Question

In: Statistics and Probability

In a usability study, two versions (A, B) of a company website were compared with respect...

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

  1. (5) Is there any difference in mean retrieval time between the two sites? Answer this question by carrying out an appropriate 10% hypothesis test, assuming that the population variances are equal. Be sure to include the null and alternative hypotheses, value of the test statistic, p-value, and conclusion in the context of the study.
  2. (3) Create a confidence interval for the mean retrieval time between the two sites and use it to test if there is a statistically significant difference between the two sites. Create this confidence interval using an appropriate confidence level to match your significance level in part a). Clearly state your conclusion.

Solutions

Expert Solution

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.


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