Question

Spotify - a popular music streaming service - recently released data summarizing trends in worldwide listening patterns for 2014.

Data was collected for a sample of 55 American users and 40 Canadian users. The sample of American users had an average streaming time of 23 hours with a standard deviation of 5 hours. The sample of Canadian users had an average streaming time of 21 hours with a standard deviation of 5.5 hours.

A hypothesis test is to be conducted at the 5% significance level to determine whether the average streaming time for American users is higher than that of Canadian users.

How many degrees of freedom would be used to conduct this hypothesis test? Assume that the population standard deviations are unequal. Express your solution as a whole number.

Answer 1

Solution:Let n1=Sample size of American users

n2=Sample size of Canadian users

=average streaming time of American users

=average streaming time of Canadian users

s1=standard deviation time of American users

s2=standard deviation time of Canadian users

=significance level

Given:n1=55, n2=40, =23, s1=5, =21, s2=5.5 and =5%

1)State the hypothesis

Null hypothesis:H0:=

Alternative hypothesis:H0:>

2)Computation of Standard error(SE)

SE=[(s12/n1)+s22/n2)] =[(52/55)+(5.52/40)]=[0.45+0.76]=1.21=1.1

Computation of degree of freedom(DF)

Assuming that the population standard deviations are unequal, the formula for DF is given by

DF=(s12/n1)+s22/n2)2 / {[(s12/n1)2/(n1-1)]+[(s22/n2)2/(n2-1)]}

=[(52/55)+(5.52/40)]2 /{[(52/55)2/(55-1)]+[(5.52/40)2 /(40-1)]}

=1.212 / {[0.2025/54]+[0.5776/39]

=78.88=79

Computation of test statistics

z= (-)/SE=(23-21)/1.1=2/1.1=1.82

From the distribution table, the z-critical value for 5% level of significance is 1.645.(two tailed)

Conclusion: Since z-observed value=1.82 is greater than z-critical value=1.645, we do not reject the null hypothesis and conclude that the average streaming time for American users is equal to that of Canadian users.