A study was conducted to determine if the salaries of librarians from two neighboring cities were equal. A sample of 15 librarians from each city was randomly selected. The mean from the first city was $28,900 with a standard deviation of $2300. The mean from the second city was $30,300 with a standard deviation of $2100. Construct a 95% confidence interval for u1 -u2. a) (-4081, 597)
b) (-2054, 238)
c) (-2871, 567)
d) (-3125, 325)

Step-by-step explanation:

The formula for determining the confidence interval for the difference of two population means is expressed as

Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)

Where

x1 = sample mean salary of city 1 librarians

x2 = sample mean salary of city 2 librarians

s1 = sample standard deviation for city 1

s2 = sample standard deviation for city 2

n1 = number of soles for city 1

n1 = number of soles for city 2

For a 95% confidence interval, we would determine the z score from the t distribution table because the number of samples are small

Degree of freedom =

(n1 - 1) + (n2 - 1) = (15 - 1) + (15 - 1) = 28

z = 2.048

x1 - x2 = 28,900 - 30,300 = - 1400

Margin of error = 2.048√(s1²/n1 + s2²/n2) = 2.036√(2300²/15 + 2100²/15)

= 1647

The upper boundary for the confidence interval is

- 1400 + 1647 = 247

The lower boundary for the confidence interval is

- 1400 - 1647 = - 3047

a

step-by-step explanation:

u take $30 minus $24. the answer/original($30)