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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: 0.2, 0.4, 0.3, 0.1

Observed frequencies: 39, 78, 64, 19

Significance level = 0.05

Short Answer

Expert verified

The test hypotheses,

H0: The variable has the given specified distribution

H1: The variable differ from the given distribution

wo do not reject the null hypothesis,H0

DonotrejectH0,H0Variable has distribution given in the problem .

Therefore, the variable has the given specified distribution.

Step by step solution

01

Step 1. Given 

The sample size is n= 39+78+64+19=200.

Level of significance ,α=0.05

02

Step 2. Calculation the goodness of fit .

Now, we want to perform the hypothesis test

The test hypotheses,

H0: The variable has the given specified distribution

H1: The variable differ from the given distribution

Calculating the Goodness of fit :

Observed
frequencies
Relative
frequencies
Expected
frequencies
Obs - Exp(Obs-Exp)2Exp
390.240-10.025
780.480-20.050
640.36040.267
190.120-10.050




(Obs-Exp)2Exp=0.392

(Obs-Exp)2Exp=0.392

The degrees of freedom of the given data is, k-1

=4-1 = 3

Critical value is X20.05for 3df is 9.815

The value of the test statistic is, X2=0.392

X2=0.392<X20.05= 7.815, because it does not fall in the rejection region So, wo do not reject the null hypothesis,H0

DonotrejectH0,H0Variable has distribution given in the problem .

Therefore, the variable has the given specified distribution.

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