5.05a Sample mean distribution: central limit theorem

7. A telephone company believes that, for young people, the average length of a telephone call on a land line is longer than on a mobile, due to the difference in price. The company collected data on the time, \(t\) minutes, of 500 calls made by young people on mobiles and the data is summarised by $$\Sigma t = 7335 , \quad \Sigma t ^ { 2 } = 172040 .$$

1. Calculate unbiased estimates of the mean and variance of \(t\). For 200 calls made on land lines by the same young people, unbiased estimates of the mean and variance of the call length were 15.9 minutes and 108.5 minutes \({ } ^ { 2 }\) respectively.
2. Stating your hypotheses clearly, test at the \(5 \%\) level whether or not there is evidence that longer calls are made on land lines than on mobiles.
   (9 marks)
3. Explain the importance of the central limit theorem in carrying out the test in part (b).

5. The number of tornadoes per year to hit a particular town follows a Poisson distribution with mean \(\lambda\). A weatherman claims that due to climate changes the mean number of tornadoes per year has decreased. He records the number of tornadoes \(x\) to hit the town last year. To test the hypotheses \(\mathrm { H } _ { 0 } : \lambda = 7\) and \(\mathrm { H } _ { 1 } : \lambda < 7\), a critical region of \(x \leq 3\) is used.

1. Find, in terms \(\lambda\) the power function of this test.
2. Find the size of this test.
3. Find the probability of a Type II error when \(\lambda = 4\).

4 The random variable \(X\) has a continuous uniform distribution on [ 0,10 ].

1. Find \(\mathrm { P } ( 3 < X < 6 )\).
2. Find each of the following.
   • \(\mathrm { E } ( X )\)
   • \(\operatorname { Var } ( X )\)
   Marisa is investigating the sample mean, \(Y\), of 8 independent values of \(X\). She designs a simulation shown in the spreadsheet in Fig. 4.1. Each of the 25 rows below the heading row consists of 8 values of \(X\) together with the value of \(Y\). All of the values in the spreadsheet have been rounded to 2 decimal places. \begin{table}[h]

   1	A	B	C	D	E	F	G	H	I	J
   1	\(X _ { 1 }\)	\(X _ { 2 }\)	\(X _ { 3 }\)	\(X _ { 4 }\)	\(X _ { 5 }\)	\(X _ { 6 }\)	\(X _ { 7 }\)	\(X _ { 8 }\)	\(Y\)
   2	6.31	2.45	3.27	3.06	4.16	1.53	0.43	7.99	3.65
   3	1.70	1.52	7.10	8.93	6.44	2.70	9.96	7.83	5.77
   4	9.15	0.52	4.95	6.99	6.52	3.15	0.81	5.35	4.68
   5	0.65	2.71	7.92	9.65	0.50	4.87	6.46	2.67	4.43
   6	3.09	6.11	3.96	0.09	0.18	4.67	0.67	6.20	3.12
   7	7.06	5.84	1.97	3.60	9.36	1.97	4.48	3.47	4.72
   8	1.46	1.57	5.45	0.37	3.76	7.56	8.48	9.12	4.72
   9	9.42	1.85	4.91	1.61	1.94	8.00	1.77	5.34	4.36
   10	2.98	5.32	2.91	4.12	9.16	1.76	9.97	6.88	5.39
   11	2.83	3.44	3.28	7.85	1.00	0.93	8.77	4.03	4.01
   12	4.51	0.59	5.84	9.87	8.65	3.94	7.18	0.23	5.10
   13	4.49	0.69	3.65	8.78	4.96	8.96	3.77	1.43	4.59
   14	6.57	8.08	4.85	6.75	7.92	0.27	9.69	4.04	6.02
   15	8.35	1.09	8.63	8.04	7.23	2.12	2.57	9.59	5.95
   16	5.24	9.53	6.08	8.21	3.61	7.07	6.65	7.63	6.75
   17	7.89	5.50	3.09	0.71	6.47	5.49	6.47	4.95	5.07
   18	8.36	7.27	2.35	9.04	0.58	2.26	3.01	7.90	5.10
   19	3.76	1.01	9.61	9.65	7.89	9.98	6.28	4.34	6.56
   20	9.94	6.84	3.38	5.53	0.26	8.53	5.72	5.12	5.66
   21	7.25	9.10	0.34	2.88	4.66	2.65	6.37	7.63	5.11
   22	7.18	7.14	5.38	0.04	4.09	6.47	4.96	4.23	4.94
   23	8.69	5.04	4.90	2.94	2.00	4.23	4.13	0.97	4.11
   24	3.46	6.33	0.48	9.35	0.23	1.18	7.97	6.37	4.42
   25	2.37	7.26	7.16	1.24	5.26	2.80	3.55	3.84	4.19
   26	2.16	8.30	7.17	3.32	2.96	1.30	9.11	0.31	4.33
   27

   \captionsetup{labelformat=empty} \caption{Fig. 4.1}
   \end{table}
3. Use the spreadsheet to estimate \(\mathrm { P } ( 3 < Y < 6 )\).
4. Explain why it is not surprising that this estimated probability is substantially greater than the value which you calculated in part (i). Marisa wonders whether, even though the sample size is only 8, use of the Central Limit Theorem will provide a good approximation to \(\mathrm { P } ( 3 < Y < 6 )\).
5. Calculate an estimate of \(\mathrm { P } ( 3 < Y < 6 )\) using the Central Limit Theorem. A Normal probability plot of the 25 simulated values of \(Y\) is shown in Fig. 4.2. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{0c58d4d7-10e9-473a-888a-b407ec90bf08-5_800_1291_306_386} \captionsetup{labelformat=empty} \caption{Fig. 4.2}
   \end{figure}
6. Explain what the Normal probability plot suggests about the use of the Central Limit Theorem to approximate \(\mathrm { P } ( 3 < Y < 6 )\). Marisa now decides to use a spreadsheet with 1000 rows below the heading row, rather than the 25 which she used in the initial simulation shown in Fig. 4.1. She uses a counter to count the number of values of \(Y\) between 3 and 6. This value is 808.
7. Explain whether the value 808 supports the suggestion that the Central Limit Theorem provides a good approximation to \(\mathrm { P } ( 3 < Y < 6 )\). Marisa decides to repeat each of her two simulations many times in order to investigate how variable the probability estimates are in each case.
8. Explain whether you would expect there to be more, the same or less variability in the probability estimates based on 1000 rows than in the probability estimates based on 25 rows.

1 It is known that the red blood cell count of adults in a particular country, measured in suitable units, has mean 4.96 and variance 0.15.

1. Find the probability that the mean red blood cell count of a random sample of 50 adults from this country is at least 5.00.
2. Explain how you can find the probability in part (a) despite the fact that you do not know the distribution of red blood cell counts.

3 A local council collects domestic kitchen waste for composting. Householders place their kitchen waste in a 'compost bin' and this is emptied weekly by the council. The average weight of kitchen waste collected per household each week is known to be 3.4 kg . The council runs a campaign to try to increase the amount of kitchen waste per household which is put in the compost bin. After the campaign, a random sample of 40 households is selected and the weights in kg of kitchen waste in their compost bins are measured. A hypothesis test is carried out in order to investigate whether the campaign has been successful, using software to analyse the sample. The output from the software is shown below.
□
Z Test of a Mean
Null Hypothesis \(\mu = 3.4\) Alternative Hypothesis \(\bigcirc < 0 > 0 \neq\) Sample Result

1. Explain why the test is based on the Normal distribution even though the distribution of the population of amounts of kitchen waste per household is not known.
2. Using the output from the software, complete the test at the \(5 \%\) significance level.
3. Show how the value of \(Z\) in the software output was calculated.
4. Calculate the least value of the sample mean which would have resulted in the conclusion of the test in part (b) being different. You should assume that the standard error is unchanged.

6 The table below shows the mean and variance of the test scores of a random samples of 70 girls who are starting an A level Mathematics course.

1. Showing your working, find a \(95 \%\) confidence interval for the population mean.
2. Explain why you can construct the interval in part (i) despite no information about the distribution of the parent population being given.
3. The same random sample of girls repeats the test. The mean improvement in score is 0.9 . The \(95 \%\) confidence interval for the improvement is \([ - 1.5,3.3 ]\). What is the sample variance for the improvement in score?

4 Shellfish in the sea near nuclear power stations are regularly monitored for levels of radioactivity. On a particular occasion, the levels of caesium-137 (a radioactive isotope) in a random sample of 8 cockles, measured in becquerels per kilogram, were as follows. \(\begin{array} { l l l l l l l l } 2.36 & 2.97 & 2.69 & 3.00 & 2.51 & 2.45 & 2.21 & 2.63 \end{array}\) Software is used to produce a 95\% confidence interval for the level of caesium-137 in the cockles. The output from the software is shown in Fig. 4. The value for 'SE' has been deliberately omitted. T Estimate of a Mean
Confidence Level 0.95 Sample
Mean 2.6025
s 0.2793
□
0.2793 N □ 8 Result T Estimate of a Mean \begin{table}[h] \end{table}

1. State an assumption necessary for the use of the \(t\) distribution in the construction of this confidence interval.
2. State the confidence interval which the software gives in the form \(a < \mu < b\).
3. In the software output shown in Fig. 4, SE stands for standard error. Find the standard error in this case.
4. Show how the value of 0.2335 in the confidence interval was calculated.
5. State how, using this sample, a wider confidence interval could be produced.

4 A machine manufactures batches of 100 titanium sheets. The thickness of every sheet in a batch is Normally distributed with mean \(\mu \mathrm { mm }\) and standard deviation 0.03 mm . You should assume that each sheet is of uniform thickness and that the thicknesses of different sheets are independent of each other. The values of \(\mu\) for three different batches, A, B and C, are 3.125, 3.117 and 3.109 respectively.

1. Determine the probability that the total thickness of 10 sheets from Batch A is less than 31.0 mm .
2. Determine the probability that, if a single sheet from Batch A is cut into pieces and 10 of the pieces are stacked together, the total thickness of the stack is less than 31.0 mm .
3. Determine the probability that, if one sheet from each of Batches A, B and C are stacked together, the total thickness of the stack is at least 9.4 mm .
4. Determine the probability that the total thickness of 10 sheets from Batch A is less than the total thickness of 10 sheets from Batch B.

5 A researcher is investigating whether doing yoga has any effect on quality of sleep in older people. The researcher selects a random sample of 40 older people, who then complete a yoga course. Before they start the course and again at the end, the 40 people fill in a questionnaire which measures their perceived sleep quality. The higher the score, the better is the perceived quality of sleep. The researcher uses software to produce a 90\% confidence interval for the difference in mean sleep quality (sleep quality after the course minus sleep quality before the course). The output from the software is shown below. Z Estimate of a Mean Confidence level □ 0.9 Sample Result
Z Estimate of a Mean

1. Explain why the confidence interval is based on the Normal distribution even though the distribution of the population of differences is not known.
2. Explain whether the confidence interval suggests that the mean sleep qualities before and after completing a yoga course are different.
3. In the output from the software, SE stands for 'standard error'.
   1. Explain what standard error is.
   2. Show how the standard error was calculated in this case.
4. A colleague of the researcher suggests that the confidence level should have been \(95 \%\) rather than \(90 \%\). Determine whether this would have made a difference to your answer to part (b).

11 The discrete random variable \(X\) has a uniform distribution over the set of all integers between 25 and \(n\) inclusive, where \(n\) is a positive integer with \(n > 25\).

1. Determine \(\mathrm { P } \left( \mathrm { X } < \frac { \mathrm { n } + 25 } { 2 } \right)\) in each of the following cases.

10 The discrete random variables \(X\) and \(Y\) have distributions as follows: \(X \sim \mathrm {~B} ( 20,0.3 )\) and \(Y \sim \operatorname { Po } ( 3 )\). The spreadsheet in Fig. 10 shows a simulation of the distributions of \(X\) and \(Y\). Each of the 20 rows below the heading row consists of a value of \(X\), a value of \(Y\), and the value of \(X - 2 Y\). \begin{table}[h] \end{table}

1. Use the spreadsheet to estimate each of the following.
   The mean of 50 values of \(X - 2 Y\) is denoted by the random variable \(W\).
2. Calculate an estimate of \(\mathrm { P } ( W > 1 )\).

1 When babies are born, their head circumferences are measured. A random sample of 50 newborn female babies is selected. The sample mean head circumference is 34.711 cm . The sample standard deviation head circumference is 1.530 cm .

1. Determine a 95\% confidence interval for the population mean head circumference of newborn female babies.
2. Explain why you can calculate this interval even though the distribution of the population of head circumferences of newborn female babies is unknown.

11 The continuous random variable \(X\) has probability density function given by \(f ( x ) = \begin{cases} a x ^ { 2 } & 0 \leqslant x < 2 , \\ b ( 3 - x ) ^ { 2 } & 2 \leqslant x \leqslant 3 , \\ 0 & \text { otherwise } \end{cases}\) where \(a\) and \(b\) are positive constants.

1. Given that \(\mathrm { E } ( X ) = 2\), determine the values of \(a\) and \(b\).
2. Determine the median value of \(X\).
3. A random sample of 50 observations of \(X\) is selected. Given that \(\operatorname { Var } ( X ) = 0.2\), determine an estimate of the probability that the mean value of the 50 observations is less than 1.9.

5. The masses, \(X\), in kg, of men who work for a large company are normally distributed with mean 75 and standard deviation 10.

1. Find the probability that the mean mass of a random sample of 5 men is less than 70 kg .
2. The mean mass, in kg , of a random sample of \(n\) men drawn from this distribution is \(\bar { X }\). Given that \(\mathrm { P } ( \bar { X } > 80 )\) is approximately \(0 \cdot 007\), find \(n\). The masses, in kg, of women who work for the company are normally distributed with mean 68 and standard deviation 6 . A lift in the company building will not move if the total mass in the lift is more than 500 kg .
3. A random sample of 3 men and 4 women get in the lift. Find the probability that the lift will not move.
4. State a modelling assumption you have made in calculating your answer for part (c).

1. A biased spinner can land on the numbers \(1,2,3,4\) or 5 with the following probabilities.

The spinner will be spun 80 times and the mean of the numbers it lands on will be calculated. Find an estimate of the probability that this mean will be greater than 3.25
(6)

1. A six-sided die has sides labelled \(1,2,3,4,5\) and 6

The random variable \(S\) represents the score when the die is rolled.
Alicia rolls the die 45 times and the mean score, \(\bar { S }\), is calculated.
Assuming the die is fair and using a suitable approximation,

1. find, to 3 significant figures, the value of \(k\) such that \(\mathrm { P } ( \bar { S } < k ) = 0.05\)
2. Explain the relevance of the Central Limit Theorem in part (a). Alicia considers the following hypotheses: \(\mathrm { H } _ { 0 }\) : The die is fair \(\mathrm { H } _ { 1 }\) : The die is not fair
   If \(\bar { S } < 3.1\) or \(\bar { S } > 3.9\), then \(\mathrm { H } _ { 0 }\) will be rejected.
   Given that the true distribution of \(S\) has mean 4 and variance 3
3. find the power of this test.
4. Describe what would happen to the power of this test if Alicia were to increase the number of rolls of the die.
   Give a reason for your answer.

1. A courier delivers parcels. The random variable \(X\) represents the number of parcels delivered successfully each day by the courier where \(X \sim \mathrm {~B} ( 400,0.64 )\)

A random sample \(X _ { 1 } , X _ { 2 } , \ldots X _ { 100 }\) is taken.
Estimate the probability that the mean number of parcels delivered each day by the courier is greater than 257

1. The concentration of an air pollutant is measured in micrograms \(/ \mathrm { m } ^ { 3 }\)

Samples of air were taken at two different sites and the concentration of this particular air pollutant was recorded. For Site \(A\) the summary statistics are shown below. For Site \(B\) there were 9 samples of air taken.
A test of the hypothesis \(\mathrm { H } _ { 0 } : \sigma _ { A } ^ { 2 } = \sigma _ { B } ^ { 2 }\) against the hypothesis \(\mathrm { H } _ { 1 } : \sigma _ { A } ^ { 2 } \neq \sigma _ { B } ^ { 2 }\) is carried out using a \(2 \%\) level of significance.

1. State a necessary assumption required to carry out the test. Given that the assumption in part (a) holds,
2. find the set of values of \(s _ { B } ^ { 2 }\) that would lead to the null hypothesis being rejected,
3. find a 99\% confidence interval for the variance of the concentration of the air pollutant at Site A.

1. Camilo grows two types of apple, green apples and red apples.

The standard deviation of the weights of green apples is known to be 3.5 grams.
A random sample of 80 green apples has a mean weight of 128 grams.

1. Find a 98\% confidence interval for the mean weight of the population of green apples. Show your working clearly and give the confidence interval limits to 2 decimal places. Camilo believes that the mean weight of the population of green apples is more than 10 grams greater than the mean weight of the population of red apples. A random sample of \(n\) red apples has a mean weight of 117 grams.
   The standard deviation of the weights of the red apples is known to be 4 grams.
   A test of Camilo's belief is carried out at the 5\% level of significance.
2. State the null and alternative hypotheses for this test.
3. Find the smallest value of \(n\) for which the null hypothesis will be rejected.
4. Explain the relevance of the Central Limit Theorem in parts (a) and (c).
5. Given that \(n = 85\), state the conclusion of the hypothesis test.

2 The volume, in millilitres, of lemonade in mini-cans may be assumed to be normally distributed with a standard deviation of 3.5. The volumes, in millilitres, of lemonade in a random sample of 12 mini-cans were as follows.

1. Construct a \(98 \%\) confidence interval for the mean volume of lemonade in a mini-can, giving the limits to one decimal place.
2. On each mini-can is printed " 150 ml ". Comment on this, using the given sample and your confidence interval in part (a).
3. State why, in part (a), use of the Central Limit Theorem was not necessary.

4 Chopped lettuce is sold in bags nominally containing 100 grams.
The weight, \(X\) grams, of chopped lettuce, delivered by the machine filling the bags, may be assumed to be normally distributed with mean \(\mu\) and standard deviation 4.

1. Assuming that \(\mu = 106\), determine the probability that a randomly selected bag of chopped lettuce:
   1. weighs less than 110 grams;
   2. is underweight.
2. Determine the minimum value of \(\mu\) so that at most 2 per cent of bags of chopped lettuce are underweight. Give your answer to one decimal place.
3. Boxes each contain 10 bags of chopped lettuce. The mean weight of a bag of chopped lettuce in a box is denoted by \(\bar { X }\). Given that \(\mu = 108.5\) :
   1. write down values for the mean and variance of \(\bar { X }\);
   2. determine the probability that \(\bar { X }\) exceeds 110 .