Practice test

Question 1

A characteristic feature of a sample median is that it:

1. Is more robust against extreme values than the mean
2. Is a measure of central tendency
3. Is based upon ranking the data
4. All of the above

Answer 1

4. All of the above

Question 2

You take a random sample of 6 people and record the number of whole years since each member of the sample left school. The data obtained is: {1, 10, 0, 0, 1, 0}. What is the sample mean, median and mode for this data?

1. Mean = 2 years, Median = 0.5 years, Mode = 10 years
2. Mean = 2 years, Median = 0.5 years, Mode = 0 years
3. Mean = 2 years, Median = 1 years, Mode = 0 years
4. Mean = 0.5 years, Median = 2 years, Mode = 0 years

Answer 2

2. Mean = 2 years, Median = 0.5 years, Mode = 0 years

Question 3

You take a random sample of 5 people and record the number of whole years that each member of the sample has been a parent. The data obtained is: {2, 11, 1, 1, 5}. What is the standard deviation of this data (to 2 d.p.)?

1. 18.00
2. 3.00
3. 3.79
4. 4.24

Answer 3

4. 4.24

(1) Calculate mean = 20/5 = 4
(2) Calculate deviance = {-2, 7, -3, -3, 1}
(3) Square deviance = {4, 49, 9, 9, 1}
(4) Calculate modified average deviance (divide by N-1 rather than N, where N is sample size – here 5) = (4+49+9+9+1)/(5-1) = 72/4 = 18
(5) Take the square root: sqrt(18) = 4.24

Question 4

Calculate the probability (expressed as a percentage to 2 d.p.) of selecting someone at random from the population with IQ lower than 79. Assume that IQ is distributed normally in the population with population mean = 100 and population standard deviation = 15

1. 8.08%
2. 91.92%
3. 21.00%
4. It is impossible to calculate this probability from the information given

Answer 4

1. 8.08%

z = (x - pop. mean) / pop. s.d.
z = (79-100)/15 = -1.4
p = 0.0808 (using the SND table)
p = 8.08% as a percentage

Question 5

You have a regular pack of cards in your hand (with no jokers). You pick a card at random and show it to a friend. The friend tells you the card is red. What is the probability of having picked a heart i) before and ii) after your friend’s revelation? (assume your friend is not lying)

1. before: 1/4; after: 1/2
2. before: 1/2; after: 1/2
3. before: 1/2; after: 1/4
4. before: 1/4; after: 1/4

Answer 5

1. before: 1/4; after: 1/2

Probability of an event = (No of outcomes consistent with that event)/(No of total possible outcomes)

Part i) Before knowing card is red: P(heart) = 13/52 = 1/4

Part ii) When know card is red: P(heart) = 13/26 = 1/2

Question 6

You take a course in woodwork and a course in needlework. At the end of the year you take an exam from each course (both marked out of 50) and score 25 in needlework and 38 in woodwork. The class mean and standard deviation for needlework are 16 and 8 respectively. The corresponding values for woodwork are 35 and 6 respectively. All other things being equal, which of the following statements is true?

1. Performance in the needlework test was best because my score was 1.125 standard deviations above the mean score, compared to only 0.5 standard deviations above the mean score in woodwork
2. Performance in the woodwork test was best because my score was higher than in the needlework test
3. Performance in the needlework test was best because my score was 1.5 standard deviations above the mean score, compared to only 1.0 standard deviations above the mean score in woodwork
4. None of the above

Answer 6

1. Performance in the needlework test was best because my score was 1.125 standard deviations above the mean score, compared to only 0.5 standard deviations above the mean score in woodwork

Question 7

A particular population of scores has mean 30 and standard deviation 10. You randomly select a sample of 16 scores from the distribution and calculate the mean. If you repeat this process many times and create a sampling distribution of the mean what are the theoretical mean and standard deviation of this sampling distribution?

1. mean = 30, standard deviation = 0.625
2. mean = 30, standard deviation = 10
3. It is impossible to estimate these quantities from the information given
4. mean = 30, standard deviation = 2.5

Answer 7

4. mean = 30, standard deviation = 2.5

standard error = population s.d. / SQRT sample size

standard error = 10 / sqrt(16) = 10/4 = 2.5

Question 8

You sample the following data {5, 7, 10, 12, 3, 5} from a Normally distributed population with a known standard deviation of 1.5 but unknown mean. Which of the following represents a 95% confidence interval for the population mean centred on the sample mean (to 1 d.p.):

1. It is impossible to calculate the confidence interval from the information provided
2. {5.8, 8.2}
3. {5.6, 8.4}
4. {5.0, 9.0}

Answer 8

2. {5.8, 8.2}

• We know the population s.d. so can calculate the standard error as 1.5 / sqrt(6) = 0.61
• We also know the sampling distribution of the mean is normal. We know the sample mean is (5+7+10+12+3+5)/6 = 42/6 = 7
• We then use the following equation
  CI95 = m - (1.96 x s.e.)
  CI95 = m + (1.96 x s.e.)
  CI95 = 7 - (0.61 x 1.96) = 5.8
  CI95 = 7 + (0.61 x 1.96) = 8.2
  CI95 = (5.8, 8.2)

Question 9

You calculate the sample mean and sample variance of a data set to be 400 and 225 respectively. What is the sample standard deviation?

1. 20
2. It is not possible to calculate this quantity from the information provided
3. 15
4. 225

Answer 9

3. 15

SD = SQRT sample variance
SD = SQRT (225) = 15

Question 10

Harry, a Clinical Psychology research student, is investigating the impact of a new exercise-based intervention for treating depression. He knows the population mean (15) and standard deviation (4) for depression scores (using a validated questionnaire) in the population of people with a depression diagnosis in the UK. He formulates a 2-tailed research hypothesis stating that after engaging in the exercise programme people will have different depression scores from the population. To test his hypothesis he takes a sample of 16 people from the UK with a depression diagnosis, who all take part in the exercise programme for a month. At the end he gives them the questionnaire and measures their mean depression score as 14. Using the table of areas under the Standard Normal Distribution, what is the conditional probability (expressed as a proportion to 4 d.p.) of having obtained a sample mean as extreme or more extreme as this if Harry assumes the null hypothesis to be true?

Answer 10

0.3174

1) s.e. = pop. s.d. / SQRT sample size
s.e. = 4 / SQRT 16 = 1

2) z = (x - mean) / s.e.
z = (14-15) / 1 = -1

3) Following a 2-tailed (conditional) probability of having obtained a z score as extreme as this (i.e. below -1 or above +1)

4) Using the table we see that the area under the SND above a z-score of 1 is 0.1587

5) WE must have the same area below -1 and so the area in 2 tails is 2 x 0.1587 = 0.3174

Question 11

Daphne, an Educational Psychologist, wants to test the hypothesis that children in year 6 in the Bigtown school district have higher levels of numeracy than the general UK population of year 6 students. She is in the fortuitous position of knowing that population numeracy test scores are normal with mean 58.31 and standard deviation 14.42. She takes a random sample of 24 students from year 6 in Bigtown, with mean numeracy score of 59.60. Calculate the z-score (to 2 d.p.) that will allow Daphne to test her hypothesis.

Answer 11

0.44

1) s.e. = pop. s.d. / SQRT sample size
s.e. = 14.42 / SQRT 24 = 2.94

2) z = (x - mean) / s.e.
z = (59.60-58.31) / 2.94 = 0.44

Question 12

A headteacher suspects that a class in her school is performing worse in maths than the national average in a key stage 2 maths exam. She has access to the population mean (57.21) and standard deviation (15.33) for this exam as well as the sample mean test score (53.47) for her set of 25 students. She formulates a 2-tailed research hypothesis stating that scores for her class are different from those in the population. What is the value of the test statistic (to 2 d.p.) that will allow the head to test her hypothesis?

Answer 12

-1.22

1) s.e. = pop. s.d. / SQRT sample size
s.e. = 15.33 / SQRT 25 = 3.066

2) z = (x - mean) / s.e.
z = (53.47 - 57.21) / 3.066 = -1.22

Question 13

An Experimental Psychologist is carrying out a study on human visual perception. She measures the minimum motion detection threshold (i.e. the lowest movement speed that can be reliably detected in a visual stimulus) in a sample of 25 people. She gets a sample mean of 1.021 millimetres per second (mm/s) and a sample s.d. of 0.112 mm/s. She does not have access to the population standard deviation for minimum motion detection thresholds. Calculate the lower end of the 95% confidence interval for the population mean centred on the sample mean (to 4 d.p.). (Remember you will need to use the table of critical values for the appropriate t-distribution)

Answer 13

0.9748

1) e.s.e.= sample. s.d. / SQRT sample size
e.s.e.= 0.112 / SQRT 25 = 0.0224

2) C value = N-1 and refer to table
25-1 = 24
Table 0.05 at v24 = 2.064

3) CI95 lower end =
sample mean - (c x e.s.e.)
1.021 - (c0.0224 x 2.064) = 0.9748

Question 14

In a particular school district, scores on a maths test (marked out of 100) for year 7 students are normally distributed with population mean and standard deviation of 60 marks and 15 marks respectively. You randomly sample 9 students from a school in the district who have taken the test. The mean score for this sample is 52.5 marks. You want to test the hypothesis that students in this school perform worse than the district average.

a) The hypothesis stated above is: [?]-tailed

• 1 tailed
• 2 tailed

b) The standard error of the mean is: [?] (to 1 d.p.)

c) The appropriate value of the z-statistic required to test the hypothesis is: [?] (to 2 d.p.)

d) Based on the hypothesis stated above, the statistic recovered in part c) and using an appropriate table, state whether you should reject the null hypothesis: [?]

• Reject
• Fail to reject

Answer 14

a) The hypothesis stated above is: 1 tailed

b) The standard error of the mean is: 5.0
s.e. = 15 / SQRT 9 = 5

c) The appropriate value of the z-statistic required to test the hypothesis is: -1.50

z = (52.5 - 60) / 5 = -1.5

d) Based on the hypothesis stated above, the statistic recovered in part c) and using an appropriate table, state whether you should reject the null hypothesis: Fail to reject

p = 0.0668
p > 0.05

Question 15

In a particular school district, scores on a maths test (marked out of 100) for year 7 students are normally distributed with population mean of 60 marks (the population standard deviation is unknown). You randomly sample 16 students from a school in the district who have taken the test. The mean score for this sample is 65 marks and the sample s.d. is 16. You want to test the hypothesis that students in this school perform better than the district average.

a) The hypothesis stated above is: [?]-tailed

• 1 tailed
• 2 tailed

b) The standard error of the mean is: [?] (to 1 d.p.)

c) The appropriate value of the z-statistic required to test the hypothesis is: [?] (to 2 d.p.)

d) Based on the hypothesis stated above, the statistic recovered in part c) and using an appropriate table, state whether you should reject the null hypothesis: [?]

• Reject
• Fail to reject

Answer 15

a) The hypothesis stated above is: 1 tailed

b) The estimated standard error of the mean is: 4.0
e.s.e. = 16 / SQRT 16 = 4

c) The appropriate value of the t-statistic required to test the hypothesis is: 1.25

t = (65 - 60) / 4 = 1.25

d) Based on the hypothesis stated above, the statistic recovered in part c) and using an appropriate table, state whether you should reject the null hypothesis: Fail to reject

v = 16 - 15
v15 1 tailed = 1.753 (p = 0.05) is closest to 1.25
but 1.753 > 1.25
so p > 0.05