Statistical Analysis for Australian Hospitals

Running head: STATISTICAL ANALYSIS FOR AUSTRALIAN HOSPITALS
Statistical Analysis for Australian Hospitals
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1STATISTICAL ANALYSIS FOR AUSTRALIAN HOSPITALS
Table of Contents
Answer 1..........................................................................................................................................3
Part A...........................................................................................................................................3
Part B...........................................................................................................................................4
Part C...........................................................................................................................................5
Part D...........................................................................................................................................5
Part E...........................................................................................................................................6
Part F............................................................................................................................................6
Answer 2..........................................................................................................................................7
Part A...........................................................................................................................................7
Part B...........................................................................................................................................7
Part C...........................................................................................................................................7
Part D...........................................................................................................................................7
Part E...........................................................................................................................................8
Part F............................................................................................................................................9
Part G...........................................................................................................................................9
Answer 3........................................................................................................................................10
Part A.........................................................................................................................................10
Part B.........................................................................................................................................10
Part C.........................................................................................................................................11

2STATISTICAL ANALYSIS FOR AUSTRALIAN HOSPITALS
Part D.........................................................................................................................................12
Part E.........................................................................................................................................12
Part F..........................................................................................................................................12
Part G.........................................................................................................................................13
Answer 4........................................................................................................................................14
Part A.........................................................................................................................................14
Part B.........................................................................................................................................14
Part C.........................................................................................................................................14
Part D.........................................................................................................................................15
Part E.........................................................................................................................................16
Part F..........................................................................................................................................16
Answer 5........................................................................................................................................17
Part A.........................................................................................................................................17
Part B.........................................................................................................................................17
Part C.........................................................................................................................................17
Part D.........................................................................................................................................17
Part E.........................................................................................................................................18
Bibliography..................................................................................................................................19

3STATISTICAL ANALYSIS FOR AUSTRALIAN HOSPITALS
Answer 1
Part A
Considering the cost of nursing staff (TotalNurseCost in Million) in only the Psychiatric wards in
the Australian hospitals, the descriptive statistics is calculated.
Descriptive Statisticsa
N Minimum Maximum Mean Std. Deviation
Total Nurse Cost in Million 22 1.60 233.73 72.4128 59.48364
Valid N (listwise) 22
a. Psychiatric Ward = Yes
Table 1: Descriptive statistics for cost of nursing staff for Psychiatric ward
The formula to calculate the 99% confidence interval for the population mean cost of nursing
staff for all Australian hospitals with Psychiatric wards is:
x ± Z s
√ n
Using the value of mean and standard deviation obtained from table 1 in the formula:
72.41 ±
Z α
2
∗59.48
√22
Z=2.58
72.41 ± 2.58∗59.48
4.69
72.41 ±2.58∗28.12
72.41 ±5.457=66.593 ,77.867

4STATISTICAL ANALYSIS FOR AUSTRALIAN HOSPITALS
Hence, the 99% confidence interval is 66.953 ≤ Z ≤ 77.867
Part B
Validity of the confidence interval is tested using Shapiro-Wilk Test. If the significance value is
greater than 0.05, then the confidence interval is valid. If the value is less than 0.05, then it does
not satisfy the normality assumptions and the conditions. The significance value is 0.007, less
than 0.05 and hence, the data is not normality distributed. The normality test outcome and the
histogram are provided below.
Tests of Normalitya
Kolmogorov-Smirnovb Shapiro-Wilk
Statistic df Sig. Statistic df Sig.
Total Staff Cost in Million .193 22 .032 .868 22 .007
a. Psychiatric Ward = Yes
b. Lilliefors Significance Correction
Table 2: Normality test
Figure 1: Histogram of normality test

5STATISTICAL ANALYSIS FOR AUSTRALIAN HOSPITALS
Part C
Null hypothesis for viability testing of the research findings is:
H0: μ≥ $100
Alternative hypothesis:
H1: μ< $100
Formula for one sample t-test:
t= x−μ
s /√ n
Part D
The mean and standard deviation values are obtained from Table 1.
Mean = 72.41, std. deviation = 59.48, n = 22
t= 72.41−100
59.48/ √ 22
t= 72.41−100
59.48/ 4.69
t=−27.59
12.682 =−2.17
Part E
In this case, degrees of freedom: 22 – 1 = 21

6STATISTICAL ANALYSIS FOR AUSTRALIAN HOSPITALS
t-value at 1% level of significance = 2.518
Thus, t value obtained in Part D, that is, -2.17 is less than the critical value of 2.518 at 21 degrees
of freedom. Hence, the null hypothesis is accepted as -2.17 falls under the acceptance region.
Here, p-value is less than 0.01 and hence, the mean cost of the nursing staff for all the Australian
hospitals with Psychiatric wards is greater than $100 million.
Part F
The outcome for One-sample Test:
One-Sample Testa
Test Value = 100
t df Sig. (2-tailed) Mean
Difference
99% Confidence Interval of the
Difference
Lower Upper
Total Nurse Cost in Million -2.175 21 .041 -27.58722 -63.4944 8.3200
a. Psychiatric Ward = Yes
Table 3: One sample test
t-value obtained in Part C matches with that obtained from the One-Sample Test. The degrees of
freedom is also same. However, the significance or p-value is different. Computed p-value is less
than or equal to 0.01 while the one-sample test yielded the value as 0.041, which is greater than
0.01. Hence, there are difference in the computed and SPSS generated p-value.

7STATISTICAL ANALYSIS FOR AUSTRALIAN HOSPITALS
Answer 2
Part A
Variables of interest are HospRegion and Obstetrics.
Part B
Null hypothesis is the number of hospitals having obstetric ward in the rural region is more than
20%.
H0: p0 ≥ 0.2
Alternate hypothesis is that the number of hospitals having obstetric ward in the rural region is
less than 20%.
H1: p1 <0.2
Part C
Conditions and assumptions for hypothesis testing are that the variables should be independent
of each other and normally distributed as well.
Part D
The data is grouped by the variable HospRegion and frequency of the hospitals is calculated to
get the number of rural hospitals with Obstetrics Ward.
Obstetrics Warda
Frequency Percent Valid Percent Cumulative
Percent
Valid Yes 14 11.8 11.8 11.8
No 105 88.2 88.2 100.0

8STATISTICAL ANALYSIS FOR AUSTRALIAN HOSPITALS
Total 119 100.0 100.0
a. Region of Hospital = Rural
Table 4: Frequencies for rural hospitals with Obstetrics ward
Total number of observations, n = 119
^p value = 0.12
Z= ^p−p0
√ p0 ( 1− p0 )
n
Putting the values in the equation, we get the following:
Z= 0.12−0.2
√ 0.2 ( 1−0.2 )
119
Z=−0.08
√ 0.16
119
Z=−0.08
0.4
119
Z=−0.08
0.003 =−26.67
Part E
Z proportion test provides the value of -26.67. Thus, the critical value of Z at 5% level of
significance results in a p-value which is less than 0.001.

9STATISTICAL ANALYSIS FOR AUSTRALIAN HOSPITALS
Part F
Formula for calculating minimum sample size:
e=Z α
2
√π ( 1−π )
n
n=( Z α
2
e )
2
π ( 1−π )
Z α
2
=1.96
Π = 0.12
And, margin of error, e = 0.06.
Hence, minimum sample size needed for estimating the true proportion of the rural hospitals
with Obstetrics ward:
n=( 1.96
0.06 )2
0.12 ( 1−0.12 )
n= ( 32.67 ) 2 0.12 ( 1−0.12 )
n=1067.33∗0.11=117.41
Hence, required n is 118.

10STATISTICAL ANALYSIS FOR AUSTRALIAN HOSPITALS
Part G
The conservative method denotes that the values are neither assumed, nor mentioned in the
proportions. In this method, the proportions used are given equal weights. Hence, π equals 0.5.
Therefore, the value of n is:
n=( Z α
2
e )
2
π ( 1−π )
n=( 1.96
0.06 )2
0.5 ( 1−0.5 )
n= ( 32.67 ) 2∗0.25
n=1067.33∗0.25
n=266.83
Thus, value of n required is 267.
Answer 3
Part A
H0 (Null hypothesis): there is no difference between the alpha wave frequencies of the confined
and non-confined brain
H1 (Alternate hypothesis): there is significant difference between the alpha wave frequencies of
the confined and non-confined brain

11STATISTICAL ANALYSIS FOR AUSTRALIAN HOSPITALS
Part B
In two-sample t test, the dependent variable is measured on continuous scale. In this case, the
independent variable contains two independent categorical groups. Moreover, the dependent
variable is normally distributed for each independent variables in the different categorical
groups. Finally, the variances are homogenous and hence, the hypothesis will be tested at 5%
level of significance.
Part C
Formula for independent two-sample t test:
t= X 1−X 2
√ S1
2
n 1 + S2
2
n 2
Descriptive Statistics
N Minimum Maximum Mean Std. Deviation
Confined 10 7 12 10.00 1.563
Nonconfined 10 8 14 10.70 1.636
Valid N (listwise) 10
Table 5: Descriptive statistics for confined and non-confined variables
Substituting the values we get:
t= 10−10.70
√ 1.536
10 +1.636
10

12STATISTICAL ANALYSIS FOR AUSTRALIAN HOSPITALS
t= −0.70
√ 3.172
10
t=−0.70
0.563
Thus, t statistics is:
t=1.24
Part D
The p-value for the two-sample t test is 0.1074.
Part E
The calculated p-value is 0.1074, which is larger than 0.05. Hence, the null hypothesis is rejected
and alternate hypothesis is accepted. Thus, there is a difference between the alpha wave
frequencies of the confined and non-confined groups.
Part F
H0 (null hypothesis): alpha wave frequencies of confined and non-confined groups are equal
H1 (alternate hypothesis): alpha wave frequencies of confined and non-confined groups are not
equal
Ranks
Solitary N Mean Rank Sum of Ranks
Alpha wave frequency Confined 10 9.45 94.50
Nonconfined 10 11.55 115.50

13STATISTICAL ANALYSIS FOR AUSTRALIAN HOSPITALS
Total 20
Table 6: Mann Whitney test ranks
Test Statisticsa
Alpha wave
frequency
Mann-Whitney U 39.500
Wilcoxon W 94.500
Z -.812
Asymp. Sig. (2-tailed) .417
Exact Sig. [2*(1-tailed Sig.)] .436b
a. Grouping Variable: Solitary
b. Not corrected for ties.
Table 7: Mann Whitney test statistics
The p-value in the Mann Whitney test is 0.417, which is larger than 0.05. Hence, the null
hypothesis is rejected and it implies that the output is statistically insignificant. The alternate
hypothesis is accepted, which indicates that the alpha wave frequencies are not equal between
the confined and non-confined groups.
Part G
The first test was performed using the parametric test and the second one is performed using
non-parametric test. Although, both the tests generated same outcome, that is, both the tests

14STATISTICAL ANALYSIS FOR AUSTRALIAN HOSPITALS
found that there is difference in the alpha-wave frequency of the confined and non-confined
groups. Thus, it can be said that there is normal distribution in the dataset.

15STATISTICAL ANALYSIS FOR AUSTRALIAN HOSPITALS
Answer 4
Part A
The question is: ‘Is the mean 2012 total nurses cost in million dollar (TotalNurseCost) less in
hospitals without obstetrics wards than those with obstetrics wards?’
H0: mean 2012 total nurses cost in million is not less in the hospitals without obstetrics wards
than those with obstetrics wards
H0: μ0 ≥ μ1
H1: mean 2012 total nurses cost in million is less in the hospitals without obstetrics wards than
those with obstetrics wards
H1: μ0 < μ1
Part B
Assumptions for validity of the test are that the variables are normally distributed and
independent of each other.
Part C
Group Statistics
Obstetrics Ward N Mean Std. Deviation Std. Error Mean
Total Nurse Cost in Million Yes 37 35.6432 49.63899 8.16060
No 122 5.6542 21.57554 1.95336
Table 8: Group statistics in t-test

16STATISTICAL ANALYSIS FOR AUSTRALIAN HOSPITALS
Formula for independent two-sample t test:
t= X 1−X 2
√ S1
2
n 1 + S2
2
n 2
t= 37−122
√ 49.64
37 + 21.58
122
t= −85
√ 1.34 +0.18
t= −85
√1.52
t=−85
1.23
Therefore, t statistics is:
t=−69.1
Part D
The degrees of freedom are 37+122-2 = 157. In this case, p-value is less than 0.00001, which is
also less than the critical value of 0.05. Hence, the null hypothesis is accepted and the alternate
hypothesis is rejected. This implies that the mean 2012 total nurse cost in million is not less in
the hospitals without obstetrics wards than those with obstetrics wards.

17STATISTICAL ANALYSIS FOR AUSTRALIAN HOSPITALS
Part E
Independent Samples Test
Total Nurse Cost in Million
Equal variances
assumed
Equal variances
not assumed
Levene's Test for Equality of
Variances
F 37.825
Sig. .000
t-test for Equality of Means
t 5.257 3.574
df 157 40.204
Sig. (2-tailed) .000 .001
Mean Difference 29.98901 29.98901
Std. Error Difference 5.70425 8.39113
95% Confidence Interval of
the Difference
Lower 18.72204 13.03259
Upper 41.25597 46.94542
Table 9: SPSS result of independent samples test
Part F
The p-values gained from the calculations, with and without SPSS, are similar to each another.
Hence, the null hypothesis is accepted and the alternate hypothesis is rejected in both the cases.
However, the value of t statistics is different. The t-test performed without SPSS has generated a
negative value, while, the t test in SPSS has yielded a positive value. The reason of this
difference is that the mean values taken in the two t-tests are different, due to the alteration of the
data in the two cases. However, to evaluate the statistical significance of a hypothesis, only the
p-value is considered.

18STATISTICAL ANALYSIS FOR AUSTRALIAN HOSPITALS
Answer 5
Part A
Variable of interest is patient’s waiting time.
Part B
Mean value = 40,
Standard deviation, SD = 6
The formula:
Z= x −μ
σ
Therefore, Z=36−40
6
Z=−4
6 =−0.67
P (Z<-0.67) = .2514
Part C
Sample data of the patients’ waiting time is normal distributed and hence, the mean and the
standard deviation are the parameters of sampling distribution.
Part D
As, mean = 40, standard deviation, S . D= 6
√25

19STATISTICAL ANALYSIS FOR AUSTRALIAN HOSPITALS
The formula is:
Z= x−μ
σ /√ n
Therefore, Z=36−40
6
√ 25
Z= −4
6/5
Z=−4
1.2 =−3.33
P (Z<-3.33) = .0004
Part E
The two answers above are different due to the reason that in the first case, only one patient was
taken as sample, while in the second case, the entire sample was taken for calculation. Thus, due
to the size of the two sample, the results are different.

20STATISTICAL ANALYSIS FOR AUSTRALIAN HOSPITALS
Bibliography
Baarda, B. and van Dijkum, C., 2019. Introduction to Statistics with SPSS. Routledge.
Carlson, K.A. and Winquist, J.R., 2016. An introduction to statistics: An active learning
approach. Sage Publications.
Chatfield, C., 2018. Statistics for technology: a course in applied statistics. Routledge.
Chen, K., 2015. An Introduction to Statistics. Performance Evaluation by Simulation and
Analysis with Applications to Computer Networks, pp.229-246.
Heumann, C. and Schomaker, M., 2016. Introduction to statistics and data analysis. Springer
International Publishing Switzerland.
Holcomb, Z.C., 2016. Fundamentals of descriptive statistics. Routledge.
Lane, D.M., Scott, D., Hebl, M., Guerra, R., Osherson, D. and Zimmer, H., 2017. An
Introduction to Statistics. Rice University.
Moore, D.S., McCabe, G.P., Alwan, L.C., Craig, B.A. and Duckworth, W.M., 2016. The practice
of statistics for business and economics. WH Freeman.
Norris, G., Qureshi, F., Howitt, D. and Cramer, D., 2014. Introduction to statistics with SPSS for
social science. Routledge.
Peck, R., Olsen, C. and Devore, J.L., 2015. Introduction to statistics and data analysis. Cengage
Learning.