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Name: A researcher claims that felons convicted of tax fraud spend less than
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Description: A researcher claims that felons convicted of tax fraud spend less than 12 months in jail, on average. She takes a random sample of 18 such cases from court files and finds that the mean is 10.4 months with a standard deviation of 3.7 months. Test the

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 researcher claims that felons convicted of tax fraud spend less than 12 months in jail, on average. She takes a random sample of 18 such cases from court files and finds that the mean is 10.4 months with a standard deviation of 3.7 months. Test the claim using a level of significance of 10%. You may assume that the amount of time served by these felons is normally distributed.

What evidence justifies the use of this test? Check all that apply:

There are two different samples being compared
np>5 and nq>5

The original population is approximately normal
The population standard deviation is not known
The population standard deviation is known
The sample size is larger than 30
The sample standard deviation is not known

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,

$S a m p l e m e a n (x) = 10.4 S a m p l e s \tan d a r d d e v i a t i o n (s) = 3.7 s a m p l e s i z e (n) = 18 L e v e l o f s i g n i f i c a n c e (α) = 0.10$

The hypothesis to be tested here is-

$H_{0} : μ = 12 H_{1} : μ < 12$

This is a left tailed test

The test statistics used for testing this hypothesis is "One-sample t-test". This is because-

The original population is assumed to be approximately normal.
Population standard deviation is not known.

The required test-statistics is obtained as-

$\begin{array}{rcl} t & = & \frac{x - μ}{\frac{s}{\sqrt{n}}} \\ = & \frac{10.4 - 12}{\frac{3.7}{\sqrt{18}}} \\ = & - 1.8347 \end{array}$

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