Pharmacy exam 2019-10-14 Degrees: Pharmacy, Biotechnology Date: October 14, 2019 Question 1 It has been measured the systolic blood pressure (in mmHg) in two groups of 100 persons of two populations $A$ and $B$. The table below summarize the results. $$ \begin{array}{lrr} \mbox{Systolic blood pressure} & \mbox{Num persons $A$} & \mbox{Num persons $B$} \newline \hline (80, 90] & 4 & 6 \newline (90, 100] & 10 & 18 \newline (100, 110] & 28 & 30 \newline (110, 120] & 24 & 26 \newline (120, 130] & 16 & 10 \newline (130, 140] & 10 & 7 \newline (140, 150] & 6 & 2 \newline (150, 160] & 2 & 1 \newline \hline \end{array} $$ Which of the two systolic blood pressure distributions is less asymmetric? Which one has a higher kurtosis? According to skewness and kurtosis can we assume that populations $A$ and $B$ are normal? In which group is more representative the mean of the systolic blood pressure? Compute the value of the systolic blood pressure such that 30% of persons of the group of population $A$ are above it? Which systolic blood pressure is relatively higher, 132 mmHg in the group of population $A$, or 130 mmHg in the group of population $B$? If we measure the systolic blood pressure of the group of population $A$ with another tensiometer, and the new pressure obtained ($Y$) is related with the first one ($X$) according to the equation $y=0.98x-1.4$, in which distribution, $X$ or $Y$, is more representative the mean? Use the following sums for the computations: Group $A$: $\sum x_in_i=11520$ mmHg, $\sum x_i^2n_i=1351700$ mmHg$^2$, $\sum (x_i-\bar x)^3n_i=155241.6$ mmHg$^3$ and $\sum (x_i-\bar x)^4n_i=16729903.52$ mmHg$^4$. Group $B$: $\sum x_in_i=11000$ mmHg, $\sum x_i^2n_i=1230300$ mmHg$^2$, $\sum (x_i-\bar x)^3n_i=165000$ mmHg$^3$ and $\sum (x_i-\bar x)^4n_i=13632500$ mmHg$^4$. Solution Group $A$: $\bar x=115.2$ mmHg, $s^2=245.96$ mmHg$^2$, $s=15.6831$ mmHg, $g_{1A}=0.4024$ and $g_{2A}=-0.2346$. Group $B$: $\bar x=110$ mmHg, $s^2=203$ mmHg$^2$, $s=14.2478$ mmHg, $g_{1B}=0.5705$ and $g_{2B}=0.3081$. Thus the distribution of the population $A$ group is less asymmetric since $g_{1A}$ is closer to 0 than $g_{1B}$ and the populaton $B$ group has a higher kurtosis since $g_{2B}>g_{2A}$. Both populations can be cosidered normal since $g_1$ and $g_2$ are between -2 and 2. $cv_A=0.1361$ and $cv_B=0.1295$, thus, the mean of group $B$ is a little bit more representative since its coef. of variation is smaller than the one of group $A$. $P_{70}\approx 125$ mmHg. The standard scores are $z_A(132)=1.0712$ and $z_B(130)=1.4037$. Thus, 130 mmHg in group $B$ is relatively higher than 132 mmHg in group $A$. $\bar y=111.496$, $s_y=15.3694$ and $cv_y=0.1378$. Thus the mean of $X$ is more representative than the mean of $Y$ since $cv_x<cv_y$. Question 2 In a symmetric distribution the mean is 15, the first quartile 12 and the maximum value is 25. Draw the box and whiskers plot. Could an hypothetical value of 2 be considered an outlier in this distribution? Solution $Q_1=12$, $Q_2=15$, $Q_3=18$, $IQR=6$, $f_1=3$, $f_2=27$, $Min=5$ and $Max=25$. Yes, because $2<f_1$. Question 3 A pharmaceutical company is trying three different analgesics to determine if there is a relation among the time required for them to take effect. The three analgesics were administered to a sample of 20 patients and the time it took for them to take effect was recorded. The following sums summarize the results, where $X$, $Y$ and $Z$ are the times for the three analgesics. $\sum x_i=668$ min, $\sum y_i=855$ min, $\sum z_i=1466$ min, $\sum x_i^2=25056$ min$^2$, $\sum y_i^2=42161$ min$^2$, $\sum z_i^2=123904$ min$^2$, $\sum x_iy_j=31522$ min$^2$, $\sum y_jz_j=54895$ min$^2$. Is there a linear relation between the times $X$ and $Y$? And between $Y$ and $Z$? How are these linear relationships? According to the regression line, how much will the time $X$ increase for every minute that time $Y$ increases? If we want to predict the time $Y$ using a linear regression model, ¿which of the two times $X$ or $Z$ is the most suitable? Why? Using the chosen linear regression model in the previous question, predict the value of $Y$ for a value of $X$ or $Z$ of 40 minutes. If the correlation coefficient between the times $X$ and $Z$ is $r=-0.69$, compute the regression line of $X$ on $Z$. Solution $\bar x=33.4$ min, $s_x^2=137.24$ min$^2$, $\bar y=42.75$ min, $s_y^2=280.4875$ min$^2$, $\bar z=73.3$ min, $s_z^2=822.31$ min$^2$, $s_{xy}=148.25$ min$^2$ and $s_{yz}=-388.825$ min$^2$. Thus, there is a direct linear relation between $X$ and $Y$ and an inverse linear relation between $Y$ and $Z$. $b_{xy}=0.5285$ min. $r^2_{xy}=0.5709$ and $r^2_{yz}=0.6555$, thus the regression line of $Y$ on $Z$ explains better $Y$ than the regression line of $Y$ on $X$ since $r^2_{yz}>r^2_{xy}$. Regression line of $Y$ on $Z$: $y=77.4095 + -0.4728z$ and $y(40)=58.4957$. $s_{xz}=-231.7967$ and the regression line of $X$ on $Z$ is $x=54.0622 + -0.2819z$. Exam Statistics Biostatistics Previous Pharmacy exam 2019-11-18 Next Pharmacy exam 2018-12-17