A business manager in a health-care facility has been investigating the cost of
ID: 3268466 • 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:
1
$1,168.00
(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.
2
$1,225.00
3
$1,195.00
4
$1,054.00
5
$1,165.00
6
$1,345.00
7
$1,240.00
8
$1,178.00
9
$1,125.00
10
$1,276.00
11
$1,423.00
12
$1,239.00
13
$1,324.00
14
$1,445.00
15
$1,345.00
16
$1,186.00
17
$1,175.00
18
$1,253.00
19
$1,435.00
20
$1,252.00
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:
1
$1,168.00
(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.
2
$1,225.00
3
$1,195.00
4
$1,054.00
5
$1,165.00
6
$1,345.00
7
$1,240.00
8
$1,178.00
9
$1,125.00
10
$1,276.00
11
$1,423.00
12
$1,239.00
13
$1,324.00
14
$1,445.00
15
$1,345.00
16
$1,186.00
17
$1,175.00
18
$1,253.00
19
$1,435.00
20
$1,252.00
Explanation / Answer
a)
using minitab
Mean of C1 = 1252.4
Median of C1 = 1239.5
Standard deviation of C1 = 105.891
b)
Assuming a level of significance of 0.05, check the validity of the health care facility's claim.
null hypothesis H0:=1225
alternative hypothesis: H1: 1225
t = X - / { s/sqrt(n)}
= 1252.4-1225 / { 105.891/ sqrt(20)}
=1.1571949
Table value of t with 19 Df at 0.05 level = 2.093
t value < 2.093 ( critical value)
failed to reject null hypothesis
accept the claim alternative hypothesis: H1: 1225
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