An experiment was planned to compare the mean time (in days) required to recover

Question

An experiment was planned to compare the mean time (in days) required to recover from a
common cold for people given a daily dose of 4 mg of Vitamin C versus those who were not given a
vitamin C supplement. Suppose that 35 adults were randomly selected for each treatment category and
that the mean recovery times and standard deviations for the two groups were as follows:

A. At ? =0.10, is there enough evidence conclude that the supplement is effective?
B. Compute a 95% confidence interval for the difference in the recovery mean time

No Vitamin Supplement
4 mg Vitamin C
Sample size
35 35 Sample mean
6.9 5.8 Sample standard deviation
2.9 1.2

Explanation / Answer

a)
Hypothesis:

H0 : mu1 = mu2
Ha : mu1 < mu2

Test statistics:
x1 = 5.8,s1 = 1.2 ,n1 =35 , x2 = 6.9 , s2 = 2.9, n2 = 35

z = (x1 -x2)/sqrt(s1^2/n1 +s2^2/n2)
= (5.8-6.9)/sqrt(2.9^2/35 + 1.2^2/35)
= -2.0735

P value = 0.0382

we can reject the null hypothesis at 0.1 significance level

There is sufficient evidence to support the claim

b)
z value at 95% = 1.96
x2 = 6.9 , s2 =2.9 , n2 = 35 , x1 = 5.8 , s1 = 1.2 , n1 = 35

CI = ( x1 -x2) +/- z* sqrt(s1^2/n1 +s2^2/n2)
= (5.8 - 6.9) +/- 1.96*sqrt(2.9^2/35 + 1.2^2/35)
= (-2.1398, -0.0602)

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