Statistical Analysis of Australian Hospitals - STA2300 Assignment

This assignment presents a comprehensive statistical analysis of data related to Australian hospitals. It covers various statistical techniques including descriptive statistics, confidence interval calculation, hypothesis testing using t-tests (both one-sample and two-sample), and non-parametric tests like the Mann-Whitney U test. The analysis explores factors such as the cost of nursing staff, the presence of obstetric wards, and patient waiting times. The document meticulously outlines the null and alternative hypotheses, assumptions, and conditions for each test. It provides detailed calculations, interpretations of p-values, and comparisons of results obtained through manual calculations and SPSS outputs. The assignment also addresses the calculation of minimum sample sizes and the application of both parametric and non-parametric tests to the same dataset, offering a thorough understanding of statistical methods in a healthcare context. The document also includes a bibliography of relevant sources.

Running head: STATISTICAL ANALYSIS FOR AUSTRALIAN HOSPITALS
Statistical Analysis for Australian Hospitals
Name of the Student:
Name of the University:
Author note:

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