Question

A researcher wished to compare the average amount of time spent in extracurricular activities by high school students in a suburban school district with that in a school district of a large city. The researcher obtained a simple random sample of 60 high school students in a large suburban school district and found the mean time spent in extracurricular activities per week to be six hours with a standard deviation of three hours. The researcher also obtained an independent simple random sample of 40 high school students in a large city school district and found the mean time spent in extracurricular activities per week to be four hours with a standard deviation of two hours. Let μ1 and μ2 represent the mean amount of time spent in extracurricular activities per week by the populations of all high school students in the suburban and city school districts, respectively. Assume two-sample t procedures are safe to use.

A. What is a 95% confidence interval for μ1 – μ2? (Use the conservative value for the degrees of freedom.)

a.2 ± 1.01 hours

b.2 ± 0.5 hours

c.2 ± 0.84 hours

d.2 ± 1.34 hours

B. What can we say about the value of the P-value? (Assume population variances are equal.)

a. P-value > 0.10

b. 0.05 < P-value < 0.10

c. 0.01 < P-value < 0.05

d. P-value < 0.01

Answer 1

A) df = Min(60,40) = 40

At 95% confidence level, the critical value is t^* = 2.021

The 95% confidence interval is

() +/- t^* * sqrt(s1^2/n1 + s2^2/n2)

= (6 - 4) +/- 2.021 * sqrt(3^2/60 + 2^2/40)

= 2 +/- 1.01

Option - a is correct.

B) The pooled variance(s_p²) = ((n1 - 1)s1^2 + (n2 - 1)s2^2)/(n1 + n2 - 2)

                                              = (59 * 3^2 + 39 * 2^2)/(60 + 40 - 2)

                                              = 7.01

The test statistic is

  

df = 60 + 40 - 2 = 98

P-value = 2 * P(T > 3.70)

            = 2 * (1 - P(T < 3.70))

            = 2 * (1 - 0.9998)

            = 0.0004

Option - d) P-value < 0.01