Exercise: Now use this House price data set (as a .csv file) and generate graphics as we did above. Graphics obtained from Rcmdr in this dataset are here.
b.2 Probability calculations
Binomial distribution
Example:
Historical records indicate that 40% of all customers who enter a
discount store make a purchase. What is the probability that two of the
next three customers will make a purchase?
This is a binomial problem. The result is 0.288 as shown in this file.
This is obtained in Rcmdr via the steps, Distributions > Discrete
Distributions > Binomial Distribution > ... Try to replicate the
results.
Exercise: Here is a more challenging problem from healthcare area involving the testing of a new drug. Find the solution using Rcmdr. (Answer: 0.74)
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Normal distribution
Example: Distribution of fuel efficiencies of a new model car, X, is normal. Assuming that X ~ N(7.13,0.27^2) find the probability Pr(6.75 < X < 7.49). Here is the result from Rcmdr.
Exercise: Weekly
demand at a grocery store for a particular brand of energy drink cans
is N(1000,100^2). How many cans should be stocked so that there is a 5%
chance of a stockout? (Answer: 1165 cans)
b.3 Confidence intervals
Example:
(Population mean) A company's financial health is (usually) measured
using its debt-to-equity ratio. A bank has collected n = 15 of its
commercial accounts in this file. Let's find the 95% confidence interval (CI) using Rcmdr. (This is done via the t-test command which includes the CI. Here is the result
Exercise: Refer to your favourite statistics text for a solved problem on the CI
for population mean. Use the steps above to check the solution.
Exercise: Let's
find a 95% CI for the true average weight of SlimPhone. The population
s.d. is known to be 0.6 gr. We take a sample of n = 5, and find the
sample mean as 70.12 gr. (Answer: [69.594,70.646])
*****
Example:
(Population proportion) The CI for population proportion is easy to
obtain. Suppose you poll 1000 people and 340 of them state that they
would vote Liberal, if the election were held today. Here is what we do
to find a 95% CI:
> prop.test(340,1000)
1-sample proportions test with continuity correction
data: 340 out of 1000, null probability 0.5
X-squared = 101.761, df = 1, p-value < 2.2e-16
alternative hypothesis: true p is not equal to 0.5
95 percent confidence interval:
0.3108142 0.3704312
sample estimates:
p
0.34
So, the sample proportion is 0.34, with a 95% CI of [0.3108,0.3704], i.e., a margin of error of about 3%.
Exercise: Refer to
your favourite statistics text for a solved problem on the CI for population
proportion. Use the steps above to check the solution.
Note: The material below will not be discussed in the Workshop. But as this material may be useful in the MFin 604 course, it will remain online.
b.4 Hypothesis testing
Hypothesis
testing in R (with one or two populations) still uses the t.test
function described above. We now discuss a problem with two populations.
Example: The data set concerns the weight losses experienced by dieters using the Atkins diet
or the conventional diet. We want to test the hypothesis that there is
no difference between the two methods. Here are the results from Rcmdr.
Exercise: Refer to your favourite statistics text for a solved problem on
hypotheis testing for population mean(s). Use the steps above to check
the solution.
Exercise: Use
the following data values to test the hypothesis that true mean is 750
vs. the hypothesis that it differs from 750: (801,814,784,836,820) What
is the p-value? (Answer: p = .0023; so reject the null)
b.5 ANOVA
The final
example involves analysis of variance (ANOVA) where we want to test the
null hypothesis that three or more population means are equal. Once
again, Rcmdr solves this problem quite admirably.
Example:
(One-way ANOVA) Let's consider a problem from agricultural testing in
an experimental farm. We want to test the null hypothesis that low (L),
medium (M) or high (H) fertilizer levels do not differ in their effects
on the average yield of a new plant. We have 18 small plots, and on six
randomly selected plots we use L, the other six M and the last six H.
Here is the data set for this example. Rcmdr uses the R function aov and finds these results.
Exercise: Refer to your
favourite statistics text for a solved problem on one-way ANOVA. Use the steps above to check the solution.
Exercise: Use this hypothetical dataset with four samples (A, B, C, and D) to do a one-way ANOVA on the equality of the means. What is the p-value? (Answer: p = 0.00146, so reject the null)
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Example:
(Two-way ANOVA) Note that the above example is a one-factor ANOVA
problem. R can of course deal with multi-factor problems, too. Consider
the data in this file where we have two factors: Shelf display height (Bottom, Middle, Top)
and shelf display width (Regular, Wide). The numbers in each cell
correspond to sample group means (sales) for the last six months from
three different stores. This is a 2 x 3 (or, 3 x 2) design.
We want to
test three hypotheses: (i) There is no interaction between the factors,
(ii) Factor 1 (height) has no effect on sales, (iii) Factor 2 (width)
has no effect on sales. These results show that (i) little or no interaction exists between height and width
(high p-value), (ii) height affects the sales (p-value near zero), (iii)
width does not affect sales (high p-value). Here, the p-values are
denoted by Pr(F>).
Exercise: Refer to your
favourite statistics text for a solved problem on two-way ANOVA. Use the steps above to check the solution.
Exercise: Use this dataset with two factors (gender and machine type possible affecting productivity) to test the
hypotheses. As before, there will be an F-ratio for each factor and also
for the interaction. It may be that there is no difference between Male
and Female, or between machines A and B. But if men do better on
machine B and women do better on machine A, then there will be
interaction. (Answer: In fact, you will notice
that there is no difference between genders and the machines, but there
is a Gender:Machine interaction with a very low p-value.)
