1. Interval estimation is:
a.
determining
the range or interval in which the value of the parameter is thought to be.
b.
the act
of generalizing from a sample to a population with calculated degree of
certainty.
c.
investigating
the validity of a claim about the value of a population characteristic.
d.
values of
the test statistic for which we reject the null in favor of the alternative
hypothesis.
2. A Type II error occurs when the null
hypothesis is:
a.
rejected
and the alternative hypothesis is actually true.
b.
accepted
and the alternative hypothesis is actually true.
c.
rejected
and null hypothesis is actually true.
d.
accepted
and the null hypothesis is actually true.
3.
The
chi-square test of independence is used to:
a.
determine
whether a frequency distribution follows a claimed distribution.
b.
determine
the relationship between two quantitative variables.
c.
determine
whether there is an association between a row variable and column variable in a
contingency table.
d.
predict
the value of a dependent variable based on the value of at least one
independent variable.
4.
The
difference between ANOVA and the t-tests is that:
a.
ANOVA does
not make any assumptions about the shape of the distribution nor about the
homogeneity of variances.
b.
ANOVA is
better than t-tests in public health research.
c.
ANOVA can
be used in situations where there are more than two means being compared.
d.
There is
no difference between ANOVA and t-tests.
5.
Suppose
we want to estimate the average weight of an adult male in New York. We draw a
random sample of 100 men and find that the sample mean is 80 kg, and the
standard deviation of the sample is 15 kg. What is the 95% confidence interval?
a. 80 ±
5.88
b. 80 ±
15.0
c. 80 ± 1.96
d. 80 ±
2.94
6. Consider the following null and
alternative hypotheses:
H0: p = 0.25
H1: p > 0.25
These
hypotheses:
a.
indicate
a two-tailed test with a rejection area in both tails.
b.
indicate
a one-tailed test with a rejection area in the left tail.
c.
indicate
a two-tailed test with a rejection area in the right tail.
d.
indicate
a one-tailed test with a rejection area in the right tail.
A sample of 30 systolic blood pressures
patients undergoing drug therapy for hypertension is recorded. We wish to know
if on the basis of these data, we may conclude that the mean systolic blood
pressure is different from 160. Use the SPSS output to answer the questions
(7, 8, 9).
|
One-Sample Test |
||||||
|
Test Value = 160 |
||||||
|
t |
df |
Sig. (2-tailed) |
Mean Difference |
95% Confidence Interval of the Difference |
||
|
Lower |
Upper |
|||||
|
SBP |
-1.804 |
29 |
.082 |
-7.200 |
-15.36 |
.96 |
7. The null hypothesis is:
a.
the mean
systolic blood pressure is equal to 160.
b.
the mean
systolic blood pressure is greater than 160.
c.
the mean
systolic blood pressure is equal to zero.
d.
the mean
systolic blood pressure is less than zero.
8. The alternative hypothesis is:
a.
the mean
systolic blood pressure is less than 160.
b.
the mean
systolic blood pressure is not equal to 160.
c.
the mean
systolic blood pressure is greater than 160.
d.
the mean
systolic blood pressure is equal to 160.
9. At α=0.05 we conclude that:
a. the null hypothesis is rejected.
b. the null hypothesis is not rejected.
c. the alternative hypothesis is rejected.
d. cannot decide.
In a trial, 15 men are randomly assigned
to a treatment (diet), and another 15 men assigned as a control group, to
determine the effectiveness of the diet on reducing weight. Use the SPSS
output to answer the questions (10, 11, 12).
10. The null hypothesis is:
a.
the mean
weight of the treatment group is not equal to the mean weight of the control
group.
b.
the mean
weight of the treatment group is greater than the mean weight of the control
group.
c.
the mean
weight of the treatment group is equal to the mean weight of the control group.
d.
the mean
weight of the treatment group is less than the mean weight of the control
group.
11. At α = 0.01 we conclude that:
a.
the alternative
hypothesis is rejected.
b.
the null hypothesis
is rejected.
c.
the null
hypothesis is not rejected.
d.
cannot
decide.
12. In the above test we assume that:
a.
the
variances of both groups are not equal.
b.
the
variance of treatment group is greater than the variance of control group.
c.
the
variance of control group is greater than the variance of treatment group.
d.
the
variances of both groups are equal.
A researcher wants to determine the
differences among 3 groups: group1 taken drug A, group 2 taken drug B and group
3 taken drug C. The following SPSS output is given to answer the questions (13,14,
15).
|
weight |
|||||
|
Sum of Squares |
df |
Mean Square |
F |
Sig. |
|
|
Between Groups |
6.268 |
.040 |
|||
|
Within Groups |
37.924 |
42 |
.903 |
||
|
Total |
44.192 |
44 |
13. The appropriate test is:
a.
the
analysis of variance.
b.
the
two-sample t-test for independent samples.
c.
the
two-sample t-test for dependent samples.
d.
the
chi-square test.
14. The F-statistic in the table above is
given by:
a. 3.471
b. 6.268
c. 6.050
d. 2.771
15. At α=0.05 we conclude that:
a.
all drugs
are the same.
b.
at least
on drug is different from the others.
c.
all drugs
are different.
d.
drug A is
better than both drug B and C.
16. From this output and at α=0.05 we
conclude that:
|
Kolmogorov-Smirnova |
Shapiro-Wilk |
|||||
|
Statistic |
df |
Sig. |
Statistic |
df |
Sig. |
|
|
weight |
1.128 |
45 |
.04 |
1.965 |
45 |
.03 |
|
a. The population random variable is b. The population random variable is c. The population random variable is not d. The population random variable has no |
The sharing of injecting equipment among
drug users was investigated to determine if there is an association between
gender and needle exchange (regular, occasional, never). Use the following
SPSS output to answer (17, 18, 19)
|
Count |
|||||
|
exchange |
Total |
||||
|
1.00 |
2.00 |
3.00 |
|||
|
gender |
1.00 |
56 |
15 |
20 |
91 |
|
2.00 |
19 |
5 |
16 |
40 |
|
|
Total |
75 |
20 |
36 |
131 |
|
Value |
df |
Asymp. Sig. (2-sided) |
|
|
Pearson Chi-Square |
4.529a |
2 |
.04 |
|
Likelihood Ratio |
4.365 |
2 |
.113 |
|
Linear-by-Linear Association |
3.736 |
1 |
.053 |
|
N of Valid Cases |
131 |
||
17. The appropriate test is:
a.
the
analysis of variance.
b.
the
two-sample t-test for Independent Samples.
c.
the
correlation analysis.
d.
the
chi-square test for independence.
18. The null hypothesis is:
a.
there is
a relationship between gender and needle exchange.
b.
men use
needle exchange more regular than women.
c.
women use
needle exchange more regular than men.
d.
there is
no relationship between gender and needle exchange.
19. At α = 0.05 we conclude that:
a.
there is
evidence of arelationship between gender and
needle exchange.
b.
there is no
evidence of arelationship between gender and
needle exchange.
c.
women use
needle exchange more regular than men.
d.
cannot
reach a decision.
The following SPSS output represent a
nurses’ assessment (X) and physicians’ assessment (Y) of the condition of 10
patients at time of admission to a trauma center. Use the following SPSS
output to answer (20, 21, 22, 23)
|
Table (1) |
||||
|
Model |
R |
R Square |
Adjusted R Square |
Std. Error of the Estimate |
|
1 |
.912a |
.831 |
.810 |
2.765 |
|
a. Predictors: (Constant), nurses |
|
Table (2) |
||||||
|
Model |
Sum of Squares |
df |
Mean Square |
F |
Sig. |
|
|
1 |
Regression |
301.745 |
1 |
301.745 |
39.473 |
.000b |
|
Residual |
61.155 |
8 |
7.644 |
|||
|
Total |
362.900 |
9 |
||||
|
Table (3) |
||||||
|
Model |
Unstandardized Coefficients |
Standardized Coefficients |
t |
Sig. |
||
|
B |
Std. Error |
Beta |
||||
|
1 |
(Constant) |
1.211 |
2.056 |
.589 |
.572 |
|
|
nurses |
1.082 |
.172 |
.912 |
6.283 |
.000 |
|
|
a. Dependent Variable: physicians |
20. In table (1), what is the meaning of (R Square)?
a.
83.1% of
the variation in nurses’ assessment is explained by variation in physicians’
assessment.
b.
83.1% of
the variation in physicians’ assessment is explained by variation in nurses’
assessment.
c.
16.9% of
the variation in physicians’ assessment is explained by variation in nurses’
assessment.
d.
95% of
the variation in physicians’ assessment is explained by variation in nurses’
assessment.
21. The sample regression line is:
a. X = 1.211 + 1.082 Y
b. Y = 2.056 + 0.172 X
c. Y = 1.211 + 1.082 X
d. X
= 1.082 X
22. From table (2) and at α =0.05 we conclude
that:
a. the linear regression model is not
appropriate for this data.
b. the relationship between physicians’
assessment and nurses’ assessment is not linear.
c. cannot reach a decision.
d. the linear regression model is suitable
for this data.
23.
The
predicted physicians’ assessment of a patient with nurses’ assessment 19 is:
a. 21.769
b. 18.394
c. 17.778
d. 19.605
24. The following scatter plot and table (1)
indicates that:
a. there is a strong negative relationship
between physicians’ assessment and nurses’ assessment.
b. there is a strong positive relationship
between physicians’ assessment and nurses’ assessment.
c. there is a weak positive relationship
between physicians’ assessment and nurses’ assessment.
d. there is no relationship between
physicians’ assessment and nurses’ assessment.
