Question

An international studies student is interested in knowing if the number of hours of television watched per week is greater in New York than in London. The international studies student takes a sample of 80 residents from London and finds that the mean television hours watched per week is 7.3 with a standard deviation of 2.3. The student also takes a sample of 80 residents from New York and finds that hours watched per week is 7.9 with a standard deviation of 2.5 hours. Test the claim a higher average hours of television are watched per week in New York than in London. Use α = .10.

Provide a 90% confidence interval estimate for the difference in means. Use α = .10.

Answer 1

Answer:

Given,

sample n1 = n2 = 80

mean x1 = 7.3 , x2 = 7.9

standard deviation s1 = 2.3 , s2 = 2.5

Ho : u1 = u2

Ha : u1 < u2

test statstic z = (x1 - x2)/sqrt(s1^2/n1 + s2^2/n2)

substitute values

= (7.3 - 7.9)/sqrt(2.3^2/80 + 2.5^2/80)

z = -1.58

P value = P(z < -1.58)

= 0.0570534 [since from z table]

= 0.057

Here p value < 0.1, so we reject Ho.

We have sufficient evidence.

Now consider,

Here at 90% CI, z value is 1.645

90% CI = (x1 - x2) +/- z*sqrt(s1^2/n1 + s2^2/n2)

substitute values

= (7.3 - 7.9) +/- 1.645*sqrt(2.3^2/80 + 2.5^2/80)

= -0.6 +/- 0.6248

= (-1.2248 , 0.0248)