Question

3. An experiment was conducted to evaluate the effectiveness of a treatment for tapeworm in the stomachs of sheep. A random sample of 24 worm-infected lambs of approximately the same age and health was randomly divided into two groups. Twelve of the lambs were injected with the drug and the remaining twelve were left untreated. After 6 months, the lambs were slaughtered and the following worm counts were recorded. Assume the counts are approximately normally distributed.

Drug-treatedsheep 18, 43, 28, 50, 16, 32, 13, 35, 38, 33, 6, 7

Untreatedsheep 40, 54, 26, 63, 21, 37, 39, 23, 48, 58, 23, 39

1. Construct a 98% confidence interval for the difference of the worm count in a lamb.
2. Please perform a statistical test and see if the drug treatment reduced the mean worm count in a lamb. Use the significance level 0.05.
3. What are your assumptions that you assumed in part (b)?

Answer 1

SOLUTION-

1.) A 98% CONFIDENCE INTERVAL FOR DIFFERENCE OF THE WORM COUNT IS GIVEN AS,

NOW, FOR DRUG SAMPLES, MEAN= 26.6 AND SD= 14.4

FOR UNDRUG SAMPLES, MEAN= 39.3 AND SD= 14.3

SAMPLES= 12

THE 98% CONFIDENCE INTERVAL CALCULATED USING CALCULATOR IS,

(-27.40, 2.07)

2.) THE HYPOTHESIS IS,

THE ASSUMPTION MADE IS THAT THE SAMPLES ARE INDEPENDENTLY DISTRIBUTED. HENCE WE PERFORM A TWO SAMPLE-T TEST AND USE MINITAB-16.

STEPS: ENTER THE DATA> STAT> BASIC STATISTICS> TWO SAMPLE-T> SELECT THE DIFFERENT SAMPLES(DRUGED SHEEPS AS SAMPLE-1)> UNDER 'OPTIONS', SET THE CONFIDENCE LEVEL AS 95.0 AND ALTERNATE AS 'LESS THAN'> OK

THE TEST STATISTIC IS T= -2.16 AND THE P-VALUE IS 0.021

AS P-VALUE<0.05, WE REJECT THE NULL HYPOTHESIS.

HENCE WE ACCEPT THE CLAIM THAT DRUG TREATMENT REDUCED THE WORM COUNT.

****IN CASE OF DOUBT, COMMENT BELOW. ALSO LIKE THE SOLUTION, IF POSSIBLE.