Chapter 8
Inferential statistics
Puzzle 1
Explain the central limit theorem.
This theorem states that when samples are large (above about 30) the sampling distribution of a parameter (e.g., the mean) will take the shape of a normal distribution regardless of the shape of the population from which the sample was drawn. For small samples, the tdistribution better approximates the shape of the sampling distribution. We also know from this theorem that the standard deviation of the sampling distribution (i.e., the standard error of the sample mean) is well approximated by the standard deviation of the sample(s) divided by the square root of the sample size (N).
Puzzle 2
Using the data in Table 8.2 (in the book), what was the mean accuracy in both the fakename group and the ownname group?
To calculate the mean accuracy in both groups, we need to add together all the scores in each group and then divide the sum by the total number of scores. If we start with the ownname group, the sum of all the scores added together was 1547 and there were 35 scores in total, so the mean accuracy in the ownname group was 1547/35 = 44.2. In the fakename group, the sum of all the scores was 1806 and there were 33 scores in total, so the mean accuracy in the fakename group was 1806/33 = 54.73.
Puzzle 3
Using the data in Table 8.2 (in the book), what was the standard error in both the fakename group and the ownname group?
Let’s start with the ownname group. First we need to calculate the sum of squares by subtracting the mean from each score, squaring each deviance and then adding up the squared deviances.
Calculating the sums of squared errors for the ownname group  

Accuracy $x$ 
Mean $\bar{X}$ 
Deviance $x\bar{X}$ 
Deviance squared $(x\bar{X})^2$ 

20  44.2  24.2  585.64  
47  44.2  2.8  7.84  
47  44.2  2.8  7.84  
63  44.2  18.8  353.44  
30  44.2  14.2  201.64  
27  44.2  17.2  295.84  
81  44.2  36.8  1354.24  
42  44.2  2.2  4.84  
35  44.2  9.2  84.64  
23  44.2  21.2  449.44  
35  44.2  9.2  84.64  
75  44.2  30.8  948.64  
24  44.2  20.2  408.04  
65  44.2  20.8  432.64  
33  44.2  11.2  125.44  
44  44.2  0.2  0.04  
53  44.2  8.8  77.44  
61  44.2  16.8  282.24  
37  44.2  7.2  51.84  
52  44.2  7.8  60.84  
43  44.2  1.2  1.44  
83  44.2  38.8  1505.44  
35  44.2  9.2  84.64  
83  44.2  38.8  1505.44  
66  44.2  21.8  475.24  
40  44.2  4.2  17.64  
42  44.2  2.2  4.84  
40  44.2  4.2  17.64  
32  44.2  12.2  148.84  
10  44.2  34.2  1169.64  
43  44.2  1.2  1.44  
40  44.2  4.2  17.64  
44  44.2  0.2  0.04  
25  44.2  19.2  368.64  
27  44.2  17.2  295.84  
Sum  11,431.60 
So, the sum of squared errors is 11,431.6. The variance is the sum of squared errors divided by the degrees of freedom ($N  1$). There were 35 scores and so the degrees of freedom were 34. The variance is, therefore,
$$ s^2 = \frac{\text{sum of squared error}}{N1} = \frac{11431.6}{34} = 336.22. $$
The standard deviation is the square root of the variance
$$ s = \sqrt{336.22} = 18.34. $$
We can now calculate the standard error using Equation (8.3) in the book
$$ \sigma_{\bar{X}} = \frac{s}{\sqrt{n}} = \frac{18.34}{\sqrt{35}} = 3.10. $$
The standard error of the ownname group is 3.10.
Now let’s calculate the standard error of the fakename group in exactly the same way as we did for the ownname group.
Calculating the sums of squared errors for the fakename group  

Accuracy $x$ 
Mean $\bar{X}$ 
Deviance $x\bar{X}$ 
Deviance squared $(x\bar{X})^2$ 

40  54.72727  14.727273  216.892562  
69  54.72727  14.272727  203.710744  
60  54.72727  5.272727  27.801653  
41  54.72727  13.727273  188.438017  
82  54.72727  27.272727  743.801653  
40  54.72727  14.727273  216.892562  
64  54.72727  9.272727  85.983471  
22  54.72727  32.727273  1071.074380  
45  54.72727  9.727273  94.619835  
78  54.72727  23.272727  541.619835  
57  54.72727  2.272727  5.165289  
64  54.72727  9.272727  85.983471  
59  54.72727  4.272727  18.256198  
89  54.72727  34.272727  1174.619835  
96  54.72727  41.272727  1703.438017  
38  54.72727  16.727273  279.801653  
52  54.72727  2.727273  7.438017  
79  54.72727  24.272727  589.165289  
63  54.72727  8.272727  68.438017  
50  54.72727  4.727273  22.347107  
48  54.72727  6.727273  45.256198  
81  54.72727  26.272727  690.256198  
52  54.72727  2.727273  7.438017  
33  54.72727  21.727273  472.074380  
46  54.72727  8.727273  76.165289  
31  54.72727  23.727273  562.983471  
27  54.72727  27.727273  768.801653  
40  54.72727  14.727273  216.892562  
39  54.72727  15.727273  247.347107  
50  54.72727  4.727273  22.347107  
61  54.72727  6.272727  39.347107  
81  54.72727  26.272727  690.256198  
29  54.72727  25.727273  661.892562  
Sum  11,846.55 
The sum of squared errors is 11,846.55 and the variance is the sum of squared errors divided by the degrees of freedom ($N  1$). There were 33 scores and so the degrees of freedom were 32. The variance is, therefore
$$ s^2 = \frac{\text{sum of squared error}}{N1} = \frac{11846.55}{32} = 370.20. $$
The standard deviation is the square root of the variance
$$ s = \sqrt{370.20} = 19.24. $$
We can now calculate the standard error using Equation (8.3) in the book
$$ \sigma_{\bar{X}} = \frac{s}{\sqrt{n}} = \frac{19.24}{\sqrt{33}} = 3.35. $$
The standard error of the fakename group is, therefore, 3.35.
Puzzle 4
What is a 95% confidence interval?
For a given statistic (e.g., the mean) calculated for a sample of observations, a 95% confidence interval is a range of values around that statistic that are believed to contain the true value of that statistic (i.e., the population value) in apprximately 95% of samples. It’s worth remembering that for a particular sample you can’t be sure whether it is one of the 95% for which the interval contains the population value or whether it is one of the 5% that does not.
Puzzle 5
Using the data in Table 8.2 (in the book and reproduced in ## Puzzle 2), calculate the 95% confidence interval for the mean in both the fakename group and the ownname group.
We have already calculated the mean and standard error for both groups (see Puzzles 2 and 3), so all we need to do is to plug these values into the following equations:
$$ \begin{aligned} \text{lower boundary of 95% CI} = \text{estimate}(1.96 \times SE) \\ \text{upper boundary of 95% CI} = \text{estimate}+(1.96 \times SE) \end{aligned} $$ For the ownname group:
$$ \begin{aligned} \text{lower boundary} = 44.2(1.96 \times 3.10) = 38.12 \\ \text{upper boundary} = 44.2+(1.96 \times 3.10) = 50.28 \end{aligned} $$
For the fakename group:
$$ \begin{aligned} \text{lower boundary} = 54.73(1.96 \times 3.35) = 48.16\\ \text{upper boundary} = 54.73+(1.96 \times 3.35) = 61.30 \end{aligned} $$
Puzzle 6
Using the 95% confidence intervals for the ownname and fakename groups, can you infer anything about whether the accuracy on the maths test was affected by whether the test was taken under their own or a fake name?
The 95% confidence interval for the fakename group ranged from 48.16 to 61.30, and for the ownname group it ranged from 38.12 to 50.28. Therefore, the intervals overlap, but only just: the upper limit of the ownname group (50.28) is only slightly larger than the lower limit of the fakename group (48.16). If we assume that both samples are ones from the 95% that produce intervals that contain the population value, then what this implies is that the population values for ‘own name’ and ‘fake name’ are very unlikely to be the same (because the intervals barely overlap). We might infer from this that the mean accuracy on the maths test was genuinely higher when taken under a fake name than when taken under their own name.
Puzzle 7
Using the data in Table 8.2 (in the book and reproduced in Puzzle 2), calculate the 99% confidence interval for the mean in both the fakename group and the ownname group.
$$ \begin{aligned} \text{lower boundary of 99% CI} &= \text{estimate}(2.58 \times SE) \\ \text{upper boundary of 99% CI} &= \text{estimate}+(2.58 \times SE) \end{aligned} $$
For the ownname group:
$$ \begin{aligned} \text{lower boundary} &= 44.2(2.58 \times 3.10) = 36.20 \\ \text{upper boundary} &= 44.2+(2.58 \times 3.10) = 52.20 \end{aligned} $$
For the fakename group:
$$ \begin{aligned} \text{lower boundary} &= 54.73(2.58 \times 3.35) = 46.09\\ \text{upper boundary} &= 54.73+(2.58 \times 3.35) = 63.37 \end{aligned} $$
Puzzle 8
Using the data in Table 8.2 (in the book), calculate the 95% confidence interval for the first 10 scores in the fakename group.
First we need to calculate the mean of the first 10 scores:
$$ \begin{aligned} \bar{X} &= \frac{\sum_{i = 1}^n x_i}{n} \\ &= \frac{40+69+60+41+82+40+64+22+45+78}{10} \\ &= \frac{541}{10} \\ &= 54.1. \end{aligned} $$
Then we need to calculate the sum of squares by subtracting the mean from each score, squaring each deviance and then adding up the squared deviances.
Calculating the sums of squared errors for the first 10 scores in the fakename group  

Accuracy $x$ 
Mean $\bar{X}$ 
Deviance $x\bar{X}$ 
Deviance squared $(x\bar{X})^2$ 

40  54.1  14.1  198.81  
69  54.1  14.9  222.01  
60  54.1  5.9  34.81  
41  54.1  13.1  171.61  
82  54.1  27.9  778.41  
40  54.1  14.1  198.81  
64  54.1  9.9  98.01  
22  54.1  32.1  1030.41  
45  54.1  9.1  82.81  
78  54.1  23.9  571.21  
Sum  3,386.90 
So, the sum of squared errors was 3386.9. The variance is the sum of squared errors divided by the degrees of freedom ($N  1$). There were 10 scores and so the degrees of freedom were 9.
The variance is, therefore
$$ s^2 = \frac{\text{sum of squared error}}{N1} = \frac{3386.9}{9} = 376.32. $$
The standard deviation is the square root of the variance
$$ s = \sqrt{376.32} = 19.40. $$
We can now calculate the standard error using Equation (8.3) in the book
$$ \sigma_{\bar{X}} = \frac{s}{\sqrt{n}} = \frac{19.40}{\sqrt{10}} = 6.13. $$
The standard error of the first 10 scores in the fakename group was 6.13.
We can now calculate the 95% confidence interval, but because we have a small sample size (less than 30), instead of using the value of z from a normal distribution, we use the value of t from the tdistribution appropriate for our sample size by using equation (8.8) in the book:
$$ \begin{aligned} \text{lower boundary of 99% CI} &= \text{estimate}(t_{n1} \times SE) \\ \text{upper boundary of 99% CI} &= \text{estimate}+(t_{n1} \times SE) \end{aligned} $$
First, if n = 10, then the degrees of freedom will be $ n1 = 9 $. The corresponding value of t(df = 9) for a 95% CI is 2.262. This gives us the following limits
$$ \begin{aligned} \text{lower boundary} &= 54.1(2.262 \times 6.13) = 40.23\\ \text{upper boundary} &= 54.1+(2.262 \times 6.13) = 67.96. \end{aligned} $$
Puzzle 9
What is a sampling distribution?
We can think of a sampling distribution like this: if we took a sample from a population and calculated a statistic (e.g., the mean), the value of this statistic would depend somewhat on the sample we took. As such the statistic will vary slightly from sample to sample. If, hypothetically, we took lots and lots of samples from the population and calculated the statistic of interest, we could create a frequency distribution of the values we got. The resulting distribution is what the sampling distribution represents: the distribution of possible values of a given statistic that we could expect to get from a given population.
Puzzle 10
What is the difference between the standard error of the mean and the standard deviation?
The standard deviation measures how representative the mean is of the observed data. A small standard deviation (relative to the value of the mean) indicates that data points are close to the mean. A large standard deviation (relative to the mean) indicates that the data points are distant from the mean. In contrast, the standard error of the mean is the standard deviation of sample means. As such, it is a measure of how representative a sample mean is likely to be of the population mean. A large standard error (relative to the sample mean) indicates that there is a lot of variability between the means of different samples and so the sample might not be representative of the population. A small standard error indicates that most sample means are similar to the population mean and so our sample is likely to be an accurate reflection of the population.