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According to Dan Lenard, an independent insurance agent in the Buffalo, N.Y. area, the following is a breakdown of the amount of life insurance purchased by males in the following age groups. He is interested in whether the age of the male and the amount of life insurance purchased are independent events. Conduct a test for independence.

Age of MalesNone<\(200,000\)200,000-\(400,000\)401,001-\(1,000,000\)1,000,001+20-294015400530-3935520201040-4920030153050+403015010

Short Answer

Expert verified

we reject the null hypothesis.

There is evidence to conclude that the age of the male and the amount of life insurance purchased are not independent events.

Step by step solution

01

Given Information

The table with observed values

AgeNone<$200k$200k-$400k$401k-$1,000k$1,000,001+Total20-294015400510030-393552020109040-49200300308050+4030151510110Total135501053555380

The table with expected values

AgeNone<$200k$200k-$400k$401k-$1,000k$1,000,001+Total20-2935.52631613.15789527.6315799.21052614.47368410030-3931.97368411.84210524.8684218.28947413.0263169040-4928.42105310.52631622.1052637.36842111.5789478050+39.0789514.4736830.3947410.1315815.92105110Total135501053555380

02

Hypotheses test

We want to test following hypotheses

H0: The age of the male and the amount of life insurance purchased are independent events

H1: The age of the male and the amount of life insurance purchased are not independent events

Since there are 4age groups and 5salary groups, the number of degrees of freedom is

(4-1)·(5-1)=12

We are using χ2distribution. Our test statistic is given by:

χ2=i=14j=15Obi,j-Exi,j2Exi,j

=125.7399

03

Graph

Using the applet for χ2distribution, we can see that p-value is 0:

Chi-Square Distribution

X~χ(df)2

04

Hypothesis rejection

Taking α=0.05, we can see that p<α. This means that we reject the null hypothesis.

There is evidence to conclude that the age of the male and the amount of life insurance purchased are not independent events.

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