Module H2 Practical 14
Tests concerning proportions
Objectives:
By the end of this practical you should be able to:
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carry out a z-test for comparing two proportions
carry out a chi-square test for comparing two proportions and
understand how this relates to the z-test.
interpret results from tests above
understand that the hypotheses for the chi-square test can also be
formulated as a test for the association between two categorical
variables
In this practical, you will begin with an example concerning a single proportion and then
move onto looking at a comparison of two proportions and the testing procedure for a chisquare test corresponding to a 2x2 tables of frequencies.
1. Farmers in a certain region believe that the chance of crop failure due to drought during
the cropping season is 1 to 10. Rainfall records in the previous 50 years show that 8 were
“drought” years. Is there evidence that the chance of crop failure is different to what the
farmers believe?
(a) First write down the null and alternative hypotheses:
H0:
H1:
(b) Calculate the test statistic for testing H0. The test statistic (using the normal
approximation) is z = (observed proportion – hypothesised proportion)/std.error, where
std. error is the standard error of the sample proportion, i.e. [((1-)/n] under the null
hypothesis.
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(c) Interpret the results from your test above.
(d) Compute a 95% confidence interval for the true chance of drought in the cropping
season. What does this interval tell you?
(e) Summarise your conclusions from the above analysis.
2. A standard farming practice (A) is to be compared with a new practice (B). Of 222
farmers adopting A, 169 had “high” crop yields, i.e. yields greater than the 25th percentile of
the national average, while out of 235 using practice B, 205 had “high” yields. Is there
evidence that the true proportions of farmers (1 and 2) getting “high” yields in the two
populations (of those using practice A and those using practice B), are significantly
different?
(a) First write down the null and alternative hypotheses in terms of 1 and 2:
H0:
H1:
(b) Calculate the difference between the two sample proportions. Next find an estimate of
the standard error of the difference between these proportions, assuming the null
hypothesis is true.
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(c) Carry out an z-test for testing the null hypothsis. Obtain the exact p-value for your zstatistic using the normdist function of Excel, i.e.
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use 1 – normdist(z,0,1,true) if your z statistic is positive, substituting for z, or
use normdist(z,0,1,true) if your z statistic is negative.
In either case, the result should be multiplied by 2 to get the two-tail p-value.
(d) Interpret the results from your test above and summarise your conclusions.
(e) Now display the data of this exercise in the form of a 2x2 table.
Totals
Totals
(f) Next, calculated the expected frequencies under the null hypothesis.
Totals
Totals
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(g) Compute the chi-square test statistic and compare it with the appropriate value from the
2 table to assess the significance of your result. Also obtain the exact p-value for the test
using the Excel function chidist(value,df).
Does your p-value here match with what you obtained as the p-value following the z test in
(c) above? Comment on this comparison.
(h) You were requested in part (a) of this question, to formulate the hypotheses in terms of
1 and 2. Re-write the hypotheses in terms of a test of association between the two
categorical variables presented in the table in part (e).
(i) Does the above change affect your test procedure? Does it affect the way you present
your conclusions? If so, rewrite your conclusions in the light of the hypotheses as written
in (h) above.
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IF YOU HAVE TIME, TRY ALSO THE FOLLOWING:
3. In a certain town, 500 people chosen at random were asked whether they approved or
disapproved of the government. Immediately after the poll, the government introduced a
measure likely to be unpopular. The same 500 people were then asked the same question
again. The interviewer set out the results in the following table and deduced that there was
no conclusive evidence of a decrease in confidence.
Approved
Disapproved
Before
250
250
After
230
270
A representative of the opposition party complained that the conclusion was wrong. From
a more detailed examination of the data, he produced the following table.
Before
Approved
After
Disapproved
Approved
215
15
Disapproved
35
235
He claimed that this showed strong evidence of a fall in government popularity, on the
basis of data on those who altered their opinion.
Discuss the two analyses and give your interpretation of the data.
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