Carry out hypothesis test

6 Last year, the mean time taken by students at a school to complete a certain test was 25 minutes. Akash believes that the mean time taken by this year's students was less than 25 minutes. In order to test this belief, he takes a large random sample of this year's students and he notes the time taken by each student. He carries out a test, at the \(2.5 \%\) significance level, for the population mean time, \(\mu\) minutes. Akash uses the null hypothesis \(\mathrm { H } _ { 0 } : \mu = 25\).

1. Give a reason why Akash should use a one-tailed test.
   Akash finds that the value of the test statistic is \(z = - 2.02\).
2. Explain what conclusion he should draw.
   In a different one-tailed hypothesis test the \(z\)-value was found to be 2.14 .
3. Given that this value would lead to a rejection of the null hypothesis at the \(\alpha \%\) significance level, find the set of possible values of \(\alpha\).
   The population mean time taken by students at another school to complete a test last year was \(m\) minutes. Sorin carries out a one-tailed test to determine whether the population mean this year is less than \(m\), using a random sample of 100 students. He assumes that the population standard deviation of the times is 3.9 minutes. The sample mean is 24.8 minutes, and this result just leads to the rejection of the null hypothesis at the 5\% significance level.
4. Find the value of \(m\).
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

2 In the past the yield of a certain crop, in tonnes per hectare, had mean 0.56 and standard deviation 0.08 Following the introduction of a new fertilizer, the farmer intends to test at the \(2.5 \%\) significance level whether the mean yield has increased. He finds that the mean yield over 10 years is 0.61 tonnes per hectare.

1. State two assumptions that are necessary for the test.
2. Carry out the test.

3 An employer has commissioned an opinion polling organisation to undertake a survey of the attitudes of staff to proposed changes in the pension scheme. The staff are categorised as management, professional and administrative, and it is thought that there might be considerable differences of opinion between the categories. There are 60,140 and 300 staff respectively in the categories. The budget for the survey allows for a sample of 40 members of staff to be selected for in-depth interviews.

1. Explain why it would be unwise to select a simple random sample from all the staff.
2. Discuss whether it would be sensible to consider systematic sampling.
3. What are the advantages of stratified sampling in this situation?
4. State the sample sizes in each category if stratified sampling with as nearly as possible proportional allocation is used. The opinion polling organisation needs to estimate the average wealth of staff in the categories, in terms of property, savings, investments and so on. In a random sample of 11 professional staff, the sample mean is \(\pounds 345818\) and the sample standard deviation is \(\pounds 69241\).
5. Assuming the underlying population is Normally distributed, test at the \(5 \%\) level of significance the null hypothesis that the population mean is \(\pounds 300000\) against the alternative hypothesis that it is greater than \(\pounds 300000\). Provide also a two-sided \(95 \%\) confidence interval for the population mean.
   [0pt] [10]

10 A certain forest contains only trees of a particular species. Dipak wished to take a random sample of 5 trees from the forest. He numbered the trees from 1 to 784. Then, using his calculator, he generated the random digits 14781049 . Using these digits, Dipak formed 5 three-digit numbers. He took the first, second and third digits, followed by the second, third and fourth digits and so on. In this way he obtained the following list of numbers for his sample. $$\begin{array} { l l l l l } 147 & 478 & 781 & 104 & 49 \end{array}$$

1. Explain why Dipak omitted the number 810 from his list.
2. Explain why Dipak's sample is not random. The mean height of all trees of this species is known to be 4.2 m . Dipak wishes to test whether the mean height of trees in the forest is less than 4.2 m . He now uses a correct method to choose a random sample of 50 trees and finds that their mean height is 4.0 m . It is given that the standard deviation of trees in the forest is 0.8 m .
3. Carry out the test at the \(2 \%\) significance level.

10 Club 65-80 Holidays fly jets between Liverpool and Magaluf. Over a long period of time records show that half of the flights from Liverpool to Magaluf take less than 153 minutes and \(5 \%\) of the flights take more than 183 minutes. An operations manager believes that flight times from Liverpool to Magaluf may be modelled by the Normal distribution.

1. Use the information above to write down the mean time the operations manager will use in his Normal model for flight times from Liverpool to Magaluf.
2. Use the information above to find the standard deviation the operations manager will use in his Normal model for flight times from Liverpool to Magaluf, giving your answer correct to 1 decimal place.
3. Data is available for 452 flights. A flight time of under 2 hours was recorded in 16 of these flights. Use your answers to parts (a) and (b) to determine whether the model is consistent with this data. The operations manager suspects that the mean time for the journey from Magaluf to Liverpool is less than from Liverpool to Magaluf. He collects a random sample of 24 flight times from Magaluf to Liverpool. He finds that the mean flight time is 143.6 minutes.
4. Use the Normal model used in part (c) to conduct a hypothesis test to determine whether there is evidence at the \(1 \%\) level to suggest that the mean flight time from Magaluf to Liverpool is less than the mean flight time from Liverpool to Magaluf.
   [0pt]
5. Identify two ways in which the Normal model for flight times from Liverpool to Magaluf might be adapted to provide a better model for the flight times from Magaluf to Liverpool. [2]

7 The lengths in mm of a random sample of 6 one-year-old fish of a particular species are as follows. \(\begin{array} { l l l l l l } 271 & 293 & 306 & 287 & 264 & 290 \end{array}\)

1. State an assumption required in order to find a confidence interval for the mean length of one-year-old fish of this species. Fig. 7 shows a Normal probability plot for these data. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{8d36bc92-07ac-40c3-9e75-26f2bc9d2fcc-07_599_753_646_246} \captionsetup{labelformat=empty} \caption{Fig. 7}
   \end{figure}
2. Explain why the Normal probability plot suggests that the assumption in part (a) may be valid.
3. In this question you must show detailed reasoning. Assuming that this assumption is true, find a 95\% confidence interval for the mean length of one-year-old fish of this species.

10 A researcher is investigating the actual lengths of time that patients spend at their appointments with the doctors at a certain clinic. There are 12 doctors at the clinic, and each doctor has 24 appointments per day. The researcher plans to choose a sample of 24 appointments on a particular day.

1. The researcher considers the following two methods for choosing the sample. Method A: Choose a random sample of 24 appointments from the 288 on that day.
   Method B: Choose one doctor's 1st and 2nd appointments. Choose another doctor's 3rd and 4th appointments and so on until the last doctor's 23rd and 24th appointments. For each of A and B state a disadvantage of using this method. Appointments are scheduled to last 10 minutes. The researcher suspects that the actual times that patients spend are more than 10 minutes on average. To test this suspicion, he uses method A , and takes a random sample of 24 appointments. He notes the actual time spent for each appointment and carries out a hypothesis test at the \(1 \%\) significance level.
2. Explain why a 1-tail test is appropriate. The population mean of the actual times that patients spend at their appointments is denoted by \(\mu\) minutes.
3. Assuming that \(\mu = 10\), state the probability that the conclusion of the test will be that \(\mu\) is not greater than 10 . The actual lengths of time, in minutes, that patients spend for their appointments may be assumed to have a normal distribution with standard deviation 3.4.
   [0pt]
4. Given that the total length of time spent for the 24 appointments is 285 minutes, carry out the test. [7]
5. In part (iv) it was necessary to use the fact that the sample mean is normally distributed. Give a reason why you know that this is true in this case.