One-tail z-test (upper tail)

Test whether the population mean has increased (H₁: μ > μ₀), using a one-tail test with positive critical value.

25 questions · Moderate -0.0

5.05c Hypothesis test: normal distribution for population mean
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CAIE S2 2022 June Q2
5 marks Moderate -0.3
2 In the past, the mean height of plants of a particular species has been 2.3 m . A random sample of 60 plants of this species was treated with fertiliser and the mean height of these 60 plants was found to be 2.4 m . Assume that the standard deviation of the heights of plants treated with fertiliser is 0.4 m . Carry out a test at the \(2.5 \%\) significance level of whether the mean height of plants treated with fertiliser is greater than 2.3 m .
CAIE S2 2023 June Q3
6 marks Standard +0.3
3 In the past, the annual amount of wheat produced per farm by a large number of similar sized farms in a certain region had mean 24.0 tonnes and standard deviation 5.2 tonnes. Last summer a new fertiliser was used by all the farms, and it was expected that the mean amount of wheat produced per farm would be greater than 24.0 tonnes. In order to test whether this was true, a scientist recorded the amounts of wheat produced by a random sample of 50 farms last summer. He found that the value of the sample mean was 25.8 tonnes. Stating a necessary assumption, carry out the test at the \(1 \%\) significance level.
CAIE S2 2021 March Q3
4 marks Moderate -0.8
3 An architect wishes to investigate whether the buildings in a certain city are higher, on average, than buildings in other cities. He takes a large random sample of buildings from the city and finds the mean height of the buildings in the sample. He calculates the value of the test statistic, \(z\), and finds that \(z = 2.41\).
  1. Explain briefly whether he should use a one-tail test or a two-tail test.
  2. Carry out the test at the \(1 \%\) significance level.
CAIE S2 2017 June Q3
8 marks Standard +0.3
3 Household incomes, in thousands of dollars, in a certain country are represented by the random variable \(X\) with mean \(\mu\) and standard deviation \(\sigma\). The incomes of a random sample of 400 households are found and the results are summarised below. $$n = 400 \quad \Sigma x = 923 \quad \Sigma x ^ { 2 } = 3170$$
  1. Calculate unbiased estimates of \(\mu\) and \(\sigma ^ { 2 }\).
  2. A random sample of 50 households in one particular region of the country is taken and the sample mean income, in thousands of dollars, is found to be 2.6 . Using your values from part (i), test at the \(5 \%\) significance level whether household incomes in this region are greater, on average, than in the country as a whole.
CAIE S2 2011 June Q3
6 marks Standard +0.3
3 Past experience has shown that the heights of a certain variety of rose bush have been normally distributed with mean 85.0 cm . A new fertiliser is used and it is hoped that this will increase the heights. In order to test whether this is the case, a botanist records the heights, \(x \mathrm {~cm}\), of a large random sample of \(n\) rose bushes and calculates that \(\bar { x } = 85.7\) and \(s = 4.8\), where \(\bar { x }\) is the sample mean and \(s ^ { 2 }\) is an unbiased estimate of the population variance. The botanist then carries out an appropriate hypothesis test.
  1. The test statistic, \(z\), has a value of 1.786 correct to 3 decimal places. Calculate the value of \(n\).
  2. Using this value of the test statistic, carry out the test at the \(5 \%\) significance level.
CAIE S2 2014 June Q3
5 marks Moderate -0.8
3 The lengths, in centimetres, of rods produced in a factory have mean \(\mu\) and standard deviation 0.2. The value of \(\mu\) is supposed to be 250 , but a manager claims that one machine is producing rods that are too long on average. A random sample of 40 rods from this machine is taken and the sample mean length is found to be 250.06 cm . Test at the \(5 \%\) significance level whether the manager's claim is justified.
CAIE S2 2019 June Q2
7 marks Moderate -0.3
2 The time, in minutes, that John takes to travel to work has a normal distribution. Last year the mean and standard deviation were 26.5 and 4.8 respectively. This year John uses a different route and he finds that the mean time for his first 150 journeys is 27.5 minutes.
  1. Stating a necessary assumption, test at the \(1 \%\) significance level whether the mean time for his journey to work has increased.
  2. State, with a reason, whether it was necessary to use the Central Limit theorem in your answer to part (i).
CAIE S2 2012 November Q2
5 marks Moderate -0.3
2 The heights of a certain type of plant have a normal distribution. When the plants are grown without fertilizer, the population mean and standard deviation are 24.0 cm and 4.8 cm respectively. A gardener wishes to test, at the \(2 \%\) significance level, whether Hiergro fertilizer will increase the mean height. He treats 150 randomly chosen plants with Hiergro and finds that their mean height is 25.0 cm . Assuming that the standard deviation of the heights of plants treated with Hiergro is still 4.8 cm , carry out the test.
CAIE S2 2012 November Q5
8 marks Standard +0.3
5 It is claimed that, on average, people following the Losefast diet will lose more than 2 kg per month. The weight losses, \(x\) kilograms per month, of a random sample of 200 people following the Losefast diet were recorded and summarised as follows. $$n = 200 \quad \Sigma x = 460 \quad \Sigma x ^ { 2 } = 1636$$
  1. Calculate unbiased estimates of the population mean and variance.
  2. Test the claim at the \(1 \%\) significance level.
OCR MEI S2 2007 June Q1
19 marks Standard +0.3
1 The random variable \(X\) represents the time taken in minutes for a haircut at a barber's shop. \(X\) is Normally distributed with mean 11 and standard deviation 3 .
  1. Find \(\mathrm { P } ( X < 10 )\).
  2. Find the probability that exactly 3 out of 8 randomly selected haircuts take less than 10 minutes.
  3. Use a suitable approximating distribution to find the probability that at least 50 out of 100 randomly selected haircuts take less than 10 minutes. A new hairdresser joins the shop. The shop manager suspects that she takes longer on average than the other staff to do a haircut. In order to test this, the manager records the time taken for 25 randomly selected cuts by the new hairdresser. The mean time for these cuts is 12.34 minutes. You should assume that the time taken by the new hairdresser is Normally distributed with standard deviation 3 minutes.
  4. Write down suitable null and alternative hypotheses for the test.
  5. Carry out the test at the \(5 \%\) level.
Edexcel Paper 3 2020 October Q5
15 marks Standard +0.3
  1. A health centre claims that the time a doctor spends with a patient can be modelled by a normal distribution with a mean of 10 minutes and a standard deviation of 4 minutes.
    1. Using this model, find the probability that the time spent with a randomly selected patient is more than 15 minutes.
    Some patients complain that the mean time the doctor spends with a patient is more than 10 minutes. The receptionist takes a random sample of 20 patients and finds that the mean time the doctor spends with a patient is 11.5 minutes.
  2. Stating your hypotheses clearly and using a \(5 \%\) significance level, test whether or not there is evidence to support the patients' complaint. The health centre also claims that the time a dentist spends with a patient during a routine appointment, \(T\) minutes, can be modelled by the normal distribution where \(T \sim \mathrm {~N} \left( 5,3.5 ^ { 2 } \right)\)
  3. Using this model,
    1. find the probability that a routine appointment with the dentist takes less than 2 minutes
    2. find \(\mathrm { P } ( T < 2 \mid T > 0 )\)
    3. hence explain why this normal distribution may not be a good model for \(T\). The dentist believes that she cannot complete a routine appointment in less than 2 minutes.
      She suggests that the health centre should use a refined model only including values of \(T > 2\)
  4. Find the median time for a routine appointment using this new model, giving your answer correct to one decimal place.
OCR MEI Paper 2 2021 November Q11
8 marks Standard +0.3
11 In 2010 the heights of adult women in the UK were found to have mean \(\mu = 161.6 \mathrm {~cm}\) and variance \(\sigma ^ { 2 } = 1.96 \mathrm {~cm} ^ { 2 }\). It is believed that the mean height of adult women in 2020 in the UK is greater than in 2010. In 2020 a researcher collected a random sample of the heights of 200 adult women in the UK. The researcher calculated the sample mean height and carried out a hypothesis test at the \(5 \%\) level to investigate whether there was any evidence to suggest that the mean height of adult women in the UK had increased. The researcher assumed that the variance was unaltered.
  1. - State suitable hypotheses for the test, defining any variables you use.
    The researcher found that the sample mean was 161.9 cm and made the following statements.
Edexcel S3 Specimen Q1
7 marks Moderate -0.3
  1. A report states that employees spend, on average, 80 minutes every working day on personal use of the Internet. A company takes a random sample of 100 employees and finds their mean personal Internet use is 83 minutes with a standard deviation of 15 minutes. The company's managing director claims that his employees spend more time on average on personal use of the Internet than the report states.
Test, at the \(5 \%\) level of significance, the managing director's claim. State your hypotheses clearly.
AQA S2 2008 January Q1
6 marks Moderate -0.3
1 David claims that customers have to queue at a supermarket checkout for more than 5 minutes, on average. The queuing times, \(x\) minutes, of 40 randomly selected customers result in \(\bar { x } = 5.5\) and \(s ^ { 2 } = 1.31\). Investigate, at the \(1 \%\) level of significance, David's claim.
AQA S2 2010 January Q1
5 marks Moderate -0.3
1 Roger claims that, on average, his journey time from home to work each day is greater than 45 minutes. The times, \(x\) minutes, of 30 randomly selected journeys result in \(\bar { x } = 45.8\) and \(s ^ { 2 } = 4.8\).
Investigate Roger's claim at the \(1 \%\) level of significance.
AQA S2 2012 June Q2
8 marks Moderate -0.3
2 The times taken to complete a round of golf at Slowpace Golf Club may be modelled by a random variable with mean \(\mu\) hours and standard deviation 1.1 hours. Julian claims that, on average, the time taken to complete a round of golf at Slowpace Golf Club is greater than 4 hours. The times of 40 randomly selected completed rounds of golf at Slowpace Golf Club result in a mean of 4.2 hours.
  1. Investigate Julian's claim at the \(5 \%\) level of significance.
  2. If the actual mean time taken to complete a round of golf at Slowpace Golf Club is 4.5 hours, determine whether a Type I error, a Type II error or neither was made in the test conducted in part (a). Give a reason for your answer.
OCR MEI Further Statistics B AS 2022 June Q3
8 marks Standard +0.3
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
Mean3.565
s1.05
N40
Result
Z Test of a Mean
Mean3.565
S1.05
SE0.1660
N40
Z0.994
p0.160
  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.
OCR MEI Further Statistics B AS 2021 November Q3
12 marks Standard +0.3
3 The weights in kg of male otters in a large river system are known to be Normally distributed with mean 8.3 and standard deviation 1.8. A researcher believes that weights of male otters in another river are higher because of what he suspects is better availability of food. The researcher records the weights of a random sample of 9 male otters in this other river. The sum of these 9 weights is 83.79 kg .
  1. In this question you must show detailed reasoning. You should assume that:
    Show that a test at the \(5 \%\) significance level provides sufficient evidence to conclude that the mean weight of male otters in the other river is greater than 8.3 kg .
  2. Explain whether the result of the test suggests that the weights are higher due to better availability of food.
  3. If the standard deviation of the weights of otters in the other river could not be assumed to be 1.8 kg , name an alternative test that the researcher could carry out to investigate otter weights.
  4. Explain why, even if a test at the \(5 \%\) significance level results in the rejection of the null hypothesis, you cannot be sure that the alternative hypothesis is true.
OCR MEI Further Statistics Major 2021 November Q5
17 marks Standard +0.3
5 A manufacturer uses three types of capacitor in a particular electronic device. The capacitances, measured in suitable units, are modelled by independent Normal distributions with means and standard deviations as shown in the table.
\cline { 2 - 3 } \multicolumn{1}{c|}{}Capacitance
TypeMean
Standard
deviation
A3.90.32
B7.80.41
C30.20.64
  1. Determine the probability that the total capacitance of a randomly chosen capacitor of Type B and two randomly chosen capacitors of Type A is at least 16 units.
  2. Determine the probability that the capacitance of a randomly chosen capacitor of Type C is within 1 unit of the total capacitance of four randomly chosen capacitors of Type B. When the manufacturer gets a new batch of 1000 capacitors from the supplier, a random sample of 10 of them is tested to check the capacitances. For a new batch of Type C capacitors, summary statistics for the capacitances, \(x\) units, of the random sample are as follows. \(n = 10\) $$\sum x = 299.6 \quad \sum x ^ { 2 } = 8981.0$$ You should assume that the capacitances of the sample come from a Normally distributed population, but you should not assume that the standard deviation is 0.64 as for previous Type C capacitors.
  3. In this question you must show detailed reasoning. Carry out a hypothesis test at the \(5 \%\) significance level to check whether it is reasonable to assume that the capacitors in this batch have the specified mean capacitance for Type C of 30.2 units.
OCR S2 Q6
10 marks Standard +0.3
6 Alex obtained the actual waist measurements, \(w\) inches, of a random sample of 50 pairs of jeans, each of which was labelled as having a 32 -inch waist. The results are summarised by $$n = 50 , \quad \Sigma w = 1615.0 , \quad \Sigma w ^ { 2 } = 52214.50$$ Test, at the \(0.1 \%\) significance level, whether this sample provides evidence that the mean waist measurement of jeans labelled as having 32 -inch waists is in fact greater than 32 inches. State your hypotheses clearly. \section*{Jan 2006}
OCR H240/02 2022 June Q11
7 marks Standard +0.3
11 In the past the masses of new-born babies in a certain country were normally distributed with mean 3300 g . Last year a publicity campaign was held to encourage pregnant women to improve their diet. Following this campaign, it is required to test whether the mean mass of new-born babies has increased. A random sample of 200 new-born babies is chosen, and it is found that their mean mass is 3360 g . It is given that the standard deviation of the masses of new-born babies is 450 g . Carry out the test at the 2.5\% significance level.
AQA Further Paper 3 Statistics 2024 June Q8
5 marks Moderate -0.3
8
16
256 2 The random variable \(T\) has an exponential distribution with mean 2 Find \(\mathrm { P } ( T \leq 1.4 )\) Circle your answer. \(\mathrm { e } ^ { - 2.8 }\) \(\mathrm { e } ^ { - 0.7 }\) \(1 - e ^ { - 0.7 }\) \(1 - \mathrm { e } ^ { - 2.8 }\) The continuous random variable \(Y\) has cumulative distribution function $$\mathrm { F } ( y ) = \left\{ \begin{array} { l r } 0 & y < 2 \\ - \frac { 1 } { 9 } y ^ { 2 } + \frac { 10 } { 9 } y - \frac { 16 } { 9 } & 2 \leq y < 5 \\ 1 & y \geq 5 \end{array} \right.$$ Find the median of \(Y\) Circle your answer. 2 \(\frac { 10 - 3 \sqrt { 2 } } { 2 }\) \(\frac { 7 } { 2 }\) \(\frac { 10 + 3 \sqrt { 2 } } { 2 }\) Turn over for the next question 4 Research has shown that the mean number of volcanic eruptions on Earth each day is 20 Sandra records 162 volcanic eruptions during a period of one week. Sandra claims that there has been an increase in the mean number of volcanic eruptions per week. Test Sandra's claim at the \(5 \%\) level of significance.
5 The continuous random variable \(X\) has probability density function $$f ( x ) = \begin{cases} \frac { 1 } { 6 } e ^ { \frac { x } { 3 } } & 0 \leq x \leq \ln 27 \\ 0 & \text { otherwise } \end{cases}$$ Show that the mean of \(X\) is \(\frac { 3 } { 2 } ( \ln 27 - 2 )\) 6 Over time it has been accepted that the mean retirement age for professional baseball players is 29.5 years old. Imran claims that the mean retirement age is no longer 29.5 years old.
He takes a random sample of 5 recently retired professional baseball players and records their retirement ages, \(x\). The results are $$\sum x = 152.1 \quad \text { and } \quad \sum ( x - \bar { x } ) ^ { 2 } = 7.81$$ 6
  1. State an assumption that you should make about the distribution of the retirement ages to investigate Imran's claim. 6
  2. Investigate Imran's claim, using the 10\% level of significance.
WJEC Further Unit 5 2022 June Q5
13 marks Standard +0.3
5. A laboratory carrying out screening for a certain blood disorder claims that the average time taken for test results to be returned is 38 hours. A reporter for a national newspaper suspects that the results take longer, on average, to be returned than claimed by the laboratory. The reporter finds the time, \(x\) hours, for 50 randomly selected results, in order to conduct a hypothesis test. The following summary statistics were obtained. $$\sum x = 2163 \quad \sum x ^ { 2 } = 98508$$
  1. Calculate the \(p\)-value for the reporter's hypothesis test, and complete the test using a \(5 \%\) level of significance. Hence write a headline for the reporter to use.
  2. Explain the relevance or otherwise of the Central Limit Theorem to your answer in part (a).
  3. Briefly explain why a random sample is preferable to taking a batch of 50 consecutive results.
  4. On another occasion, the reporter took a different random sample of 10 results.
    1. State, with a reason, what type of hypothesis test the reporter should use on this occasion.
    2. State one assumption required to carry out this test.
CAIE S2 2016 June Q2
5 marks Moderate -0.3
In the past, the mean annual crop yield from a particular field has been 8.2 tonnes. During the last 16 years, a new fertiliser has been used on the field. The mean yield for these 16 years is 8.7 tonnes. Assume that yields are normally distributed with standard deviation 1.2 tonnes. Carry out a test at the 5\% significance level of whether the mean yield has increased. [5]
AQA Paper 3 2022 June Q17
6 marks Moderate -0.3
The number of working hours per week of employees in a company is modelled by a normal distribution with mean of 34 hours and a standard deviation of 4.5 hours. The manager claims that the mean working hours per week of the company's employees has increased. A random sample of 30 employees in the company was found to have mean working hours per week of 36.2 hours. Carry out a hypothesis test at the 2.5% significance level to investigate the manager's claim. [6 marks]