In: Statistics and Probability
Problem #1 - Using Excel
A healthcare consultant wants to compare the patient satisfaction ratings of two hospitals. The consultant collects ratings from 20 patients for each of the hospitals as listed in column A and column B of this data sheet.
The consultant wants determine whether there is a difference in the patient ratings between the hospitals. The population standard deviations for the ratings are known to be 8 for Hospital A and 10 for Hospital B.
Use a 0.05 significance level and fill out each of the following blank cells to reach your conclusion. Use the Excel "Data Analysis" ToolPak to calculate the necessary statistics.
Hospital A | Hospital B |
81 | 89 |
77 | 64 |
75 | 85 |
74 | 78 |
86 | 89 |
90 | 85 |
62 | 87 |
73 | 77 |
91 | 82 |
98 | 79 |
81 | 69 |
85 | 78 |
77 | 65 |
78 | 71 |
83 | 77 |
90 | 88 |
78 | 73 |
76 | 68 |
71 | 75 |
80 | 80 |
Step 1 | Develop Hypotheses | |
H0: | ||
Ha: | ||
Step 2 | Specify the Significance Level (a) | |
a= | ||
Step 3 | Compute the Test Statistic | |
Test Statistic = | ||
Step 4 | Determine the Critical Value | |
Critical Value = | ||
Step 5 | Compute the p-value | |
p-value | ||
Step 6 | Make your decision whether to reject H0 or not | |
p-value vs. a | ||
Test statistic vs. Critical value | ||
Decision | ||
Step 7 | Interpret the statistical conclusion | |
Given that,
mean(x)=80.3
standard deviation , sigma1 =8
number(n1)=20
y(mean)=77.95
standard deviation, sigma2 =10
number(n2)=20
null, Ho: u1 = u2
alternate, H1: μ1 != u2
level of significance, alpha = 0.05
from standard normal table, two tailed z alpha/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=80.3-77.95/sqrt((64/20)+(100/20))
zo =0.82
| zo | =0.82
critical value
the value of |z alpha| at los 0.05% is 1.96
we got |zo | =0.821 & | z alpha | =1.96
make decision
hence value of | zo | < | z alpha | and here we do not reject
Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.82 ) =
0.41184
hence value of p0.05 < 0.41184,here we do not reject Ho
------------------------------------------------------------------------------
null, Ho: u1 = u2
alternate, H1: μ1 != u2
test statistic: 0.82
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.41184
no evidence to support that there is a difference in the patient
ratings
between the hospitals