Question

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 mu 1 - mu 2.

Answer 1

Solution:

1) H0: equal salaries in city 1 and 2
H1: different salaries in city 1 and 2
2) need to conduct a t-test of the difference in means
one issue is whether to pool the two std deviations (variances) to compute a pooled overall
variance. Technically, you are supposed to first test whether the variances are equal or not - if equal - pool them; if not equal , use separate variances; I'm going to pool them ans compute
sp^2=[(n1-1)s1^2 + (n2-2)s2^2]/(n1 + n2 -2) = [14x2300^2 + 14x2100^2]/28 = 4,850,000
and thus pooled sp=2202.3
then compute test statistic (28,900 - 30,300)/[2202.3sqrt(1/15 + 1/15)] = -1.74
3) for 28 degrees of freedom, critical value is 2.048 for two-sided test at alpha=.05
accept H0
4) conclude no statistical evidence that salaries from these 2 cities are different, i.e., data consistent with equal salaries ffrom 2 cities

95% confidence interval for μ1 - μ2.
(-3125, 325)

[OR]

Answer:

the confidence interval range is (-3124.9,325)

x1-x2=-1400

sp=sqrt((2300^2+2100^2)15)=804.15

t(0.025,14)=2.145

lower= -1400-2.15*804.15=-3145.9

upper= -1400+2.15*804.15=325

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