Question

A distribution and the observed frequencies of the values of a variable from a simple random sample of the population are provided. Use the​ chi-squaregoodness-of-fit test to​ decide, at a significance​ level of 0.10, whether the distribution of the variable differs from the given distribution. (You need to add the expected column based on the distribution)

a.       Specify which Chi-Square test you are doing (Goodness-of-Fit or Test for Independence):

b.       Formulate the null (H0) and alternate (HA) hypotheses:

c.       Specify the level of significance (alpha, a):

d.       What is the value of the test statistic?

e.       What is the critical value?

f.        Do you reject H0 or fail to reject H0?

g.       To validate your results, we’ll also check our p-value. What is the p-value?

h.       Based on your p-value, do you reject or do not reject?

i.         State your summary statement of the conclusion in non-technical terms.

frequency	distribution
15	0.3
9	0.2
5	0.2
18	0.2
3	0.1

Answer 1

Solution:-

a) We are using chi square test of Goodness-of-Fit.

b)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: The distribution of the variable does not differs from the given distribution.

Alternative hypothesis: The distribution of the variable differs from the given distribution.

c) Formulate an analysis plan. For this analysis, the significance level is 0.10. Using sample data, we will conduct a chi-square goodness of fit test of the null hypothesis.

Analyze sample data. Applying the chi-square goodness of fit test to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.

DF = k - 1 = 5 - 1
D.F = 4
(Ei) = n * pi

d)

X2 = 9.8

e)

X2Critical = 7.78

Rejection region is X2 > 7.78.

where DF is the degrees of freedom, k is the number of levels of the categorical variable, n is the number of observations in the sample, Ei is the expected frequency count for level i, Oi is the observed frequency count for level i, and X2 is the chi-square test statistic.

Interpret results. Since the X2 -value (9.8) lies in the rejection region, hence we cannot accept the null hypothesis.

f) Reject null hypothesis.

g)

The P-value is the probability that a chi-square statistic having 4 degrees of freedom is more extreme than 9.8.

We use the Chi-Square Distribution Calculator to find P(X2 > 9.8) = 0.044.

Interpret results. Since the P-value (0.044) is less than the significance level (0.05), we cannot accept the null hypothesis.

h) Reject null hypothesis.

i) From the above test we have sufficient evidence in the favor of the claim that the distribution of the variable differs from the given distribution.