Question

7. A randomized, double-blind, placebo-controlled study evaluated the effect of the herbal remedy Echinacea purpurea in treating upper respiratory tract infections in 2- to 11-year olds. Each time a child had an upper respiratory tract infection, treatment with either echinacea or a placebo was given for the duration of the illness. One of the outcomes studied was “severity of symptoms.” A severity scale based on four symptoms was monitored and recorded by the parents of subjects for each instance of upper respiratory infection. The peak severity of symptoms in the 337 cases treated with echinacea had a mean score of 6.0 (standard deviation 2.3). The peak severity of symptoms in the placebo group (n2 = 370) had a mean score of 6.1 (standard deviation 2.4). Test the mean difference for significance. Discuss your findings. Is it possible to use statcrunch? if so, how?

Answer 1

H0: Null Hypothesis: 1 = 2

HA: Alternative Hypothesis: 1 2

Test statistic is:

t = (1 - 2)/SE

= (6.0 - 6.1)/0.1771 = - 0.5647

ndf = n1 + n2 - 2 = 337 + 370 - 2 = 705

Take = 0.05

Two Tail Test

From Table, critical values of t = 1.9633

Since the calculated value of t = - 0.5647 is greater than critical value of t = - 1.9633, Fail to reject H0.

Conclusion:

The mean difference is not significant.

It is possible to use StatCruch as per the following steps:

1. Stat > T Stats > Two Sample

2. Choose With Summary option to enter the Sample mean, Sample standard deviation and Sample size for both samples

3. Select the Pool variances option.

4. Select the Hypothesis test option

5. Click Compute to view the results