Question

3. You are testing two different landing pages for a retail website. You test the pages with 100 visitors in each group and the first landing page results in an average session length of 2.6 min (sd = 3.8 min) and the second produces an average session length of 3.6 min (sd = 4.5 min). A one-minute increase in session length seems like a lot and you need to decide whether to report this to your boss, who is the head of digital marketing. a. [5 pt] What is the confidence interval for the difference in average session length? b. [2 pt] Is this a statistically significant difference? C. [1 pt] Should you tell your boss? d. [bonus 1 pt] What would you do next?

Answer 1

Here we have < =2.6,5 =3.8, n =100 X =3.6, Sz=4.5, n2 =100 Level of significance: a =0.05 Confidence level: CL =0.95 Since we can assume that variances are equal so degree of freedom of t test will be: df = n, +n, -2 =198 The pooled standard deviation is: -1)s : +(n-1)s 2 =4.1647 V nun-2 Critical value: t =1.972 The standard error is: i =+ 1 SE = =4.1647*sqrt((1/100)+(1/100)) yn n2 = 0.589 Margin of error: ME = T, SE =1.1615 Confidence interval is: ( -7,) ME =(2.6-3.6) + 1.1615 =(-2.1615, 0.1615) Excel function for critical value: =TINV(0.05,198) Hence, the required confidence interval is (-2.1615,0.1615).

(b)

Since confidence interval does not contain so difference is no statistically significant.

(c)

You should tell your boss that there is no significant difference in session length.

(d)

You can use either page.

To be more sure, you can took other greater samples.