Blood
pressure. A study found a mean systolic blood pressure of 
= 124.6 mm Hg in
35 individuals. The standard deviation s = 10.3 mm Hg.
(A) Calculate the estimated standard the error of the mean.
(B) How many people would you have to study to decrease the standard error
of the mean to 1 mm Hg? [Recall that se = s / 
n.
Rearrange this formula to solve for n. Then plug-in assumptions
for se and s to derive sample size requirement.]
Published
report A study published in the American Journal of Public Health
(Langenberg, 2005) addressed the statistical relation between tall
stature, cardiovascular mortality, and employment grade. Results were
reported in a table with the column heading �Mean Height, cm. (SE).� The
table entry for �Stroke in the Low Employment Grade� was 173.2 (0.2)
based on n = 1243. From this
table, you are supposed to understand that x-bar = 173.2 and the
standard error of the mean = 0.2.
What is the standard deviation of the data in this sample? [Rearrange the
formula for the sem to solve for s. Then plug the values of n
and sem into the formula.]
t
curve.
This exercise is intended to help you become familiar with the characteristics
of t distributions.
(A) Sketch a t curve. To the eye, this curve will look
like a z curve (i.e., have mean 0, points of inflection approximately 1 unit above and below
the mean, and so on). Label the horizontal axis with tick marks that at 1-unit
intervals.
(B) Use the t
table to determine the t quantile with 9 degrees of freedom
and cumulative probability 0.90 (i.e., t9,.90). Place this value
on
the horizontal axis of the curve and shade the region under the curve to its
right. The area in the right tail = 1 - 0.90 = 0.10.
(C) Use the symmetry of the t curve to determine the t quantile that cuts off the bottom 10% of the
curve (i.e., t9,.10). Shade the region to the left of
this point.
(D) What is the combined area of the shaded regions of the
curve you just sketched?
t quantiles. Use your
t table to
determine the following t
quantiles:
(A) t19,.95 [This is the t quantile with 19 degrees of freedom
and cumulative probability 0.95.]
(B) t24,.975
(C) t35,.975
(D) t674,.99 [A t distribution with
this many degrees of
freedom is nearly the same as a z distribution; use the row
in the t table for z.]
(E) t19,.05 [Use
your knowledge of the symmetry of the t distribution to determine the
mirror image of t19,.95.]
(F) t19,.025 [This is the mirror image of t19,.975.]
Approximating
the areas beyond a t quantile. Sometimes you will need to determine
the area under the curve to the right or left of a t quantile
that does not appear in the body of the t table. For example, you may need to determine the area in the tail beyond a tstatistic
of 2.65 with 8
degrees of freedom. Even though this t quantile does not appear in
the table, you can still derive its approximate probability by bracketing it between landmarks
that are listed in the t table. In this case, a tstatistic
of 2.65 with 8 df is bracketed between t8,.975
(2.31) and t8,.99 (2.90). This shows it to have a cumulative probability
that is a little bigger than 0.975 and a little smaller
than 0.99.
(A) Sketch the t8 distribution curve (see exercise 3 for
instruction), showing t8,.975
and t8,.99 on the horizontal axis of the curve.
Wedge " 2.65" between these landmarks.
(B) What is the approximate size of the area under
the curve to the right of 2.65 under this curve?
(C) Use StaTable or
other software to determine area under the curve (exact probabilities) beyond 2.65 on a t distribution
with 8 degrees of freedom.
More t
probabilities.
(A) Sketch the probability (area under the curve) of observing a t quantile with 9 df that greater than
2.82. Include t quantile landmarks on the horizontal axis of the
sketch that bracket the 2.82. What is Pr(T9 >
2.82)?
(B) What is the probability of seeing a t quantile with 9 df
that is less than -2.98?
t critical values for a confidence interval. You have a SRS of n = 28 individuals. What is the value of the t quantile (critical value) would you use to calculate a 95% confidence interval for �?
t for confidence. You have a SRS of n = 28 and want to calculate a 90% confidence interval. What t quantile would you use (from the t table) for your calculation?
Red wine (based on Nigdikar et al. 1998; Moore, 2003, pp. 416, 643). Drinking moderate amounts of wine may reduce the risk of coronary artery disease in some individuals. One possible reason for this is that red wine contains polyphenols, and polyphenols help serum cholesterol profiles. In an en experiment involving 9 men, the subjects drank half a bottle of red wine each day for two weeks. Level of polyphenols in blood samples were measured at the beginning and end of the experiment. Percent change in polyphenols levels are {3.5, 8.1, 7.4, 4.0, 0.7, 4.9, 8.4, 7.0, 5.5}. Calculate a 95% confidence interval for the mean percent change in polyphenols if all men drank this amount of red wine.
Calcium in sound teeth. A dental researcher measures the calcium content of sound teeth (% of tooth content that is calcium). A sample of 5 teeth shows the following values {33.4, 36.2, 34.8, 35.2, 35.5}.Provide a 99% confidence interval for the mean percent calcium content of sound teeth. [You may use your calculator to find the mean and standard deviation. Please calculate the confidence interval by hand, showing all work.]
Boy height. A SRS of n = 26 boys between the ages of 13 and 14 reveals a mean height of 63.8 inches with a standard deviation of 3.1 inches. Assume height in the population varies according to a Normal distribution. Calculate a 95% CI for the mean height of all boys in this age range.
Vector control in an African
village. A study of insect vector control in
an African village found that the
mean sprayable surface area of 100 houses
was 249 square feet with standard deviation = 39.82 square feet. (Data
are fictitious but realistic; see Osborn, 1979, p. 6 for full data
set.)
(A) Determine the
95% confidence interval for the mean sprayable surface of houses in the village.
(B) Would it be correct to say that
95% of all the houses in the village have sprayable surfaces between the lower
confidence limit and upper confidence limit? Explain your response.
Respiratory
function in furniture workers.
Forced expiratory
volume (FEV) is a measure of respiratory health in which you forcibly blow through a tube.
The rate of air expelled (liters per second) is
measured as an index
of lung function. FEV
in seven
workers at a
furniture manufacturing plant are {3.94, 1.47, 2.06, 2.36, 3.74, 3.43, 3.78}.
Calculate a 90% CI for the mean FEV for the population of furniture
workers.
COPD (Rosner, 1990, p. 177). Skin-fold
thickness taken at the triceps region
averages 1.35 cm (standard deviation = 0.50 cm) in a sample of 40 healthy male
controls with normal respiratory function. In 32 men with chronic obstructive pulmonary
disease, skin thickness at the triceps region averages 0.92 cm (standard deviation = 0.40 cm).
(A) Calculate 95% confidence intervals for the skin fold thickness in the healthy population.
(B) Calculate 95% confidence intervals for the skin fold thickness in the population of men with
chronic obstructive pulmonary disease.
(C) Plot the above confidence intervals in side-by-side fashion on graph paper.
Compare the intervals. Interpret your results.
Body weight, high school girls.
Body weight expressed as a percentage of
ideal in 9 high school girls expressed are: {114, 100, 104, 94, 114, 105, 103, 105, 96}.
(A) Plot the data as a stemplot using split-stem
values. Are there any major departures from Normality in the data?
(B) Assume these 9 girls represent a SRS from their school.
Calculate a 95% confidence for population mean � of this variable in the
school. Show all work.
(C) What is the margin of error of your estimate? (Numerical value.)
(D) How large a sample would be needed to reduce the margin
of error of the 95% confidence interval down to 3?
(A) The mean time before the appears of symptoms in the treated group
(induction time) was 81.9 days (se = 2.2 days). Solve the
formula for the standard error for sample standard deviation s. (se =
s / n.)
Use this to determine the standard deviation in this group.
(B) The mean induction time in the control group was 102.8 days (se =
3.8 days). What was the standard deviation of the data in this group?