Question

In: Statistics and Probability

Problem #1 - Using Excel A healthcare consultant wants to compare the patient satisfaction ratings of...

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

Solutions

Expert Solution

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


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