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statistics for nursing a practical approach

Questions and Answers of Statistics For Nursing A Practical Approach

Show that the asymptotic BP of the ????-trimmed mean is ????.
Verify that the breakdown points of the SD, the MAD and the IQR are 0, 1/2 and 1/4, respectively
Show that(a) if the sequence ????m in (2.93) converges, then the limit is a solution of (2.19)(b) if the sequence in (2.94) converges, then the limit is a solution of (2.67).
Verify numerically that the constant c at the end of Section 2.5 that makes the bisquare scale consistent for the normal is indeed equal to 1.56.
Let ????̂ be a location M-estimator. Show that if the distribution of the xi is symmetric about ????, so is the distribution of ????̂, and that the same happens with trimmed means.
Let [a, b] where a and b depend on the data, be the shortest interval containing at least half of the data.(a) The Shorth (“shortest half”) location estimator is defined as the midpoint????̂ =
Verify (2.37), (2.51) and (2.56)
Show that the M-scale (2.49) with ????(t) = I(|t| > 1) is the hth order statistic of the |xi| with h = n − [n????].
Define the sample ????-quantile of x1,..., xn – with ???? ∈ (1∕n, 1∕1∕n) – as x(k), where k is the smallest integer ≥ n???? and x(i) are the order statistics (1.2). Let????(x) = ????I(x
Show that if ???? = ????′ vanishes identically outside an interval, there is no density verifying (2.14).
If x ∼ N(????, ????2), calculate MD(x), MAD(x) and IQR(x).
Show using (2.46) that L-estimators are shift and scale equivariant [recall that the order statistics of yi = −xi are y(i) = −x(n−i+1)!] and also fulfill C1-C2-C3 of Section 2.4.
Show that if ???? is odd, then the M-estimator ????̂ satisfies conditions C1-C2-C3 at the end of Section 2.4.
Compute the ????-trimmed means with ???? = 0.10 and 0.25 for the data of Example 1.2
Verify (2.30) [hint: use ????′(x)=−x????(x) and integration by parts]. From this, find the values of k which yield variances 1/???? with ???? = 0.90, 0.95 and 0.99 (by using an equation solver,
Show that if x = ????0 + u where the distribution of u is symmetric about 0, then????0 is a solution of (2.22).
Show that if ???? is a solution of (2.19), then ???? + c is a solution of (2.19) with xi + c instead of xi.
For what values of ???? does the the Student distribution have moments of order k?
Verify (2.11) using (2.27).
For the data of Example 1.2, compute the mean and median, the 25% trimmed mean and the M-estimator with previous dispersion and Huber’s ???? with k = 1.37. Use the latter to derive a 90% confidence
Show that in a sample of size n from a contaminated distribution (2.6), the number of observations from H is random, with binomial distribution Bi(n, ????).
Show that the median of the exponential distribution is ???? log 2, and hence Med(x)∕ log 2 is a consistent estimate of ????.
The interquartile range (IQR) is defined as the difference between the third and the first quartiles.(a) Calculate the IQR of the N(????, ????2) distribution.(b) Consider the sample interquartile
Show for ti defined in (1.3) that |ti| < (n − 1)∕√n for all possible datasets of size n, and hence for all datasets |ti| < 3 if n ≤ 10.
Consider the situation of the former problem.(a) Show that if n is even, the maximum change in the sample median when x0 ranges from −∞ to +∞ is the distance from Med(x) to the next order
Show that if a value x0 is added to a sample x ={x1, ..., xn}, when x0 ranges from −∞ to +∞, the standard deviation of the enlarged sample ranges between a value smaller than SD(x) and infinity.
The critical F2,3(0.95) = 9.55. Moving F first becomes significant in 1984.Had Dr. Shipman been stopped in 1984, 15 years of patient murders would have been prevented, representing 178 patients.
It appears that the heart exhibits greater efficiency when the HR is in the reference range of about 70-80 beats/min, with inferior efficiency less than 65 beats/min and efficiency diminishing
The upper critical value is surpassed on observation 8.The test indicates that H1 should be accepted: The population death rate for intubated patients is 50%. Table 25.8 Completion of Table 25.5
(1) H0" population death rate - 0.052. Hi" population death rate= 0.50. (2) Any difference found will not be used in clinical decisions but will tell the investigator that a different baseline death
From Table III, the critical X 2 value for c~ - 0.05 for 1 df is 3.84. The calculated 0.63 is much less than 3.84. Conclusion: No difference between the survival patterns has been shown. (The actual
130 +/~1xl + f12X2 + f13X3 -+- f14X4 + f15X5 9 (1) Yes. (2) Not very. (3) Drop DRE and age. Reduced model: (1 to 3) The model is still a significant predictor and is still not a very useful one. (4)
Model." y =/~0 + fllXl -]- f12X2. Data: Enter data for the triple-hop distance using the operated leg into the y position, enter data using the nonoperated leg into the Xl position, and enter the
Model: SBP - b0 + bl (SVmR) + bz(SVmR) 2. Data: The square of SVmR would be calculated. The values for SBP, SVmR, and SVmR 2 would be entered. Interpretation: The final predictive equation is SBP =
r 2 = 0.5329. The two-tailed 95% critical t for 24 df is 2.064. The calculated t is ~/24 x 0.5329/0.4671 = 5.2327, which is much larger than the critical t. The population correlation coefficient is
Se = 14.00, and Sb = 18.04. The two-tailed 95% critical t for 24 df is 2.064. Calculated t to test the slope is 5.23, which is significantly greater than the critical t; the slope of the regression
r = 2.2746/(0.1552 x 20.0757) = 0.7300. A correlation coefficient of 0.73 is rather high; there is clearly an association between the variables, rs =0.7050, which is close to r.
bl = 94.3817, and b0 = 8.9388. The slope-mean form is y - 25.6538 =94.3817(x - 0.1771), and the slope-intercept form is y = 8.9388 + 94.3817x.If SVmR = 0.3, the change in SBP is predicted to be
One possible MOE might be a 10-year postoperative weighted sum of measures of visual correction, v; glare, g; night vision, n [patient rating from no interference (0) to nonfunctional vision at night
Pd = 0.00767. ns = 4172. nd= Pd • ns = 32. Pm = 25/4172 -- 0.00595.The additional proportion detected by screening is Pd --Pm -- 0.00767 -0.00599 -- 0.00168. NNT = 1/0.00168 = 595 entering inmates
The dependent variable y is the postoperative hop distance. There are two independent variables, Xl and x2, which represent hop time and distance on the uninjured leg. The geometric picture would be
Although we could use the means of the two variables for the b and d constants, the plot is drawn such that we can see the intercept and need to use onlyd. By drawing a line along a ruler with the
The TGF curve appears to be a straight-line fit. The PDGF curve appears to be well fit by a section of a parabola, opening upward. The figures are repeated here with those respective fits. TGF 60,000
The predicted required sample size will be the same as if the r were positive, because of the symmetry property. In Table 22.1, r = 0.23 corresponds to n -- 74.
Jr is central; therefore, the surgeon uses Eq. (22.13). p = 0.7, w = 0.1, and, for 95% confidence, he retains the 1.96 given in Eq. (22.13). Substitution yields 1.962p(1 -p) 3.8416 x 0.7(1 - 0.7) _
The one-sided test uses Zl-~ = 1.645. Pm = (0.23 + 0.13)/2 - 0.18, which is not near 0 or 1; therefore, the binomial form, Eq. (22.10), is used.Substitution yields 0.87]/0 n1n2=1.6452 x 0.18 x 0.82 +
Jr = 0.18. Because n is not near 0 or 1, the binomial form, Eq. (22.8), will be used. A one-sided test is used (the training will not increase the number of enhancements); therefore, Zl-oe = 1.645.
By substituting in Eq. (22.1), we find(gl_od2)2o "2 1.962 x 242 n -- = = 22.1. d 2 102 A sample of at least 23 patients is needed; however, per the explanation in Section 22.2, a few more patients
Either trauma type could evoke a greater heart rate; thus, a two-tailed test is used. z1-,~/2 = 1.96, and Zl-~ = 0.84. From Eq. (22.2), nl = n2 = (Zl-cd2 -+- Zl-/~ )2 (6.132 + 6.342)/62 -- 16.9.The
The posttrauma heart rate can be either greater than or less than the healthy rate; thus, a two-sided test is used. z1-~/2 = 1.96, and Z l-~ = 0.84. From Eq. (22.2),)2 2/d2 (1.96 + 0.84)2(9.1)2/62
Follow the steps in equivalence testing. (1) A = 2. (2) HOL: 8 = --2;Hog: 8 = 2; and Hi: -2 < 8 < 2. (3) c~ = 0.05. (4) Row 6 of Table 21.1 indicates Table II to find the critical value. From Table
Mean nonoperated time is 2.54 seconds, so we take A = 0.25 second. The mean of differences between leg times is d = 0.1600, and the standard deviations of differences is s = 0.2868. From the second
From Eq. (19.4) and then Eq. (19.5), calculateFrom Table III, the critical X~af - 7.81. The calculated M- 0.1587 is much smaller, so we do not have evidence to reject H0; variability of parasite
H0" aL 2 -- a 2, and Hi" a 2 ~ a 2. From Table V, the critical F-value for ot - 0.05 with 7,7 df is 3.79. By substituting in Eq. (19.2), we find F =S2L/S 2 -- 1.122/0.882 -- 1.620. Because 1.62 <
H0: a 2 = 0 "2, and Hi" a 2 > 0 "2. From Table III, the critical X 2 value for ot - 0.05 with 15 dfis 25.00. Substituting in Eq. (19.1), we find X2 - df x s 2 15 x 1.2302~ 2 -- 0.52 = 90.77.Because
From the formulas in Table 17.4, calculate SST- (n- 1)s 2 - 19 x 4799.358 = 91,187.805; SSM = 5[(291.4- 314.9) 2 + (323.2- 314.9) 2 +(274.8 - 314.9) 2 + (371.2 - 314.9) 2 = 27,234.20; and SSE - SST -
(Unequal Variances) By substituting in Eq. (17.10) and then Eq. (17.8), we findFrom Table II, the critical value of t for a two-tailed 5% a for 33 df is 4-2.03. The value -0.874 is not outside the
(Equal Variances) From Table 17.3, the sample sizes are small, only a's are estimated, and the standard deviations are not significantly different(see Exercise 19.2); therefore, we choose a
The critical value of 9 df was 2.26, and the t statistic was 4.06. Moving values to the left 4.06 units to standardize the curve resets the critical value at -1.80./~ is the area left of-1.80, or, by
Substitution in Eq. (17.2) (with renamed elements) yields t = (d - O)/sa =1.58/(1.23/~) = 4.06. From Table II, the critical value for a 5% twotailed t with 9 dfis 2.26. Because 4.06 is much larger,
Follow the answer to Exercise 16.2 until the rank sums T = 43.5 or 92.5 are found. Calculate/z = nl(nl + n2 + 1)/2 = 8 x 17/2 - 68; o -2 -~nln2(nl + n2 + 1)/12 = 8 x 8 x 17/12 = 90.67; a = 9.52; and
Follow the answer to Exercise 16.1 until the smaller rank sum T = 7 is found. Calculate/z = n(n + 1)/4 = 8 x 9/4 - 18; o -2 -- (2n + 1)lz/6 -17 x 18/6 = 51; and then z = (T- tz)/a = (7- 18)/v/~ =
The ranks for each patient according to PSA levels, with rank sums at the bottom, are as follows:From Table III, the critical chi-square for k- 1 = 2 df for c~ = 0.05 is 5.99. Because 1.50 2 Rank
Frequency plots per group are as follows:From Table III, the 5 % critical value for 2 dfis 5.99, which is much larger than H; ESR is not different for the three treatments. From a statistical
From Table VIII, the probability that a signed rank sum is 7, given n = 8, is 0.149. This p-value is too large to infer a difference; the patient's median temperature is not shown to be different
The data differences (98.6 - each datum) are 0.5, 2.8, 1.1, 1.4, 0.9, -0.7, -0.6, and 0.5. The unsigned ranks are 1.5, 8, 6, 7, 5, 4, 3, and 1.5; the signed ranks are 1.5, 8, 6, 7, 5, -4, -3, and
The data tally appears in Table 15.27. X2df -- (Ib - cl - 1)2/(b +c) -(4 - 1)2/8 = 1.125. From Table III, the critical value of X 2 for c~ - 0.05 with 1 df is 3.84. Because 1.125 < 3.84, you conclude
We judge that introducing better testing can only reduce the rate; therefore, the test is one-sided. From Table I, a critical value of z at a - 0.05 will be -1.645. We note that )~ = nn = 5, so we
If Zrb is the true population proportion of boys presenting, H0: Zrb = zr and Hi: Zrb r n. Then, Pb = 0.68. Using the normal approximation, a = 0.0707. From Eq. (15.8), z = (IPb -- nl)/a --
n = 16, no = 11, and zr = 0.50. From Table VI, the probability of 11/16 occurring by chance alone is 0.105, which is too large to infer evidence of a nonrandom phenomenon.
L - 1.5506, SEL - 0.7327, and x 2 = (1.5506/0.7327) 2 - 4.4787. From Table III, the p-value for this result is between 0.025 and 0.05, a significant result. The actual p-value is 0.034, which is
Sensitivity - 0.981, the chance of detecting appendicitis using magnetic resonance imaging. Specificity = 0.979, the chance of correctly not diagnosing appendicitis. Accuracy - 0.980, the overall
ell = 19 x 100/195 - 9.7436, and so on. X a - (8 - 9.7436)2/9.7436 +... =4.92. df - 2. From Table III, 4.92 for 2 dflies between 0.05 and 0.10. A computer evaluates the p as 0.085. Because p > 0.05,
p - 0.041 for the FET, indicating a significant difference between the success of the two treatments with a risk for error less than 5 %; MB is better than CS. X a - 5.132, yielding p - 0.023, also
1 + r = 1.267, 1 - r = 0.733, and the exponential term is 0.5718. Substitution in Eq. (14.13) yields P[-0.0123 < p < 0.5076] = 0.95. The confidence interval crosses zero, indicating that the
p is not near 0 or 1, so cr = v/p(1 -p)/n = ~/0.261 x 0.739/226 = 0.0292.Substitute in Eq. (14.9) to obtain P[p - 1.9&r - 1/2n < Jr < p + 1.9&r +1/2n] = P[0.261-1.96 • < Jr < 0.261+ 1.96 • 1/452]
s 2 = 36,000,000 and df = 18. From Table 14.3, the intersection of the 0.025 column with 18 df is 31.53 for the right tail and 8.23 for the left tail.Substitution in Eq. (14.8) yields P[20,551,855 <
The surgeon calculates Sm - s/~/~ = 4.57/~2-6 = 1.02. From Table II, t for 95% "Two-tailed 1 - ~ (except both tails)" for 19 df is 2.093. The surgeon substitutes these values in Eq. (14.7) to obtain
To substitute in Eq. (14.3), the z-values must be found from Table 14.1 (or Table I). Under the column for "Two-tailed 1 -or," 0.900 (90%) yields z =1.645 and 0.990 (99%) yields 2.576. Substituting m
zr - 0.01 and n - 70; therefore, the expected number of AIDS cases is k = 0.7. no = 5 cases occurred. In Table 13.6, the intersection of the column )~ - 0.7 and no - 5 shows the probability to be
In Table 13.5, which displays the n = 6 block of Table VI, the probability 0.114 lies at the intersection of the column zr = 0.10 and the row no = 2.The probability that there would be two or more
For samples so small, numerator and denominator dfs both 3, the F ratio as shown in Table 13.4 would have to be larger than 9.28. For these types of assay, F - 0.0586/0.0120 - 4.88. As 4.88 < 9.28,
X e - 17 x 576.48/259.85 = 37.71. From the row for 17 df in Table 13.3 we see that a X 2 value of 37.71 lies between 35.72, associated with an area under the curve in the right tail of 0.5%, and
For 7 df, a two-tailed u - 0.05 is obtained by m + 2.365 • s, or the interval from 235.8 to 669.7.
z = (126- 120)/5 = 1.20. Your patient is 1.2cr above the mean. From Table 13.1, the area in the right tail is 0.115. Therefore, 11.5% of the population have greater DBP.
Find a medical article in which meta-analysis is used and evaluate it using the guidelines of Section 10,6.
From Table 10.1, what test for age bias for DB14 data would you choose?For sex bias?
Might sex and/or age differences in the independent variables have biased the outcomes for (a) DB2? (b) DB147 What can be done to rule out such bias?
Is the study represented by DB6 an RCT? Why or why not?
From the discussion in Section 10.3, what sort of study gave rise to (a)DB2? (b) DB14? What are the independent and dependent variables in each?
Choose a medical article from your field. Evaluate it using the guidelines given in Section 10.1.
Give another example each of a medical phenomenon that could be detected using cross-correlation without lag, cross-correlation with lag, and (lagged) autocorrelation.
Give one example each of a medical phenomenon that could be detected using cross-correlation without lag, cross-correlation with lag, and (lagged)autocorrelation.
Sketch a rough survival graph from the life table survival results of Exercise 9.9. Table 9.3 Survival Data of 370 Women in Rochester, Minnesota, Having Adult-Onset Diabetes Mellitus Who Were Older
Data 35,66 on the survival of women with untreated breast cancer give rise to a life table with format and essential data as in Table 9.4. Complete the table. What is the probability that a woman
Sketch a rough survival graph from the life table survival results in Exercise 9.7.
Table 9.3 gives the basic life table data 42 for the survival of 370 women with diabetes mellitus. Complete the table. What is the estimate of probability that a woman with diabetes survives more
Suppose you are working with James Lind in 1747, attempting to isolate the cause of scurvy. What: After a long period at sea, a number of crew members are suffering from scurvy, characterized by
A 2003 article 15 addresses the occurrence of acute gastroenteritis caused by Norwalk-like viruses among crew members on the U.S. Navy vessels Peleliu (2800 crew members) and Constellation (4500 crew
In a Canadian study including the effect of caffeine consumption on fecundability, 1~ 2355 women drank caffeine-containing drinks and 89 did not. Of those who consumed caffeine, 236 conceived within

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