40 \%\) of their weight removed, so that their weights, \(Y\), are modelled by \(Y = 0.6 X\).

1. Find the probability that a randomly chosen unprepared leek weighs less than 300 grams.
2. Find the probability that a randomly chosen prepared leek weighs more than 175 grams.
3. Find the probability that the total weight of 4 randomly chosen prepared leeks in a pack is less than 600 grams.
4. What total weight of prepared leeks in a randomly chosen pack of 4 is exceeded with probability 0.975 ?
5. Sandie is making soup. She uses 3 unprepared leeks and 2 onions. The weights of onions are modelled by the Normal distribution with mean 150 grams and standard deviation 18 grams. Find the probability that the total weight of her ingredients is more than 1000 grams.
6. A large consignment of unprepared leeks is delivered to the supermarket. A random sample of 100 of them is taken. Their weights have sample mean 252.4 grams and sample standard deviation 24.6 grams. Find a \(99 \%\) confidence interval for the true mean weight of the leeks in this consignment.

3 Engineers in charge of a chemical plant need to monitor the temperature inside a reaction chamber. Past experience has shown that when functioning correctly the temperature inside the chamber can be modelled by a Normal distribution with mean \(380 ^ { \circ } \mathrm { C }\). The engineers are concerned that the mean operating temperature may have fallen. They decide to test the mean using the following random sample of 12 recent temperature readings.

1. Give three reasons why a \(t\) test would be appropriate.
2. Carry out the test using a \(5 \%\) significance level. State your hypotheses and conclusion carefully.
3. Find a 95\% confidence interval for the true mean temperature in the reaction chamber.
4. Describe briefly one advantage and one disadvantage of having a 99\% confidence interval instead of a 95\% confidence interval.

4 The distance a particular football player runs in a match is modelled by a normal distribution with standard deviation 0.3 kilometres. A random sample of \(n\) matches is taken.
The distance the player runs in this sample of matches has mean 10.8 kilometres.
The sample is used to construct a \(93 \%\) confidence interval for the mean, of width 0.0543 kilometres, correct to four decimal places. 4

1. Find the value of \(n\) 4
2. Find the \(93 \%\) confidence interval for the mean, giving the limits to three decimal places.
   4
3. Alison claims that the population mean distance the player runs is 10.7 kilometres. She carries out a hypothesis test at the 7\% level of significance using the random sample and the hypotheses $$\begin{aligned} & \mathrm { H } _ { 0 } : \mu = 10.7 \\ & \mathrm { H } _ { 1 } : \mu \neq 10.7 \end{aligned}$$ 4 (c) (i) State, with a reason, whether the null hypothesis will be accepted or rejected. 4 (c) (ii) Describe, in the context of the hypothesis test in part (c)(i), what is meant by a Type II error. \includegraphics[max width=\textwidth, alt={}, center]{9be40ed6-6df8-426a-8afd-fefc17287de6-06_2488_1730_219_141}

7 A club secretary collects data about the time, \(T\) minutes, needed to process the details of a new member. The mean of \(T\) is denoted by \(\mu\). The variance of \(T\) is denoted by \(\sigma ^ { 2 }\). The results of a random sample of 40 observations of \(T\) are summarised as follows. \(\mathrm { n } = 40 \quad \Sigma \mathrm { t } = 396.0 \quad \Sigma \mathrm { t } ^ { 2 } = 4271.40\)

1. Determine a 99\% confidence interval for \(\mu\).
2. The secretary discovers that over a long period the value of \(\sigma ^ { 2 }\) is in fact 10.0 . The secretary collects an independent random sample of 50 observations of \(T\) and constructs a new 99\% confidence interval for \(\mu\) based on this sample of size 50 , but using \(\sigma ^ { 2 } = 10.0\). Find the probability that this new confidence interval contains the value \(\mu + 1.6\).