HSA 523 Health Data Analysis
Dr. Robert Jantzen
Answer key for Homework 7
NOTE: click here for an MS-Excel spreadsheet that will conduct the following one sample
tests: the z test for a population proportion and the t test for a population mean.
1. The proportion of short-term acute hospitals in the US that are investor owned for-profits
is .18, or 18%. A much smaller number (.098 or 9.8%) of the 235 hospitals sampled by
Loubeau and Jantzen were for-profits.
A. What are the data requirements for conducting a z test for a population proportion?
The data requirements include a random sample and n*ps >= 5 & n*(1-ps) >=5.
B. Is the Jantzen/Loubeau sample large enough to test hypotheses about the population
proportion?
Yes. 23 of the hospitals were for profit (>=5) and 212 were non profits (>=5).
C. Test whether the sample could have been drawn from a population of hospitals that had
an investor owned proportion of .18 (18%). Show all work, including the hypotheses, sample
statistic, critical statistic and decision rule. What can you conclude about the
representativeness of the sample?
This is a z test for the population proportion.
Ho: population proportion = .18
Ha: population proportion not equal .18
Sample z (from calculator) = -3.28 and Critical z = 1.96 (two-tailed value)
Since the sample z (in absolute value) is >= critical z, we reject the Ho. There is sufficient
evidence that the population proportion that this sample was drawn from is not .18 meaning
that this sample is not representative of the population of hospitals.
D. What does the estimated p value (also called the sig. level) of the test show?
The p value is .001 meaning that there’s only a .001 probability (.1% chance) of finding a
sample proportion that differs from the null hypothesis value as much as this sample does,
when the null hypothesis is true (that’s why we reject the Ho). It’s the probability of
rejecting the null when it’s true.
2. Assume that JCAHO regulations require that at least 90% of a hospital's technical staff
must meet current continuing educational requirements. Also assume that a random sample
of 40 employees finds that 6 have not met those requirements.
A. Is there sufficient data for conducting a z test for a population proportion.
Yes. A random sample is required and that there are at least 5 employees who meet the
requirement and 5 who don’t.
B. Test whether the JCAHO regulation is being violated, i.e., that more than 10% of all
employees don't meet the requirement. Show all work, including the hypotheses, sample
statistic, critical statistic and decision rule.
This is a z test for the population proportion.
Ho: population proportion <= .10
Ha: population proportion > .10
Sample z (from calculator) = 1.05 and Critical z = 1.65 (one-tailed value)
Since the sample z (in absolute value) is < critical z, we cannot reject the Ho. There is
insufficient evidence to state that the population proportion in violation is > .1.
C. What does the estimated p value (also called the sig. level) of the test show?
The p value is .146 meaning that there’s a .146 probability (14.6% chance) of finding a
sample proportion that differs from the null hypothesis value as much as this sample does,
when the null hypothesis is true (that’s why we can’t reject the Ho). It’s the probability of
rejecting the null when it’s true.
3. A manufacturer of radon detectors placed a random sample of twelve detectors in a
chamber that exposed them to 105 picocuries per liter of radon. The detector readings were
as follows:
91.9, 97.8, 111.4, 122.3, 105.4, 95.0, 103.8, 99.6, 96.6, 119.3, 104.8, 101.7
A. Create a data file using SPSS containing the above detector readings and find the mean
and standard deviation for the sample readings.
Descriptive Statistics
N
Minimum
radon readings (hw7)
12
Valid N (listwise)
12
91.90
Maximum
122.30
Mean
Std. Deviation
104.1333
9.39742
B. What are the data requirements for conducting a t-test for the population mean (sigma
unknown)?
The requirements include a random sample and either the numbers are normally distributed
or the sample size is >=30 with #s that aren’t highly skewed.
C. Using the estimated sample mean and standard deviation, conduct a one-sample t-test to
assess whether the manufacturer’s detectors were accurately measuring the true radon count?
Show the hypothesis, sample t statistic, critical t value and decision rule.
Ho: population mean = 105
Sample t =- .32
Ha: population mean not equal 105
Critical t (2 tailed at 5%) = 2.20
Since the |Sample t| is < |Critical t|, we can’t reject the null hypothesis. There’s insufficient
evidence to say that the population mean is not 105 (hence they are correctly calibrated).
D. What does the estimated p value (also called the sig. level) of the test show?
The p-value of .75 is the probability of rejecting the null hypothesis when it’s true. It shows
the probability of observing samples whose means differ at least as much as this one does
from the null value, when the null is true.
E. Use SPSS to calculate the appropriate sample t statistic by clicking on Analyze *
Compare Means * One Sample T-test, and then moving the variable name into the Test
Variable box, typing in the appropriate Test Value and clicking on OK. Does the SPSS
sample t agree with the one that you calculated?
SPSS generates the same sample t (of- .32).
4. Assume that the CDC is concerned about exposure to lead at a particular company and
randomly samples 30 employees blood. Also assume that the average blood lead level (BLL)
for those sampled is 8 ug/dl, with a standard deviation of 2. Also assume that "normal"
BLLs average 6 ug/dl.
A. What are the data requirements for conducting a t-test for the population mean (sigma
unknown)?
The requirements include a random sample and either the numbers are normally distributed
or the sample size is >=30 with #s that aren’t highly skewed.
B. Using the estimated sample mean and standard deviation, conduct a one-sample t-test to
assess whether the population mean BLL level for all company employees is greater than the
"normal" level of 6. Show the hypothesis, sample t statistic, critical t value and decision rule.
Ho: population mean <= 6
Sample t =5.48
Ha: population mean > 6
Critical t (1 tailed at 5%) = 1.70
Since the |Sample t| is > |Critical t|, we reject the null hypothesis. There’s sufficient
evidence that the population mean BLL for this company’s employees is greater than 6 (and
the sample mean agrees with the alternative hypothesis, which is needed in a one-tailed test
in order to reject Ho). We’re 95% sure with a 5% chance of erroneously rejecting the null.
C. What does the estimated p value (also called the sig. level) of the test show?
The p-value of .000003 is the probability of rejecting the null hypothesis when it’s true. It
shows the probability of observing samples whose means differ at least as much as this one
does from the null value, when the null is true. Samples like this are extremely rare if the
null was true.
#1 Z Test of Hypothesis for the
Population Proportion
#3 t Test of Hypothesis for the Population
Mean (sigma unknown)
Data
Null Hypothesis
p=
Level of Significance
Number of Successes
Sample Size
Data
Null Hypothesis
=
Level of Significance
Sample Size
Sample Mean
Sample Standard
Deviation
0.18
0.05
23
235
Intermediate Calculations
Sample Proportion
0.09787234
Standard Error
0.025061626
Sample Z Test Statistic
3.277028354
One-Tail Test
Absolute Critical Z Value
p - value
1.644853476
0.000524589
Two-Tail Test
Absolute Critical Z Value
p - value
1.959962787
0.001049178
105
0.05
12
104.1333
9.39742
Intermediate Calculations
Standard Error of the
Mean
2.712801483
Degrees of Freedom
11
Sample t Test Statistic
-0.319485228
One-Tail Test
Absolute Critical T
Value
p - value
1.795883691
0.377672052
Two-Tailed Test
Absolute Critical T
Value
p - value
2.200986273
0.755344105
#3 One-Sample Test
Test Value = 105
95% Confidence Interval
of the Difference
t
radon readings (hw7)
df
-.319
Sig. (2-tailed)
11
.755
Mean
Difference
-.86667
Lower
-6.8375
Upper
5.1042
#2 Z Test of Hypothesis
for the Population
Proportion
#4 t Test of Hypothesis for the
Population Mean (sigma
unknown)
Data
Null Hypothesis
p=
Level of Significance
Number of Successes
Sample Size
Data
Null Hypothesis
=
Level of Significance
Sample Size
Sample Mean
Sample Standard Deviation
6
0.05
30
8
2
0.1
0.05
6
40
Intermediate Calculations
Sample Proportion
Standard Error
Sample Z Test Statistic
0.15
0.047434165
1.054092553
Intermediate Calculations
Standard Error of the Mean
Degrees of Freedom
Sample t Test Statistic
0.365148372
29
5.477225575
One-Tail Test
Absolute Critical Z Value
p - value
1.644853476
0.145920298
One-Tail Test
Absolute Critical T Value
p - value
1.699127097
3.36957E-06
Two-Tail Test
Absolute Critical Z Value
p - value
1.959962787
0.291840597
Two-Tailed Test
Absolute Critical T Value
p - value
2.045230758
6.73915E-06