# Course Work Example on Descriptive Medical Statistics

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Descriptive statistics (n=40)

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The table below illustrates the descriptive statistics for the sample of 40 patients. Health plans require patients admitted to hospitals suffering from coronary heart diseases not to stay in hospitals for more than 6.5 days. Other hand, people claim that 6.5 days are too low for a patient suffering from the disease to be discharged. To investigate and statistically proof whether the allegations are true, the number of days taken by coronary heart disease patients while in the hospital for 40 patients were recorded. The data was used to determine the average number of days and standard deviation that was used to test the hypothesis as illustrated below.

Column1

Mean 7.7

Standard Error 0.413242339

Median 8

Mode 8

Standard Deviation 2.613574034

Sample Variance 6.830769231

Kurtosis -0.490362171

Skewness -0.035014338

Range 11

Minimum 2

Maximum 13

Sum 308

Count 40

From the above descriptive statistics table, mean=7.7, standard error=0.413242339, mean=8, , standard deviation= 2.613574034, median=8,sample variance= 6.830769231, kurtosis=0.490362171, skewness=0.035014338, range=11, minimum=2, maximum=13, sum=308 and sample size=40.here we are much concerned with the mean and standard deviation.

The mean=7.7

Standard deviation=2.614

n=44

hypothesis

null hypothesis: patient hospitalized with coronary heart disease requires no more than 6.5 days of hospital care.

Alternative hypothesis: patient hospitalized with coronary heart disease requires more than 6.5 days of hospital care.

The t-test is used as test statistic since the population means is unknown and n=40 hence small population.

t statistic=sampe mean-hypothesized mean standard deviatio=7.7-6.52.614=0.4591

t value=t (0.05,40-1) = 1.697, it is one-tailed since it tests the inequality that is greater than 6.5 days.

Since t statistic=0.4591 is less than critical value=1.697, meaning that the test statistic falls in the acceptance region. Hence, we do not reject the hypothesis and make a conclusion that patient hospitalized with coronary heart disease requires less or equal to 6.5 days of hospital care. From the test, it is statistically evidenced that patients admitted to hospitals suffering from coronary heart diseases not to stay in hospitals for more than 6.5 days. The believe that patients admitted to hospitals suffering from the disease that they should take more that 6six and half days is wrong.

Question two

There is a claim that there exists an association between performance and shift. To verify this assertion Chi-square test of independence was used. We wish to establish the connection between performance and shift we start by stating the hypothesis as shown below.

The hypothesis to be tested

Null hypothesis: Performance does not depend on shift.

Alternative Hypothesis: Performance depends on shift.

In this question, we wish to use Pearson chi-square test of independence to check whether there exists a dependence between the two categorical variables that is performance and shift.

Chi-Square Test for Association between performance and shift.

Day Evening Night total

Timely 100(85.16%) 80(78.06%) 40(56.77%) 220

Tardy 20(34.84%) 30(31.94%) 40(23.23%) 90

Total 120 110 80 310

Pearson Chi-Square = 26.142, Degrees of Freedom = 2, P-value = 0.000

From the above chi-square test of independence table, the probability of rejecting the null hypothesis when is true, P-value = 0. 000 is less than 0.05 hence we reject the null hypothesis and make a conclusion that Performance depends on shift. There is 0.0% chance of rejecting the null hypothesis when it is true that there is no association between performance and shift.

References

Cohen, Barry (2014). Explaining psychological statistics. Hoboken, NJ: John Wiley & Sons, 2013.

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