Biostatistics for Healthcare Research: Hypothesis Testing Analysis

This document presents a comprehensive solution to a biostatistics assignment focused on hypothesis testing within the context of healthcare research. The assignment includes three main questions. Question 1 involves a detailed step-by-step analysis of hypothesis testing, calculating necessary statistics (SEM, t-value), establishing critical values, determining the region of rejection, and making decisions about the null hypothesis based on the t-value. Question 2 presents a research question about the average physical and mental health component scores for a female population, comparing them to a national norm. It provides the null and alternative hypotheses, t-values for both health components, statistical significance findings, and a summary of the results derived from SPSS output, including one-sample statistics and a one-sample t-test. The findings indicate that both physical and mental health scores are significantly lower than the national norm. Question 3 addresses the power of a test, explaining how it can be increased by focusing on a high 1-β value during sampling to reduce the risk of Type II error. The document provides a complete analysis and interpretation of statistical concepts relevant to healthcare research.

Running head: BIOSTATISTICS FOR HEALTHCARE RESEARCH
BIOSTATISTICS FOR HEALTHCARE RESEARCH
Name of the Student
Name of the University
Author note

1BIOSTATISTICS FOR HEALTHCARE RESEARCH
Question 1.
Step 1: Specify the hypotheses (symbolically)
Null Hypothesis: μ = 55
Alternative hypothesis: μ ≠ 55
α = 0.05 (two-tailed)
Step 2: Calculate the necessary statistics
SEM = SD /√ (n)
=8 /√ 64
= 1
t = (57-55)/1=2
Step 3: Establish the critical value and region of rejection. (Drawing the picture helps
you understand the process)
Degree of freedom or Df=N-1=64-1=63
Alpha values is α = 0.05
For the critical value can be found for Df of 63 and α = 0.05 in the following two tail
distribution table
S= √ ∑ ( X- X )2
(n-1) = √ 148 .90
9 =4 . 07
t=X −μ
SEM
S x=s
√ n =4 . 07
3 . 16 =1. 29
t c =(9 . 90−6 .75 )
1. 29 =2 . 44
t α =2 .262
2. 44>2 .262

2BIOSTATISTICS FOR HEALTHCARE RESEARCH
The two tail distribution table:
DF A = 0.2 0.10 0.05 0.02 0.01 0.002 0.001
∞ ta = 1.282 1.645 1.960 2.326 2.576 3.091 3.291
1 3.078 6.314 12.706 31.821 63.656 318.289 636.578
2 1.886 2.920 4.303 6.965 9.925 22.328 31.600
3 1.638 2.353 3.182 4.541 5.841 10.214 12.924
4 1.533 2.132 2.776 3.747 4.604 7.173 8.610
5 1.476 2.015 2.571 3.365 4.032 5.894 6.869
6 1.440 1.943 2.447 3.143 3.707 5.208 5.959
7 1.415 1.895 2.365 2.998 3.499 4.785 5.408
8 1.397 1.860 2.306 2.896 3.355 4.501 5.041
9 1.383 1.833 2.262 2.821 3.250 4.297 4.781
10 1.372 1.812 2.228 2.764 3.169 4.144 4.587
11 1.363 1.796 2.201 2.718 3.106 4.025 4.437
12 1.356 1.782 2.179 2.681 3.055 3.930 4.318
13 1.350 1.771 2.160 2.650 3.012 3.852 4.221
14 1.345 1.761 2.145 2.624 2.977 3.787 4.140
15 1.341 1.753 2.131 2.602 2.947 3.733 4.073
16 1.337 1.746 2.120 2.583 2.921 3.686 4.015
17 1.333 1.740 2.110 2.567 2.898 3.646 3.965
18 1.330 1.734 2.101 2.552 2.878 3.610 3.922
19 1.328 1.729 2.093 2.539 2.861 3.579 3.883
20 1.325 1.725 2.086 2.528 2.845 3.552 3.850

3BIOSTATISTICS FOR HEALTHCARE RESEARCH
21 1.323 1.721 2.080 2.518 2.831 3.527 3.819
22 1.321 1.717 2.074 2.508 2.819 3.505 3.792
23 1.319 1.714 2.069 2.500 2.807 3.485 3.768
24 1.318 1.711 2.064 2.492 2.797 3.467 3.745
25 1.316 1.708 2.060 2.485 2.787 3.450 3.725
26 1.315 1.706 2.056 2.479 2.779 3.435 3.707
27 1.314 1.703 2.052 2.473 2.771 3.421 3.689
28 1.313 1.701 2.048 2.467 2.763 3.408 3.674
29 1.311 1.699 2.045 2.462 2.756 3.396 3.660
30 1.310 1.697 2.042 2.457 2.750 3.385 3.646
60 1.296 1.671 2.000 2.390 2.660 3.232 3.460
120 1.289 1.658 1.980 2.358 2.617 3.160 3.373
8 1.282 1.645 1.960 2.326 2.576 3.091 3.291
As per the findings from the table the t value for DF 60 is 2. Therefore, for value 63
the critical value will be less than 2.
Hence the region of rejection will start from the value less than 2.
Therefore, the null hypothesis has been rejected and the alternative hypothesis has
been accepted. Hence, there is a significant difference between maximal legal speed
limit and mean of real speed.
Step 4: Make a decision regarding the null hypothesis.

4BIOSTATISTICS FOR HEALTHCARE RESEARCH
As per the hypothesis testing rule, to reject the null hypothesis the t value must fall under the
region of rejection. In other words to reject null hypothesis the t value must be better than the
critical value. For this case the t value is 2. The critical value is slightly less than 2.
Hence it can be consider that since the critical value is less than 2 the t value is greater than
critical value. Therefore, the t value for this case falls under the region of rejection.
From the above analysis, the Null hypothesis is rejected and alternative hypothesis is
accepted. Therefore,
Question 2.
a. Research Question
Is the average physical health component score for female population satisfies the national
norm which is 50?
b. Hypotheses (non-directional)
Null hypothesis H0: Average physical health component score for female population is not
significantly different form the national norm which is 50
Alternative Hypothesis HA: Average physical health component score for female population
is significantly different from the form the national norm which is 50
c. Values of t for two analysis
The t value for physical health component score is -13.469 and the t value for mental health
component score is -8.770

5BIOSTATISTICS FOR HEALTHCARE RESEARCH
d. Statistical Significance
From the significant 2-tailed value it has been found that for physical health component score
the significant value is lower than 0.05. Therefore, the values are significant for physical
health component score.
For mental health component score the significant value is lower than 0.05. Therefore, the
values are significant for mental health component score.
The level of significance for both physical health component score and mental health
component score is 95% confidence interval.
e. Summary of result
From the SPSS results it has been found that the mean values of physical health component
score as well as the mental health component score is lower than 50. At the same time, the
difference are significant as per the probability testing value. Therefore, it can be sated that
the physical health component score as well as the mental health component score are
significantly lower than the national norm which is 50. Therefore, the chosen women
population has significantly lower mental and physical health condition and that should be
the major concern of national health care services.
f. SPSS Data
One-Sample Statistics

6BIOSTATISTICS FOR HEALTHCARE RESEARCH
N Mean Std. Deviation Std. Error Mean
SF12: Physical Health
Component Score,
standardized
893 45.11422 10.840059 .362749
SF12: Mental Health
Component Score,
standardized
893 46.82843 10.806477 .361625
One-Sample Test
Test Value = 50
t df
Sig. (2-
tailed)
Mean
Difference
95% Confidence Interval of the
Difference
Lower Upper
SF12: Physical Health
Component Score,
standardized
-13.469 892 .000 -4.885780 -5.59772 -4.17384
SF12: Mental Health
Component Score,
standardized
-8.770 892 .000 -3.171570 -3.88131 -2.46184
Question 3.
The power of a test is the probability of rejecting the null hypothesis. It is denoted by 1-β.
Increasing the power can be possible through considering the 1-β high enough while
executing sampling. It will increase the value of 1-β and reduce the value of probability of
type to error β. As a result it will decrease the risk of type II error.