Statistics 2010:
Statistics for the Social and Behavioral Sciences
Name: __________________________ Section #: _____________
Please read the scenario below and answer the following problems.
Dr. Hans is interested in studying the life expectancy of elderly persons living in the assisted living facility that he manages. Life expectancy in the United States is 78.7 years with a standard deviation of 16.51 years. Dr. Hans reviews the records of the past 30 residents who have passed away. Dr. Hans would like to know if the life expectancy of residents living in his facility is more or less than the population. Use an alpha (probability of a Type I error, p -level) of .01. The age of each of the residents at death (in numerical order) is recorded below.
______________________________________________________________________________
Resident #1 = 80 Resident #2 = 81 Resident #3 = 82
Resident #4 = 83
Resident #7 = 84
Resident #10 = 85
Resident #5 = 83
Resident #8 = 84
Resident #11 = 87
Resident #6 = 83
Resident #9 = 85
Resident #12 = 88
Resident #13 = 88
Resident #16 = 89
Resident #19 = 89
Resident #22 = 90
Resident #14 = 88
Resident #17 = 89
Resident #20 = 90
Resident #23 = 91
Resident #15 = 89
Resident #18 = 89
Resident #21 = 90
Resident #24 = 91
Resident #25 = 92 Resident #26 = 93 Resident #27 = 93
Resident #28 = 94 Resident #29 = 95 Resident #30 = 95
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1. Identify the four assumptions for a z -test. Briefly report if you have met each assumption and explain how you know.
Assumption: DV is scale; yes; life expectancy is treated as ratio. A zero indicates the absolute absence of life expectancy.
Assumption: Random selection; no; it is a convenience sample. Only the past 30 residents were selected.
Assumption:
Population is normally distributed; we don’t know, but we can assume because the sample size is at least 30
Assumption:
Parameters for 1 population; yes; µ = 78.7; σ = 16.51
2. What is the full (write out) null and research hypotheses for the study?
Null Hypothesis ( H
0
): there is no difference in life expectancy for residents at Dr. Hans’ facility and those in the overall U.S. population
Research Hypothesis ( H
1
): there is a difference in life expectancy for residents at Dr.
Hans’ facility and those in the overall U.S. population
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3. At the 1% probability rate of committing a Type 1 error (also called alpha [α] or p -level), what would be the critical value(s)? z crit =
+/- 2.58
4. What is the mean for Dr. Hans’ residents?
Mean = 88
5. Calculate the standard error for the population distribution of means: 3.01
6. Calculate the mean for the population distribution of means. 78.7
7. Draw a z -distribution (normal distribution of means ) that includes both z -statistics (in units of
1) and raw scores (life expectancy) on the x -axis. Draw a line(s) for the critical value(s).
Z crit
= -2.58 Z crit
= 2.58
Z stat
= 3.09
Z = -3
69.67
Z = -2
72.68
Z = -1
75.69
Z = 0
µ = 78.7
Z = 1
81.71
Z = 2
84.72
Z = 3
87.73
8. Calculate the zstatistic for Dr. Hans’ sample of residents. Draw a line on the distribution in problem 7 that represents the zstatistic based on Dr. Han’s sample z stat =
3.09
9. Using the z -table in the back of the book (Appendix B-1) and the z -statistic, calculate the percentile for Dr. Hans’ sample of residents in comparison to the population.
Percentile: 100 percentile
10. Based on the results of the z -test, what would be your statistical decision? Reject the null hypothesis; there appears to be a difference between the life expectancy of Dr. Hans’ residents and the overall population
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11. Based on the data collected for the sample, what is the effect size? Explain what the effect size tells you.
d = 0.56; This is a medium effect; Individual residents in Dr. Hans’ facility have a 0.56 standard deviation increase in life expectancy on average over individuals in the general population
12. Calculate the 99% confidence interval for the sample data? Explain what the confidence interval tells you.
[80.23 , 95.77 ] There is a 99% chance that the true population mean life expectancy for
Dr. Hans’ residences is between 80.23 years and 95.77 years
13. Dr. Hans would like to use the findings of the study to advertise that the elderly care at his facility results in an increased life for residents. Provide a recommendation on whether Dr. Hans should advertise the findings ( practical decision ). Your response should consider the z-test results, effect size, test assumptions, sampling technique, scenario, and any additional insights you might have.
There may be several possible answers, but here is an example:
No, Dr. Hans should not use the findings in advertisements. Even though we rejected the null with our hypothesis test indicating that life expectancy at his facility is higher than the overall population at a greater than chance level and also had a medium effect size indicating a moderate increase in life expectancy for individuals, Dr. Hans used a convenience sample which introduces bias in the results. Those who are admitted to resident living facilities tend to be older than the general population already. A better comparison would be with a population made up of only assisted living persons. Otherwise the advertisement would be deceitful.
Here may be another:
Yes, Dr. Hans should use the findings in advertisements. We rejected the null with our hypothesis test indicating that life expectancy at his facility is higher than the overall population at a greater than chance level. We also had a medium effect size indicating a moderate increase in life expectancy for individuals. Both of these results are correct statements. Although Dr.
Hans used a convenience sample, a truly random sample would have been difficult to obtain if he wished to infer to residents who would be admitted in the future (they wouldn’t have an equal chance of being in the sample). Given the standard of living at advanced ages and the difficulty of caring for loved ones, it may make sense that admitting someone into a living facility with focused 24-hour care would increase life expectancy.
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