Download hsa523.hw 6key

HSA 523 Health Data Analysis
Dr. Robert Jantzen
Homework 6 Answer Key
NOTE: click here for a link to an MS-Excel spreadsheet that will calculate confidence
intervals and sample sizes for means and proportions. Click here for a link to an MSExcel spreadsheet that will calculate critical z & t scores for differing significance levels.
1. If you use the large sample size formula to compute a confidence interval (CI) for a
population proportion, what is the appropriate z critical value for each of the following
confidence levels:
a. For 99%, z= 2.58 b. For 95%, z= 1.96 c. For 90%, z= 1.645
2. The use of the large sample size confidence interval for a population proportion
requires a sufficient sample size. For each of the following combinations of n (sample
size) and ps (sample proportion), test whether the large sample size confidence interval
should be used:
n (ps) and n (1- ps) must be >= 5 to use the large sample confidence interval calculation.
a. n=50 and ps=.3 OK
b. n=50 and ps=.05 Not OK
c. n=15 and ps=.45 OK
d. n=100 and ps=.01 Not OK
3. A study of 39 bartenders who had respiratory problems before smoking was banned
in bars in San Francisco found that 21 were symptom free two months after the ban was
instituted. Assuming that the sample was a random one, is the sample large enough for
constructing a confidence interval (CI)) for the population proportion of bartenders with
respiratory problems who were symptom free 2 months after the smoking ban. Construct
the 95% CI and interpret.
Sample Size
Number of Successes
Confidence Level
39
21
95%
Intermediate Calculations
Sample Proportion
0.538461538
Z Value
-1.95996279
Standard Error of the
Proportion
0.079826849
Interval Half Width
0.156457654
Confidence Interval
Interval Lower Limit
0.382003884
Interval Upper Limit
0.694919193
The sample is large enough because n(ps) and n(1-ps) are both >= 5. The 95% CI is .382
to .695 which means we are 95% confident that the true proportion of all bartenders who
would be symptom free after the smoking ban lies in that interval.
4. Suppose you want to estimate the fraction of the population of drivers who use
seatbelts, with 95% confidence, to within .02 or 2% points. What is the required sample
size?
Need a sample size of 2401.
Data
Estimate of True
Proportion
Sampling Error
Confidence Level
0.5
0.02
95%
Intermediate Calculations
Z Value
Calculated Sample Size
1.95996279
2400.90883
Result
Sample Size Needed
2401
5. What is the appropriate critical t value for estimating a population mean for each of
the following confidence levels and sample sizes?
Can use either the calculator or the t table at n-1 degrees of freedom to find these
answers.
a. 95% and n=17 t= 2.01
b. 90% and n=12 t= 1.80
c. 99% and n=24 t= 2.81
d. 90% and n=25 t= 1.71
e. 95% and n=10 t= 2.26
6. The following summary statistics were obtained from a random sample of 47 students
who did not return to college after the first semester and a random sample of 257 who did:
Nonpersisters (47)
Persisters (257)
Number of hours worked per
week during first semester:
Mean
Standard Deviation
25.6
14.4
18.1
15.3
a. Compute 99% confidence intervals for the mean number of hours worked for both
groups. Interpret each CI.
b. Do the two CIs suggest that the 2 groups means are different or the same? Why?
Data
Sample Standard
Deviation
Sample Mean
Sample Size
Confidence Level
Data
Sample Standard
Deviation
Sample Mean
Sample Size
Confidence Level
14.4
25.6
47
99%
15.3
18.1
257
99%
Intermediate Calculations
Standard Error of the Mean
2.100455878
Degrees of Freedom
46
t Value
2.687011147
Interval Half Width
5.643948357
Intermediate Calculations
Standard Error of the Mean
0.954387778
Degrees of Freedom
256
t Value
2.595170372
Interval Half Width
2.476798885
Confidence Interval
Interval Lower Limit
Interval Upper Limit
Confidence Interval
Interval Lower Limit
Interval Upper Limit
19.96
31.24
15.62
20.58
The 99% CIs show the intervals where we’re 99% confident that the true population
means lie within. Since they overlap, the two population means might be the same.
7. If you wanted to know what the average number of days per week current Iona
College students drink alcohol, how could you find out?
a. Describe who you would sample and what question(s) you would ask.
We could conduct a random sample and ask questions concerning alcohol use, like how
many days did you drink alcohol last week, etc.
b. How many students would you survey, assuming you wanted your estimate to be
within 1 of the actual mean frequency with a 95% confidence level? Describe in detail
your sample size design.
Since the range is 7 days (max-min), then the standard deviation can be approximated as
7/6 or 1.17. The needed sample size is 6.
Data
Population Standard
Deviation
Sampling Error
Confidence Level
1.17
1
95%
Intemediate Calculations
Z Value
-1.95996279
Calculated Sample Size
5.258566556
Result
Sample Size Needed
6
8. The SPSS file Stafserv.sav contains scope of service information for a random sample
of 50 US hospitals. Specifically, the file lists how many specialized services (as reported
by the American Hospital Directory) each hospital provided in 1996.
i. Double-click here to load the file into SPSS. If the SPSS program doesn't start up
automatically, download and save the Stafserv.sav file by right-clicking on the highlighted
file name, and then clicking on Save Target As and "maneuvering" the Save In screen to
a folder like C:\Temp. Note: this file might be too big to save to a floppy or your U:
drive on the Iona network. After saving the file, start up SPSS , then click on Open an
Existing File, then OK, then the down-triangle next to the Look In box, and then on the
appropriate drive where you downloaded the file to (namely C:\Temp) button. SPSS will
then display files on the selected drive, and you can then click on your file name and
Open to open it. Use the file and SPSS to:
a. generate a 95% confidence interval for the average number of specialized services
provided by US private hospitals. In order to generate the confidence interval, use the
Analyze * Descriptive Statistics * Explore sequence of commands and move the
stafserv variable into the Dependent List box. Click on Statistics and note that the
Descriptive - Confidence interval for the mean button is already clicked on with a 95%
confidence level. Then click on Continue * OK. Interpret the confidence interval.
b. Using the sample's estimated mean and standard deviation for the number of
licensed beds, compute the 95% confidence interval for the population mean. Does your
estimate agree with the SPSS calculation?
a. The 95% CI shows where the population mean # of services lies, with a 95%
certainty.
b. The SPSS and Excel CI calculations agree.
SPSS Descriptives
Statistic
# of staffed services
Mean
95% Confidence
Interval for Mean
23.3400
Lower Bound
Upper Bound
25.1975
23.3556
Median
23.0000
Std. Deviation
42.719
6.53596
Minimum
11.00
Maximum
36.00
Range
25.00
Interquartile Range
.92432
21.4825
5% Trimmed Mean
Variance
Std. Error
9.00
Skewness
-.017
.337
Kurtosis
-.641
.662
Data
Sample Standard
Deviation
Sample Mean
Sample Size
Confidence Level
6.536
23.34
50
95%
Intermediate Calculations
Standard Error of the Mean
0.924329984
Degrees of Freedom
49
t Value
2.009574018
Interval Half Width
1.857509521
Confidence Interval
Interval Lower Limit
Interval Upper Limit
21.48
25.20