Answered: ars of age. x1: Rate of hay fever per…

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

A random sample of n1=16 communities in western Kansas gave the following information for people under 25 years of age.

x1: Rate of hay fever per 1000 population for people under 25

98 107 111 116 151 134 126 96 122 111 121 133 142 123 95 91

A random sample of n2=14 regions in western Kansas gave the following information for people over 50 years old.

x2: Rate of hay fever per 1000 population for people over 50

81 112 103 102 78 76 112 113 104 119 92 106 97 106

Assume that the hay fever rate in each age group has an approximately normal distribution. Do the data indicate that the age group over 50 has a lower rate of hay fever? Use a = 0.05 State the null and alternate hypotheses.

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

Given:

$n_{1} = 16$

x₁: Rate of hay fever per 1000 population for people under 25

98 107 111 116 151 134 126 96 122 111 121 133 142 123 95 91

Now, mean and standard deviation

$x_{1} = \frac{\sum x_{i}}{n_{i}} = \frac{1877}{16} = 117.3125$

$\begin{array}{rcl} s_{1} & = & \sqrt{\frac{1}{(n_{1} - 1)} \sum {(x_{i} - x)}^{2}} \\ = & \sqrt{\frac{1}{16 - 1} {(98 - 117.3125)}^{2} + . . . . . . . + {(91 - 117.3125)}^{2}} \\ = & 17.5451 \end{array}$

Now,

$n_{2} = 14$

x₂: Rate of hay fever per 1000 population for people over 50

81 112 103 102 78 76 112 113 104 119 92 106 97 106

$\begin{array}{rcl} x_{2} & = & \frac{\sum x_{i}}{n} = \frac{1401}{14} = 100.0714 \\ s_{2} & = & \sqrt{\frac{1}{(n_{2} - 1)} \sum {(x_{i} - x)}^{2}} \\ = & \sqrt{\frac{1}{14 - 1} {(81 - 100.0714)}^{2} + . . . . . . . . + {(106 - 100.0714)}^{2}} \\ = & 13.6239 \end{array}$

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