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., t_{9,.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., t_{9,.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) t_{19,.95} [This is the t quantile with 19 degrees of freedom
and cumulative probability 0.95.]
(B) t_{24,.975}
(C) t_{35,.975}
(D) t_{674,.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) t_{19,.05} [Use
your knowledge of the symmetry of the t distribution to determine the
mirror image of t_{19,.95}.]
(F) t_{19,.025} [This is the mirror image of t_{19,.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 t_{statistic
}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 t_{statistic
}of 2.65 with 8 df is bracketed between t_{8,.975}
(2.31) and t_{8,.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 t_{8} distribution curve (see exercise 3 for
instruction), showing t_{8,.975}
and t_{8,.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(T_{9} >
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 µ?
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 µ 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?
P-value from t_{stat}. A one-sample t
statistic for
H_{0}: µ = 0 based on n = 16 is t_{stat} = 2.44.
(A) How many degrees of freedom are associated with this test statistic?
(B) Provide the t quantiles from the t table that bracket the t_{stat}.
(C) What are the right-tail probabilities for the bracketing t quantiles?
(D) What is the one-sided P-value for this problem?
(E) What is the
two-sided P-value?
Critical value. You take a SRS of n = 21 from a Normal population to test H_{0}: µ = 0 versus H_{1}: µ > 0. What values of the t_{stat} will give a P-value that is less than or equal to 0.01?
Beware
a = 0.05. Two trials looked at red wine consumption in lowering overall cholesterol levels in hypercholesterolemic men. These fictitious studies were done under identical conditions. In each trial, men consumed eight ounces of red wine per day.
(a) In trial A, 25 subjects lowered their cholesterol by an average of 5.percent. The standard deviation of the change was 11.9 percent. In testing
H_{0}: μ = 0, t_{stat} = 2.10, df = 24, and
P = 0.0464. Is this study significant at
alpha = 0.05?
(b) In trial B, 25 different subjects lowered their cholesterol by 5% with standard deviation 12.2 percent. The tstat = 2.05, df = 24, and P = 0.0514. Is this result significant at
alpha = 0.05?
(c) Is it reasonable to come to a different conclusion for trial A and trial B?
Red wine.
Test the data in exercise 9 for a significant change in polyphenols
levels. (Note: No mean change implies µ = 0.) Conduct a flexible
significance test with a two-sided alternative. Show all testing steps
(except for step B).
Menstrual cycle length. Menstrual cycles length (days) in a random sample of 9 women are {31, 28, 26, 24, 29, 33, 25, 26, 28}. Assume the population is Normal, Test whether the mean length of the menstrual cycle in this population is a lunar month. (A lunar month is 29.5 days.) Show all hypothesis testing steps, including statements of the null and alternative hypothesis.
Behavioral problems in stressed adolescents. Because there is evidence that stress in a child's life may lead to behavioral problems later in life, it might be expected that a sample of children who have been subjected to an unusual amount of stress would show an unusually high level of behavioral problems. On the other hand, children with high stress levels could feel that they have had enough going on in their lives without complicating matters further, so they might actually show an unusually low number of behavioral problems. To investigate this question, we selected at a SRS of 5 children, each of whom is under a high level of social stress. We examine these children and assign a score of 50 represents an average amount of behavioral problems. Data are {48, 62, 53, 51, 58 }. Test whether these scores are significantly different from the established norm of 50. Show all hypothesis testing steps.
Cholesterol in Asian immigrants (Rosner, 2000, p. 223). We want to compare fasting serum cholesterol levels of recent Asian immigrants to that of the overall U. S. population. Assume cholesterol levels in 20- to 39-years old women in the United States is Normal with µ = 190 mg/dl. Blood tests are preformed on 100 female Asian immigrants in this age range. The mean cholesterol level in this sample is 181.52 mg/dl (standard deviation = 40 mg/dl). Conduct a two-sided test to determine whether the recent immigrants have lower average cholesterol levels than their native counterparts. Show all hypothesis testing steps.
Lowering elevated heart
rate. A new calcium channel blocking agent is tested in 9 patients with unstable angina. Resting heart rate after 48 hours
of treatment decreased an average of 5 beats per minute
(standard deviation = 2.5). Is this drop
significant. (Use a two-sided alternative.)
SIDS. A
sample of birth weights (in grams)
of 10 infants who had died of Sudden Infant Death Syndrome (SIDS) in a large
metropolitan area was {2998, 3740, 2031, 2804, 2454, 2780, 2203, 3803, 3948,
2144}. The
mean weight of all births in this metropolitan was 3300 grams. Is the mean birth weight of SIDS cases significantly different from that of the rest of the
population? [Same as illustrative example in biostat-text.]
Everley's syndrome. Determination of plasma calcium concentration in the 18 patients between the age of 20- and 44-years with Everley's syndrome gave a mean of 3.2 mmol/l, with standard deviation 1.1. (Everley's syndrome is a rare congenital disorder. High calcium concentration is thought to provide a useful diagnostic sign and clues to the efficacy of treatment.) Published reports show that healthy people have mean plasma calcium concentrations that average 2.5 mmol/l . Is the mean calcium level in the Everley's syndrome patients abnormally high? (Source: http://bmj.bmjjournals.com/collections/statsbk/7.shtml).