Question

Perform a goodness-of-fits test to determine if a uniform distribution between 0 and 10 is a good fit to the 150 data observations summarized in the table below. Use α=0.05.

Range of data values	0.0 < X < 2.0	2.0 < X < 4.0	4.0 < X < 6.0	6.0 < X < 8.0	8.0 < X < 10.0
Number of observations in this range	28	35	29	31	27

Answer 1

Solution:-

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

Null hypothesis: The uniform distribution between 0 and 10 is a good fit to the 150 data observations.

Alternative hypothesis: The uniform distribution between 0 and 10 is not good fit to the 150 data observations.

Formulate an analysis plan. For this analysis, the significance level is 0.05. 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
(E_i) = n * p_i

X² = 1.333

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, E_i is the expected frequency count for level i, O_i is the observed frequency count for level i, and X² is the chi-square test statistic.

The P-value is the probability that a chi-square statistic having 4 degrees of freedom is more extreme than 1.3333.

We use the Chi-Square Distribution Calculator to find P(X² > 1.3333) = 0.856

Interpret results. Since the P-value (0.856) is greater than the significance level (0.05), we cannot accept the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that the uniform distribution between 0 and 10 is a good fit to the 150 data observations.