Are nursing salaries in Tampa, Florida, lower than those in Dallas, Texas? Salar
ID: 3182599 • Letter: A
Question
Are nursing salaries in Tampa, Florida, lower than those in Dallas, Texas? Salary data show staff nurses in Tampa earn less than staff nurses in Dallas (The Tampa Tribune, January 15, 2007). Suppose that in a follow-up study of 40 staff nurses in Tampa and 50 staff nurses in Dallas you obtain the following results. a. Formulate hypothesis so that, if the null hypothesis is rejected, we can conclude that salaries for staff nurses in Tampa are significantly lower than for those in Dallas. Use alpha = .05 and df = 87. Show all steps b. What is your conclusion?Explanation / Answer
Solution:-
The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1 - 2 = 0
Alternative hypothesis: 1 - 2 0
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = sqrt[(60002/40) + (70002/50] = sqrt(3.33 + 9) = 1371.1309
DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
= 87 (given)
t = [ (x1 - x2) - d ] / SE = [ (56,100 - 59.400) - 0 ] / 1371.1309 = 40.8718
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.
Since we have a two-tailed test, the P-value is the probability that a t statistic having 87 degrees of freedom is more extreme than 40.8718.
We use the t Distribution Calculator to find P(t < 40.8718)
The P-Value is < 0.00001.
The result is significant at p < 0.05.
Interpret results. Since the P-value (0.00001) is very less than the significance level (0.05), we cannot accept the null hypothesis. That is accepting the alternate hypothesis, which concludes that the salaries for staff nurses in Tampa are significantly lower for those in Dallas.
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