A business manager in a health-care facility has been investigating the cost of
ID: 3295302 • Letter: A
Question
A business manager in a health-care facility has been investigating the cost of maintaining patients within its plan. The health care facility, in its advertising campaigns, has claimed that the cost of patient care runs about $1,225 per month. A sample of 20 cases for the last month reported the following: (a) Determine and interpret the Mean, Median, and the Standard Deviation. (b) Assuming a level of significance of 0.05, check the validity of the health care facility's claim. (Remember to identify the null hypothesis and the alternative hypothesis. Please show all work for full credit) A t-table is attached for your ready reference.Explanation / Answer
Solution:-
Data set - 1168,1225,1195,1054,1165,1345,1240,1178,1125,1276,1423,1239,1324,1445,1345,1186,1175,1253,1435,1252
(a)
Total numbers = 20
Mean = Sum / total no.s = 1252.4
Median = Middle value of the given data set = (1239+1240)/2 = 1239.5
Standard deviation = sqrt((x - mu)2) / (n-1))
= 105.891
(b)
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: = 1225 (Claim)
Alternative hypothesis: 1225
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n) = 105.891 / sqrt(20) = 23.6779
DF = n - 1 = 20 - 1 = 19
t = (x - ) / SE = (1252.4 - 1225)/23.6779 = 1.1572
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
Since we have a two-tailed test, the P-value is the probability that the t statistic having 19 degrees of freedom is less than -1.1572 or greater than 1.1572.
We use the t Distribution Calculator to find P(t < -1.1572), and P(t > 1.1572).
The P-Value is 0.261532.
The result is not significant at p < 0.05.
Interpret results. Since the P-value (0.261532) is greater than the significance level (0.05), we cannot reject the null hypothesis.
Conclusion. Fail to reject the null hypothesis. We have sufficient evidence to prove the claim that cost of patient care runs about $1,225 per month.
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