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determine the x2 test statistic the degrees of freedom, the p value, and test the hypothesis at the a

User RowanC
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1 vote

Answer:

The null hypothesis will not be rejected at 5% significance level.

Explanation:

Consider a Chi-square test for goodness of fit.

The hypothesis can be defined as:

H₀: The observed frequencies are same as the expected frequencies.

Hₐ: The observed frequencies are not same as the expected frequencies.

The test statistic is given as follows:


\chi^(2)=\sum\limits^(n)_(i=1)((O_(i)-E_(i))^(2))/(E_(i))

The information provided is:

Observed values:

Half Pint: 36

XXX: 35

Dark Night: 9

TOTAL: 80

The expected proportions are:

Half Pint: 40%

XXX: 40%

Dark Night: 20%

Compute the expected values as follows:

E (Half Pint)


=(40)/(100)*80=32

E (XXX)


=(40)/(100)*80=32

E (Dark night)


=(20)/(100)*80=16

Compute the test statistic as follows:


\chi^(2)=\sum\limits^(n)_(i=1)((O_(i)-E_(i))^(2))/(E_(i))


=((36-32)^(2))/(32)+((35-32)^(2))/(32)+((9-16)^(2))/(16)\\\\=3.844

The test statistic value is, 5.382.

The degrees of freedom of the test is:

n - 1 = 3 - 1 = 2

The significance level is, α = 0.05.

Compute the p-value of the test as follows:

p-value = 0.1463

*Use a Chi-square table.

p-value = 0.1463 > α = 0.05.

So, the null hypothesis will not be rejected at 5% significance level.

User Simon Price
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