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An urn contains a very large number of marbles of four different colors: red, orange, yellow, and green. A sample of 12 marbles drawn at random from the urn revealed 2 red, 5 orange, 4 yellow, and 1 green marble. Test the hypothesis that the urn contains equal proportions of the differently colored marbles at 0.05 level of significance. We cannot reject the hypothesis at the 0.05 significance level.

2 Answers

5 votes

Final answer:

The chi-square goodness-of-fit test is used to test if an urn contains equal proportions of differently colored marbles. The expected frequencies are calculated assuming equal proportions, and the chi-square statistic is compared to the critical value at the 0.05 significance level. The hypothesis is accepted or rejected based on this comparison.

Step-by-step explanation:

To test the hypothesis that an urn contains equal proportions of red, orange, yellow, and green marbles, we use the chi-square goodness-of-fit test. Given the sample of 12 marbles which resulted in 2 red, 5 orange, 4 yellow, and 1 green marble, we can calculate the expected frequencies if the marbles were in equal proportions. Since there are four colors, we would expect to draw 3 of each color if the proportions were equal (12 marbles ÷ 4 colors = 3 marbles of each color expected).

The chi-square test statistic is calculated using the formula: X^2 = Σ[(O - E)^2 / E], where O is the observed frequency and E is the expected frequency. Plugging the values into the formula, we get X^2 = ((2-3)^2 / 3) + ((5-3)^2 / 3) + ((4-3)^2 / 3) + ((1-3)^2 / 3).

To determine whether the observed sample is significantly different from the expected frequencies, we compare the calculated chi-square value to the critical value from the chi-square distribution table at a 0.05 level of significance with 3 degrees of freedom (since there are four categories - 1). If the calculated value is less than the critical value, we cannot reject the null hypothesis of equal proportions. If it is higher, we reject the null hypothesis.

Without the actual calculations here, the conclusion whether the hypothesis can be rejected or not would depend on the comparison between the calculated chi-square value and the critical value from the chi-square distribution table.

User Witzar
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Answer:

We can reject the hypothesis at the 0.05 significance level.

Step-by-step explanation:

given that an urn contains a very large number of marbles of four different colors: red, orange, yellow, and green. A sample of 12 marbles drawn at random from the urn revealed 2 red, 5 orange, 4 yellow, and 1 green marble.

H0: All colours are equally likely

Ha: atleast two colours are not equally likely

(Two tailed chi square test at 5% significance level)

color Red Orange Yellow Green total

Observed O 2 5 4 1 12

Expected E 3 3 3 3 12

(O-E)^2/E 0.3333 5.3333 0.3333 5.3333 11.3333

chi square = 11.3333

df =3

p value = 0.010055

Since p < 0.05 we reject H0

We can reject the hypothesis at the 0.05 significance level.

User Chrickso
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