Chapter 11: Q. 38 (page 754)

Preventing strokes Aspirin prevents blood from clotting and so helps prevent strokes.
The Second European Stroke Prevention Study asked whether adding another anticlotting
drug named dipyridamole would be more effective for patients who had already had a
stroke. Here are the data on strokes during the two years of the study
a. Summarize these data in a two-way table.
b. Do the data provide convincing evidence of a difference in the effectiveness of the four
treatments at the $α = 0.05$ significance level?

Short Answer

Expert verified

(a) To develop the two-way table, let the medicines be given within the columns and let the successes/failures be given within the columns.

(b)There is convincing evidence of a difference in the effectiveness of the four treatments at the $α = 0.05$ significance level.

Step by step solution

Part (a) Step 1: Given information

We have been given the data on the usage of different kinds of drugs used for stroke patients.

Part (a) Step 2: Explanation

To develop the two-way table, let the medicines be given within the columns and let the successes/failures be given within the columns. We then already know the number of patients who encompasses a stroke for each treatment and the push aggregates (which are the number of patients). The number of disappointments is at that point the push total decreased by the number of victories. The column aggregates is the entirety of all checks within the column.

	Stroke	No Stroke	Total
Placebo	$250$	$1399$	$1649$
Aspirin	$206$	$1443$	$1649$
Dipyridamole	$211$	$1443$	$1654$
Both	$157$	$1493$	$1650$
Total	$824$	$5778$	$6602$

Part (b) Step 1: Given information

We have been given data on the usage of different kinds of drugs used for stroke patients.

Part (b) Step 2: Explanation

Let us to begin with decide the row/ column summations of each drive within the result of portion( a), which is the sum of all values within the comparing row/ column.

We refer to the table drawn in the previous part (a)

Thenullthesiscountriesthat the variables areindependent, while theindispensablethesiscountriesthat they aren'tindependent.

$H o$ : Treatment and stroke are independent.

Ha : Treatment and stroke are dependent.

The anticipated frequentness E are the product of the column and row aggregate, divided by the table aggregate.

Part (b) Step 3: Explanation

$E_{1} 1 = (r_{1} \times c_{1}) / n = (1649 \times 824) / 6602 \approx 205.81 E_{1} 2 = (r_{1} \times c_{2}) / n = (1649 \times 5778) / 6602 \approx 1443.19 E_{2} 1 = (r_{2} \times c_{1}) / n = (1649 \times 824) / 6602 \approx 205.81 E_{2} 2 = (r_{2} \times c_{2}) / n = (1649 \times 5778) / 6602 \approx 1443.19 E_{3} 1 = (r_{3} \times c_{1}) / n = (1654 \times 824) / 6602 \approx 206.44 E_{3} 2 = (r_{3} \times c_{2}) / n = (1654 \times 5778) / 6602 \approx 1447.56 E_{4} 1 = (r_{4} \times c_{1}) / n = (1650 \times 824) / 6602 \approx 205.94 E_{4} 2 = (r_{4} \times c_{2}) / n = (1650 \times 5778) / 6602 \approx 1444.06$

When performing a chi-square test for a two-way table, we have the taking after $3$ conditions: Irregular, $10 %$ and Expansive counts Random: Fulfilled, accepting that the patients were haphazardly relegated to a treatment $10 %$ . Fulfilled, since the $6602$ patients who had a stroke are less than $10 %$ of all patients who had a stroke. Large tallies: Fulfilled, since all anticipated checks are at slightest $5$ . Thus we note that all conditions are fulfilled.

Whenperformingaki-squaretestfor a two-waytable, we've the taking afterconditionsIrregular, andextensivecountsRandom Fulfilled,acceptingthat thecaseswereaimlesslyrelegatedto a treatment.Fulfilled, since thecaseswhohada stroke arelowerthan of allcaseswhohada stroke.LargecensusesFulfilled, since allanticipatedchecksare atfewest.thereforewenotethat allconditionsarefulfilled.

$(χ^2 & = \sum ((O - E)^2) / E = ((250 - 205.81)^2) / 205.81 + ((1399 - 1443.19)^2) / 1443.19 + ((206 - 205.81)^2) / 205.81 + ((1443 - 1443.19)^2) / 1443.19 + ((211 - 206.44)^2) / 206.44 + ((1443 - 1447.56)^2) / 1447.56 + ((157 - 205.94)^2) / 205.94 + ((1493 - 1444.06)^2) / 1444.06 \approx 24.2428)$

Theki-squaresubtotals are thesquaredcontrasts between thewatchedandanticipatedfrequentness, partitioned by theanticipatedfrequence. Theregardof thetest- statistic is at thatpointthetotalof theki-squaresubtotals
Thedegreesoffreedomis theproductof thenumberof row and thenumberofcolumns, bothdroppedby one.

The is theliabilityofgettingtheregardof thetestdimension, or aregardmoreextraordinary. Theis thenumber( orinterim) within thecolumntitleof theki-squarevehicletablewithin thereferencesectioncontainingthe x-value within thedrive.

$d f (r - 1) (c - 1) - (4 - 1) (2 - 1) = 3$

The $P - v a l u e$ is the likelihood of getting the esteem of the test measurement, or a esteem more extraordinary. The $P - v a l u e$ is the number (or interim) within the column title of the chi-square conveyance table within the reference section containing the x-value within the push $d f = 3$ .

$P < 0.0005$

If the $P - v a l u e$ is less than or break even with to the importance level, at that point the null theory is rejected.

$P < 0.05$

Still, at that point the null proposition is rejected, If the is lower than or break indeed with to the significance position.

Soreject
There's satisfying substantiation of a difference in the effectiveness of the four treatments at the significance position.