In: Statistics and Probability
In a usability study, two versions (A, B) of a company website were compared with respect to the time it takes to retrieve certain information from the site. One hundred subjects were randomly selected from the population, and 50 subjects were randomly assigned to each version. Sample means and standard deviations of retrieval time for the two versions are provided below.
Version |
Mean |
Standard Deviation |
A |
209 |
37 |
B |
225 |
41 |
A) H0:
H1:
The pooled variance(sp2) = ((n1 - 1)s1^2 + (n2 - 1)s2^2)/(n1 + n2 - 2)
= (49 * (37)^2 + 49 * (41)^2)/(50 + 50 - 2)
= 1525
The test statistic t = ()/sqrt(sp2/n1 + sp2/n2)
= (209 - 225)/sqrt(1525/50 + 1525/50)
= -2.05
df = 50 + 50 - 2 = 98
P-value = 2 * P(T < -2.05)
= 2 * 0.0215 = 0.0430
Since the P-value is less than the significance level(0.0430 < 0.10), so we should reject the null hypothesis.
So at 10% significance level, there is sufficient evidence to conclude that there is a difference in mean retrieval time between two sites.
b) At 90% confidence interval the critical value is t* = 1.661
The 90% confidence interval for is
() +/- t* * sqrt(sp2/n1 + sp2/n2)
= (209 - 225) +/- 1.661 * sqrt(1525/50 + 1525/50)
= -16 +/- 12.9728
= -28.9728, -3.0272
Since the interval does not contain 0, so we can conclude that there is a statistically significant difference between the two sites.