FORK 1002 Statistics Exercise set 1: Basic statistics
1. Exercises on descriptive statistics.
Consider the following sample of observations from a study of the eﬀect ofViagra on a variable called “libido”:
Source: Page 350 in Andy Field (2009), Dis-covering Statistics using SPSS, 3rd. Edition,
xi = 1: Placebo. xi = 2: Low dose of Viagra. xi = 3: High dose of Viagra.
(a) Let the sum of the libido values be equal to 52, that is,
(b) What is the sample mean of libido for those who received placebo?
(c) What is the sample mean of libido for those who received viagra (that is,
i − y)2 = 43.73, where y is the sample mean of yi. Compute
the sample variance and sample standard deviation
(g) What are the maximum and minimum values of libido, and what is the
The table below lists the possible returns associated with an investment, andthe probability associated with each possible return. A negative return meansa loss.
(a) Compute the weighted return using the probabilities as weights
(b) Compute the weighted variance using the probabilities as weights
(c) Compute the weighted standard deviation
Let xi and yi be deﬁned as in exercise 1 (the Viagra sample), and compute thefollowing sums:
iyi, where a5 = 0, 2, a6 = 0, 1, a7 = 0, 4, a8 = 0, 3
4. Exercises on the normal distribution.
Let z denote the value of a test expression that is normally distributed:
(a) What is the probability that z is less than: i) 0? ii) 1? iii) 1.9? iv) -1.2?
(b) What is the probability that z is greater than: i) 0.5? ii) 1.3? iii) 1.6?
(c) What is the probability that z lies between: i) -0.3 and 0.2? ii) -1.7 and
5. Exercises on the t-distribution.
Let T (df ) denote the value of a test expression that is t-distributed with dfdegrees of freedom:
(a) What is the probability that: i) T (9) is greater than 1.1? ii) T (15) is
greater than 1.341? iii) T (40) is greater than 1.684? iv) T (80) is greaterthan 1.99?
(b) What is the probability that: i) T (30) is less than 1.055? ii) T (50) is less
than 2.403? iii) T (60) is less than -1.045? iv) T (20) is less than 0?
(c) What is the probability that: i) T (13) lies between -0.870 and 1.35? ii)
T (27) lies between -1.703 and 1.703? iii) T (40) lies between -0.851 and0.681?
6. Exercises on conﬁdence intervals.
A 100 · (1 − α)% conﬁdence interval for a population mean is usually com-puted as
= y + tα/2(df ) · s/ nL = y − tα/2(df ) · s/ n,
where U and L are the upper and lower bounds of the interval, respectively,df = n − 1 is the degrees of freedom, s is the sample standard deviation andn is the number of observations
(a) Compute a 90% interval for the population mean of libido
(b) Compute a 95% interval for the population mean of libido
(c) Compute a 99% interval for the population mean of libido
7. Exercise on testing the diﬀerence between two means.
The hypothesis that Viagra has an eﬀect on libido can be tested with thetest expression
where y is the mean libido for those who received placebo, y is the mean
libido for those who received viagra, n1 and n2 are the sample sizes, ands2 = 1.7 and s2 = 2.77 are the sample variances. Suppose the test expression
is t-distributed with df = nmin − 1, where nmin is the smallest of n1 and n2. Perform the test using a 1% signiﬁcance level:
(a) Deﬁne the null and alternative hypotheses
(b) Obtain the critical values and the rejection area
(c) Compute the value of the test expression
Suppose your null hypothesis is H0 : µ = 0:
(a) If the alternative hypothesis is H1 : µ > 0, and if the value of a normally
distributed test expression is 0.64, then what is the p-value of the test?
(b) If the alternative hypothesis is H1 : µ < 0, and if the value of a normally
distributed test expression is 0.15, then what is the p-value of the test?
(c) If the alternative hypothesis is H1 : µ ̸= 0, and if the value of a normally
distributed test expression is -0.31, then what is the p-value of the test?
(d) If the alternative hypothesis is H1 : µ > 0, and if the value of a t(19)-
distributed test expression is 1.729, then what is the p-value of the test?
(e) If the alternative hypothesis is H1 : µ < 0, and if the value of a t(19)-
distributed test expression is -1.729, then what is the p-value of the test?
(f) If the alternative hypothesis is H1 : µ < 0, and if the value of a t(35)-
distributed test expression is -0.682, then what is the p-value of the test?
(g) If the alternative hypothesis is H1 : µ ̸= 0, and if the value of a t(21)-
distributed test expression is 2.518, then what is the p-value of the test?
(a) Load the dataset viagra.xls. [Hint for SPSS: File → Open → Data]
(b) Compute the sample mean, sample standard deviation and sample range
of the variable libido. [Hint for SPSS: Analyze → Descriptive Statistics → Descrip-tives., put libido into the “Variable(s)” box, make sure “Mean”, “Std. Deviation”,
“Range”, “Maximum” and “Minimum” are ticked via “Options”, then press “Con-
(c) Make a new variable equal to dose − 1 and call it dosenew [Hint for SPSS:
Transform → Compute Variable., write “dosenew” in the “Target Variable” box,write “dose-1” in the “Numeric Expression” box, press “OK”]
(d) Compute a 90% conﬁdence interval for the mean of libido. [Hint for SPSS:
Analyze → Compare Means → One Sample T Test, click on “Options”, write “90” inthe “Confidence Interval Percentage” box, press “Continue”, press “OK”]
(e) Make a bar diagram of libido [Hint for SPSS: Graphs → Chart Builder., choose
“Bar” in the “Choose From” box, drag the chosen bar type into the “Chart preview.”
box, drag libido into the “x-axis” area, press “OK”]

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