Question

problem with a phone line that prevents a customer from receiving or making calls is upsetting to both the customer and the telecommunications company. The file Phone contains samples of 20 problems reported to two different offices of a telecommunications company and the time to clear these problems (in minutes) from the customers’ lines:

Central Office I Time to Clear Problems (minutes)

1.48 1.75 0.78 2.85 0.52 1.60 4.15 3.97 1.48 3.10

1.02 0.53 0.93 1.60 0.80 1.05 6.32 3.93 5.45 0.97

Central Office II Time to Clear Problems (minutes)

7.55 3.75 0.10 1.10 0.60 0.52 3.30 2.10 0.58 4.02

3.75 0.65 1.92 0.60 1.53 4.23 0.08 1.48 1.65 0.72

Perform a hypothesis test to determine if there’s evidence in this data of a difference in the mean waiting time between the two offices by answering the following questions:

(a) What are the null and alternate hypotheses for this test?

(b) Assuming that the population variances from both offices are equal, what is the value of the test statistic?

(c) Using the critical value approach, is there evidence of a difference in the mean waiting time between the two offices? (Use α = 0.05.) (Answer this using complete sentences and be sure to include what decision rule you used to arrive at your conclusion.)

(d) Find the p-value in (c) and interpret its meaning.

(e) What assumption (other than equal variances) is necessary in (c)?

(f) Assuming that the population variances from both offices are equal, construct and interpret a 95% confidence interval estimate of the difference between the population means in the two offices.

Answer 1

= 2.21

s1 = 1.72

n1 = 20

= 2.01

s2 = 1.89

n2 = 20

a)

  

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

                                             = (19 * (1.72)^2 + 19 * (1.89)^2)/(20 + 20 - 2)

                                             = 3.265

The test statistic is t = ()/sqrt(s_p²/n1 + s_p²/n2)

                                = (2.21 - 2.01)/sqrt(3.265/20 + 3.265/20)

                                = 0.35

c) df = 20 + 20 - 2 = 38

At = 0.05, the critical values are +/- t_0.025,38 = +/- 2.024

Since the test statistic value is not greater than the positive critical value(0.35 < 2.024), so we should not reject H₀.

At 0.05 significance level, there is not sufficient evidence to conclude that there is a difference in the mean waiting time between the two offices.

d) P-value = 2 * P(T > 0.35)

                 = 2 * (1 - P(T < 0.35))

                 = 2 * (1 - 0.6359)

                 = 0.7282

e) Assume that the populations of both offices are normally distributed.

f) At 95% confidence level, the critical value is t^* = 2.024

The 95% confidence interval is

() +/- t^* * sqrt(s_p²/n1 + s_p²/n2)

= (2.21 - 2.01) +/- 2.024 * sqrt(3.265/20 + 3.265/20)

= 0.2 +/- 1.1565

= -0.9565, 1.3565

We are 95% confident that the difference between the population means in the two offices lies within the limits between -0.9565 and 1.3565.