Q. 12.23 In each of Exercises 12.18-12.23... [FREE SOLUTION]

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Chapter 12: Q. 12.23 (page 491)

In each of Exercises 12.18-12.23, we have provided a distribution and the observed frequencies of the values of a variable from a simple random sample of a population. In each case, use the chi-square goodness-of-fit test to decide, at the specified significance level, whether the distribution of the variable differs from the given distribution.
Distribution: $0.5, 0.3, 0.2$
Observed frequencies: $147, 115, 88$
Significance level $= 0.01$

Short Answer

Expert verified

The variable has the given specified distribution.

Step by step solution

Step 1. Given Information

We are given the sample size is,

$n = 9 + 7 + 1 + 12 + 21 = 50$

Level of significance,

$伪 = 0.01$

Step 2. Calculation the goodness of fit .

Performing the hypothesis test,

The test hypotheses,

$H_{0}$ :The variable has the given specified distribution

$H_{a}$ :The variable differ from the given distribution

Calculating the Goodness of fit,

Observed frequencies	Relative frequencies	Expected frequencies	Obs-Exp	$\frac{(O b s - E x p)^{2}}{E x p}$
9	0.2	10	-1	0.1
7	0.1	5	2	0.8
1	0.1	5	-4	3.2
12	0.3	15	-3	0.6
21	0.3	15	-6	2.4

Step 3. Finding the distribution of variables differs or not.

We get,

$鈭� \frac{(O b s - E x p)^{2}}{Exp} = 7.1$

The degrees of freedom of the given data is,

$= k - 1 = 5 - 1 = 4$

Critical value is $蠂_{0.10}^{2}$ for $4 d f$ is $7.779$ ,

The value of the test statistic is, $蠂^{2} = 7.1$

$蠂^{2} = 7.1 < 蠂_{0.10}^{2} = 7.779$ ,

because it does not fall in the rejection region.

So, we do not reject the null hypothesis, $H_{0}$

Do not reject $H_{0}; H_{0}$ Variable has distribution given in the problem.

Therefore, the variable has the given specified distribution.

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Short Answer

Step by step solution

Step 1. Given Information

Step 2. Calculation the goodness of fit .

Step 3. Finding the distribution of variables differs or not.

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