Question

1. (7) A new drug is developed to treat headache, and we want to compare its performance with an existing drug. The performance measure here is the time from taking the drug to people start to feel better. A random sample of volunteers were collected, in one occasion of headache, these people use one of the drug, and then in another occasion of headache (there is a long time between occasions), these volunteers use the other drug. Assuming all variables are Normal, use the SPSS outputs to answer questions.

                                                                

      Paired Samples Test

	Paired Differences mean	Std. Dev.	Std. Error	t	df	Sig. (2-tailed)
New– old	.4	.750	.250	1.98	8	.058

Independent Samples Test

Group Statistics
	drug	N	Mean	Std. Deviation	Std. Error Mean
time	old	22	3.09855	.416654	.088831
	new	24	3.31996	.334813	.068343

Independent Samples Test

	Levene's Test for Equality of Variances		t-test for Equality of Means
	F	Sig.	t	df	Sig. (2-tailed)
Equal variances assumed	.043	.0388	-1.863	28	.0794
Equal variances not assumed			-1.680	24.9	.0888

1.            Which t test will you use?

          

2.           What are the hypotheses if you intend to show that the new drug is better?

3. What is the t and P-value for this test?

4. Will H_o be rejected at 5% level of significance?

5. What is your final conclusion?

Answer 1

a) -

We will use independent sample t-test, since sample for the different drugs is different. Both data set don't contain same number of data points, so we can't use paired sample t-test.

b) -

Hypothesis -

Null hypothesis : H₀ - ₁ = ₂ i.e. There is no significant difference between effects of two drugs.

Alternative hypothesis : H₁ - ₁ > ₂ i.e. New drug has better effect than first drug.

c) -

First, we have to check that variances are equal or not. If p-value for F-test is less than 0.05, then we can say that variances can be assume equal.

Here, p-value = 0.0388 < 0.05, hence we can say that variances can't be assume equal. Hence, we will choose t-value for equal vaiances not assumed.

So, t = -1.680, p-value = 0.0888.

d) -

We know that, if p-value is less than 0.05, then we will reject H_0.

Here, p-value = 0.088 > 0.05, hence we do not reject H₀ at 5% level of significance.

e) -

There is sufficient evidence that two drugs effects are not significantly different. i.e. New drug is not better than old drug.