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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: 85, 215, 130, 70

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 reject the null hypothesis,H0

localid="1651872607346" ∴RejectH0,H0Variable has distribution given in the problem .

Therefore, the variable differ from given specified distribution.

Step by step solution

01

Step 1. Given 

The sample size is n= 85+215+130+70=500.

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
850.2100-152.25
2150.4200151.125
1300.3150-202.67
700.150208




localid="1651872624059" ∑(Obs-Exp)2Exp=14.0417

localid="1651872628579" ∑(Obs-Exp)2Exp=14.0417

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

=4-1 = 3

Critical value is localid="1651872634369" X20.05for 3df is 7.815

The value of the test statistic is, localid="1651872639841" X2= 14.0417

localid="1651872644534" X2=14.0417 > localid="1651872649178" X20.05= 7.815, because it fall in the rejection region So, wo reject the null hypothesis,H0

localid="1651872654495" ∴RejectH0,H0Variable has distribution given in the problem .

Therefore, the variable differ from given specified distribution.

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