Salary data provided by www.salary.com showed that staff nurses in Tampa, Florid
ID: 3434674 • Letter: S
Question
Salary data provided by www.salary.com showed that staff nurses in Tampa, Florida, earn less than staff nurses in Dallas, Texas (the Tampa Tribune, January 15, 2007). Suppose that in a follow-up study of 42 staff nurses in Tampa and 22 staff nurses in Dallas you obtain the following results
Tampa: n=42 xbar=55000 s =6,500
Dallas: n =22 xbar=59300 s= 7,200
Does the data indicate the average nursing salary in Tampa is smaller than the average nursing salary in Dallas at ?=0.05?
What is The null and Alternative Hypothesis?
What is the the critical Value(s)?
What is the t/z stat?
What is the P-Value(s)?
Construct a 98% confidence interval for the difference in true means. State the C.I. in terms of Tampa-Dallas.....What is the lower and upper endpoint?
Explanation / Answer
What is The null and Alternative Hypothesis?
Let mu1 be the mean for Tampa
Let mu2 be the mean for Dallas
Null hypothesis: mu1=mu2
Alternative hypothesis: mu1<mu2
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What is the the critical Value(s)?
The degree of freedom =n1+n2-2=42+22-2 = 62
It is a left-tailed test.
Given a=0.05, the critical value is t(0.05, df=62) =-1.67 (from student t table)
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What is the t/z stat?
t =(xbar1-xbar2)/sqrt(s1^2/n1+s2^2/n2)
=(55000-59300)/sqrt(6500^2/42+7200^2/22)
=-2.35
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What is the P-Value(s)?
The p-value= P(t with df=62 <-2.35 ) =0.0110 (from student t table)
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Construct a 98% confidence interval for the difference in true means. State the C.I. in terms of Tampa-Dallas.....What is the lower and upper endpoint?
Given a=1-0.98 =0.02, t(0.01, df=62) =2.39 (from student t table)
So the lower bound is
(xbar1-xbar2)- t*sqrt(s1^2/n1+s2^2/n2)
=(55000-59300)-2.39*sqrt(6500^2/42+7200^2/22)
=-8682.452
So the upper bound is
(xbar1-xbar2)+ t*sqrt(s1^2/n1+s2^2/n2)
=(55000-59300)+2.39*sqrt(6500^2/42+7200^2/22)
=82.45198
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