Question

2. A clinical lab wants to evaluate the resistance power of its new medicine. It has provided the new medicine to a random sample of 300 rats and retained the old drug for the remaining (200 rats). Three months after the dosage of the new drug, the group with the new medicine has a resistance of 40.2 with a s.d of 6.5 while the group with the old drug had an average of 35 with a s.d of 4.5.

NOTE: Follow all the steps in hypothesis testing. You can use Graphpad for this one.

1. What is your Null and Alternate Hypothesis. (0.5 point)

2. Did the new medicine result in significantly higher resistance of the dogs? State the value of the test-statistic, the p-value and the 95% confidence interval. (1 point)

(c)What do you mean by saying that this is the confidence interval in this case? (0.5 point)

Answer 1

a)

Null Hypothesis - H0 : Average resistance power of new drug is same as old drug ( Mu1 = Mu2 )

Alternate Hypothesis - H1 : Average resistance power of new drug is NOT same as old drug ( Mu1 Mu2 )

b) We apply t-test for two independet sample as per the following formula:

....................... (1)

Given: Let '1' indicate 'New' drug and '2' indicate 'Old' drug

x-bar1=40.2

s1=6.5

n1=300

x-bar2=35

s2=4.5

n2=200

Substitute the above values in the (1) formula:

t-test statistic = (40.2-35)/sqrt((6.5^2/300)+(4.5^2/200))

t-test statistic value = 10.57

t-critical value for 498 degrees of freedom at alpha of 0.05 = 1.965 ( from t-table )

Since t-test statistic value > t-critical value ==> We Reject H0

Conclusion : New medicine result in significanlty higher resistance of dogs.( it should be rats )

p-value = 0.0000 ( from the table for t-test statistic value = 10.57 )

95% Confidence Intervals :

Lower bound : (40.2-35)-1.965*(sqrt((6.5^2/300)+(4.5^2/200)) = 4.233

Upper bound : (40.2-35)+1.965*(sqrt((6.5^2/300)+(4.5^2/200)) = 6.166

c)

95% confidence interval ( 4.233 , 6.166) indicates the range of likely values for (μ1-μ2)