Answered: In order to investigate whether a…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

In order to investigate whether a particular medicine has the side effect of raising blood pressure, the blood pressure was measured before and after treatment on 25 patients.

The following measure was obtained, x_iis for blood pressure for patient nr i before treatment and there y_ifor blood pressure after treatment,

i = 1, 2, ..., 25, and z_i = y_i − x_i .

We got ¯x = 149.1 ¯y = 150.8, ¯z = 1.7, s_x = 8.2, s_y = 9.4, s_z = 1.6,

where s_x and s_{y are} the standard deviations for the x- and y-values, and then s_z is the standard deviation for the z-values. How much higher is the average blood pressure after treatment compared to the blood pressure before treatment? Answer the question by calculating a 95% confidence interval under the normal distribution assumption.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Expert Solution

Step 1

The confidence interval is the interval in which the population parameter is likely to fall. For small samples, the confidence interval for population parameter at $α %$ level of significance and (n-1) degree of freedom is given by:

$E (t) \pm t_{α %} \times S . E (t)$

Here, t is the estimator of a population parameter, $t_{α %}$ is tabulated t value at $α %$ level of significance and (n-1) degree of freedom and S.E(t) is the standard error of the estimator of the population parameter.

Let $x_{i}; (i = 1, 2, . . ., n_{1})$ and $y_{j}; (j = 1, 2, . . ., n_{2})$ be two samples with mean $\bar{x}$ and $\bar{y}$ and standard deviation $s_{x}$ and $s_{y}$ , respectively. The confidence interval for the difference of means is calculated as follows: