Question

A business manager in a health-care facility has been investigating the cost of maintaining patients within its plan. A sample of 15 cases for the last month reported the following:

1	$1,200.00	4	$1,174.00	7	$1,131.00	10	$1,120.00	13	$1,150.00
2	$1,016.00	5	$1,165.00	8	$1,178.00	11	$1,293.00	14	$1,240.00
3	$1,120.00	6	$1,254.00	9	$1,234.00	12	$1,018.00	15	$1,079.00

In its advertising campaigns, the business manager has claimed that the cost of out-of-pocket patient care runs no more than $1,120 per month.

(a) Determine the mean. Median, mode, and standard deviation.

(b) Identify the research problem:

(c) One-tail or two-tails problem, why (discuss):

(d) The Null and Alternate Hypotheses (express mathematically):

(e) Discuss the use of z or t statistic:

(f) Computations (you may use EXCEL, SPSS or any other suitable tool):

(g) Conclusion in terms of Hypothesis:

(h) Discuss validity of claim:

Answer 1

Part (a)

Calculation of mean, median and standard deviation: They have been calculated using excel functions AVERAGE (Select all the values); MEDIAN (Slect all the values): STDEV.S(Select all the values) respectively. The results are summarized below:

No.	Value ($)
1	$1,200.00
2	$1,016.00
3	$1,120.00
4	$1,174.00
5	$1,165.00
6	$1,254.00
7	$1,131.00
8	$1,178.00
9	$1,234.00
10	$1,120.00
11	$1,293.00
12	$1,018.00
13	$1,150.00
14	$1,240.00
15	$1,079.00
Mean	$1,158.13
Median	$1,165.00
Std Dev	$81.10651

Part (b)

In its advertising campaigns, the business manager has claimed that the cost of out-of-pocket patient care runs no more than $1,120 per month.

The research problem will be: If the cost of out-of-pocket patient care ≤ $1,120 per month.

Part (c)

The hypothesis being tested is whether mean is significantly lower than $ 1,120. The hypothesis being tested is not of the form if mean is greater than one value and lower than some other value. Thus, it's a one tail test.

Part (d)

H_o: μ ≤ $ 1,120 versus H_a: μ > $1,120

Part (e)

Sample mean, s = $1,158.13 (Calculated in part a)1,120 -

Population mean, μ = $ 1,120

Population std dev, s = $81.10651 (Calculated in part a)

n = number of samples = 15 which is less than 30, hence sample size is small.

Hence, we should select test statistics as t.

Part (f)

Confidence level of level of significance is not given in the question.

We will therefore stick to the standard assumption of  Level of significance α = 0.05

Test statistics, t (α / 2; n-1) = t (0.025; 14) = 2.145 (from the t table)

Calculated value of t = (x - µ) / (s / √n) = (1,120 - 1,158.13) / (81.10651 / √15 ) = $ 1.8209 < 2.145 = value of t statistics from t table.

Part (g)

Since calculated value of t < value of t statistics, the null hypothesis can't be rejected.

Hence the null hypothesis can't be rejected. Hence, we have no reason to reject the hypothesis that H_o: μ ≤ $ 1,120

Part (h)

So, the claim in the advertisement appears to be valid. We have no reason to believe that the calim is invalid.