STAT 5002 - Statistics Assignment: Systolic Blood Pressure Analysis

This statistics assignment analyzes data collected on systolic blood pressure to determine the risk of kidney disease within a population. The study employed a simple random sampling method to collect data from 30 participants, 15 males and 15 females. The assignment presents descriptive statistics, including mean and standard deviation, and calculates a 95% confidence interval for the mean systolic blood pressure. Furthermore, it conducts a one-sample t-test to test the hypothesis that the mean systolic blood pressure is 127 mmHg. The assignment also compares the mean and standard deviation between the male and female groups and performs t-tests to evaluate the equality of population standard deviations and means. The results indicate an acceptance of the null hypothesis in several tests, suggesting that the mean systolic blood pressure is 127 mmHg, and that the population standard deviations and means for the two groups are equal. The assignment provides a comprehensive statistical analysis, including hypothesis testing and confidence intervals, to draw conclusions about the population's systolic blood pressure.

Statistics
Statistics
Student name:
Instructor:
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Statistics
Introduction
A survey was conducted with the main objective being to determine the extent of risk that a
population was to getting kidney disease. Since hypertension is one of the major risks of getting
kidney disease, a sample of 30 was sample and their systolic blood pressure measured and
recorded. A simple random sampling method was employed to sample 15 males and 15 females.
However, this sampling method has some bias but less compared to other methods of sampling.
One of the biases occurs when the sample obtained is not inclusive of the larger population. The
study claims that the mean systolic BP is 127 for the population.
Data analysis and results
The mean and standard deviation of systolic BP
DESCRIPTIVE STATISTICS
Mean 124.7666667
Median 123.5
Standard Deviation 9.031299343
Table 1
From the table above, it can be observed that the mean systolic BP is 124.76 mmHg. The
standard deviation is 9.03 mmHg.
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95% confidence interval for the mean
Confidence Level(95.0%) 3.372343
Mean 124.7667
Lower bound 121.3943
Upper bound 128.139
Table 2
The 95% confidence interval values are as shown in table 2 above. It can be concluded that 95%
of the times, we are confident the mean systolic BP will always lie between 121.77 mmHg and
128.14 mmHg.
One sample t-test
Hypothesis
H0: Mean systolic BP = 127mmHg
H1: Mean systolic BP ≠ 127mmHg
t-Test: Two-Sample Assuming
Unequal Variances
Variable 1 Variable 2
Mean 124.766667 127
Variance 81.5643678 0
Observations 30 30
Hypothesized Mean Difference 0
df 29
t Stat -1.354453
P(T<=t) one-tail 0.09302588
t Critical one-tail 1.69912703
P(T<=t) two-tail 0.18605176
t Critical two-tail 2.04522964
Table 3
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From the table above, it can be observed that the p-value for the two tailed test is 0.18. This is
greater than the level of significance which is 0.05. This means that the null hypothesis is
accepted hence concluding that the mean systolic BP is 127mmHg.
Mean and standard deviation for the two groups (males and females)
MALE FEMALE
Mean 126.53 Mean
123.0
0
Standard Deviation 8.93 Standard Deviation 9.09
Table 4
The mean systolic BP for the males (126.53 mmHg) is higher than that of their female
counterparts (123 mmHg).
Test for population standard deviation
Hypothesis
H0: Population standard deviations for the two populations are not unequal
H1: Population standard deviations for the two populations are unequal
t-Test: Two-Sample Assuming Unequal Variances
male female
Mean 126.5333 123
Variance 79.69524 82.57143
Observations 15 15
Hypothesized Mean Difference 0
df 28
t Stat 1.074275
P(T<=t) one-tail 0.145935
t Critical one-tail 1.701131
P(T<=t) two-tail 0.29187
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t Critical two-tail 2.048407
Table 5
From the table above, it can be observed that the p-value for the two tailed test is 0.29. This is
greater than the level of significance which is 0.05. This means that the null hypothesis is
accepted hence concluding that the population standard deviations for the two populations are
not unequal. Therefore we will assume equality of standard deviation in the next question.
Test for equality of population means
Since there are only two variables, a t-test is going to be employed.
Hypothesis
H0: The population means for the two populations are equal.
H1: The population means for the two populations are unequal.
t-Test: Two-Sample Assuming
Equal Variances
Male Female
Mean 126.5333333 123
Variance 79.6952381 82.57143
Observations 15 15
Pooled Variance 81.13333333
Hypothesized Mean Difference 0
df 28
t Stat 1.074275281
P(T<=t) one-tail 0.145935124
t Critical one-tail 1.701130934
P(T<=t) two-tail 0.291870248
t Critical two-tail 2.048407142
Table 5
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From the table above, it can be observed that the p-value for the two tailed test is 0.29. This is
greater than the level of significance which is 0.05. This means that the null hypothesis is
accepted hence concluding that the population means for the two populations are equal.
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