Question

2) You have developed a series of questions to measure job satisfaction of bus drivers and taxi drivers in New York City. You selection random samples of 135 bus drivers and 120 taxi drivers. The average job satisfaction of bus drivers is 10.6 with a standard deviation of 2.8. For taxi drivers these statistics are 8.2 and 3.5.

a. What are your estimates of the average job satisfaction for the entire populations of NYC bus drivers and taxi drivers? Use the 95% confidence level.

b. Can you conclude that there is a difference in the job satisfaction of bus drivers and taxi drivers? Explain the evidence that supports your conclusion.

c. Reconstruct the confidence intervals using the 90% confidence level.

d. Does this change your conclusion about evidence of a difference in job satisfaction? Why (or why not)?

Answer 1

For bus drivers,

mean  job satisfaction m1 = 10.6

standard deviation of job satisfaction s1 = 2.8

Sample size, n = 135

Standard error of sample mean, se1 = s1 / = 2.8 / = 0.2409856

z value for 95% confidence interval is 1.96

Margin of error, E1 = se1 * z = 0.2409856 * 1.96 = 0.4723318

95% confidence interval of job satisfaction of bus drivers is,

(m1 - E1, m1 + E1)

= (10.6 - 0.4723318, 10.6 + 0.4723318)

= (10.13, 11.07)

Thus, point estimate of ob satisfaction of bus drivers is 10.6 and 95% confidence interval of job satisfaction of bus drivers is,

(10.13, 11.07)

For taxi drivers,

mean job satisfaction m2 = 8.2

standard deviation of job satisfaction s2 = 3.5

Sample size, n = 120

Standard error of sample mean, se2 = s2 / = 3.5 / = 0.3195048

Margin of error, E2 = se2 * z = 0.3195048 * 1.96 = 0.6262294

95% confidence interval of job satisfaction of bus drivers is,

(m2 - E2, m2 + E2)

= (8.2 - 0.6262294, 8.2 + 0.6262294)

= (7.57, 8.83)

Thus, point estimate of ob satisfaction of taxi drivers is 8.2 and 95% confidence interval of job satisfaction of taxi drivers is,

(7.57, 8.83)

b.

Since 95% confidence interval of job satisfaction of bus and taxi drivers do not overlap, there is a significant difference in the job satisfaction of bus drivers and taxi drivers

c.

For 90% confidence interval, z = 1.645

For bus drivers,

Margin of error, E1 = se1 * z = 0.2409856 * 1.645 = 0.3964213

95% confidence interval of job satisfaction of bus drivers is,

(m1 - E1, m1 + E1)

= (10.6 - 0.3964213, 10.6 + 0.3964213)

= (10.20, 11.00)

Thus, point estimate of ob satisfaction of bus drivers is 10.6 and 90% confidence interval of job satisfaction of bus drivers is,

(10.20, 11.00)

For taxi drivers,

Margin of error, E2 = se2 * z = 0.3195048 * 1.645 = 0.5255854

95% confidence interval of job satisfaction of bus drivers is,

(m2 - E2, m2 + E2)

= (8.2 - 0.5255854, 8.2 + 0.5255854)

= (7.67, 8.73)

Thus, point estimate of ob satisfaction of taxi drivers is 8.2 and 90% confidence interval of job satisfaction of taxi drivers is,

(7.67, 8.73)

d)

Since 90% confidence interval of job satisfaction of bus and taxi drivers still do not overlap, there is a significant difference in the job satisfaction of bus drivers and taxi drivers. Our conclusion will not change.