Question

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

Answer 1

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.