Broyles Textbook Practice Exercise Case Study

Broyles Textbook Practice Exercise [Name of the Writer]
[Name of the Institution]

Broyles Textbook Practice Exercise

Objective
The objective is to solve the textbook exercises i.e. “Practice Exercise 1” (page 157) and "Practice Exercise 11" (page 180). For the data set listed, we use Excel to extract the mean and standard deviation for the sample of lengths of stay for cardiac patients as in the Practice Exercise 1 and also apply the test to solve the problem Similarly we also use Excel to solve the another exercise.
Practice Exercise 1

Table 1: Descriptive Statistics
Lengths of Stay

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
-0.4903 is less than 3, flatter than a normal distribution with a wider peak. The probability for extreme values is less than for a normal distribution, and the values are wider spread around the mean. Similarly the negative skewness i.e. -0.035014 shows that the Left skewed distribution, most values are concentrated on the right of the mean, with extreme values to the left. Also Length of Stay have smaller standard deviation i.e. 2.61357 therefore smaller standard deviation indicates that there is less dispersion & less variability in the lengths of stay for cardiac patients. Also we see that the average value of 7.7 of the Lengths of stay for cardiac patients. Also we see that maximum 13 no of Lengths of Stay and minimum 2 no of Lengths of Stay (Weiss,
Ha: Performance of the laboratory is independent of shift.
Level of Significance

α=0.05

Test Statistics

T=0.14466

Critical Value

T-critical=4.3026

Critical Region

If T ≥ T-critical ---- Reject Ho
If T≤ T-critical ---- Do not reject Ho
Conclusion

Since our T test statistics value does not fall in critical region therefore there is no enough evidence to reject Ho & therefore conclude that Performance of the laboratory is not independent of shift or there is no difference between the means, also the mean of timely is 73.33 and the mean of tardly is 30 as we compare that the mean of timely is greater than the mean of tardly, and also the variance of timely is 933.33 and variance of tardly is 100 therefore as we compare that the variance of timely is less than the variance of tardly so the variance of timely shows the less variation (Howell, 2012).
References

Howell, D. (2012). Statistical methods for psychology. Cengage Learning.
Lowry, R. (2014). Concepts and applications of inferential statistics.
Weiss, N. A., & Weiss, C. A. (2012). Introductory statistics. Pearson