Answer:
The null and alternative hypotheses are:
[tex]H_{0}:[/tex] The bag of colored candies follows the distribution stated
[tex]H_{a}:[/tex] The bag of colored candies does not follow the distribution stated
Under the null hypothesis, the test statistic is:
[tex]\chi^{2} =\sum \frac{(O-E)^{2}}{O}[/tex]
Where:
[tex]O[/tex] is the observed frequencies
[tex]E[/tex] is the expected frequencies
The calculation are done in the attachment
[tex]\therefore \chi^{2} =14.490[/tex]
Now, we have to find the chi-square critical value at 0.05 significance level for df = n - 1 = 6 - 1 = 5. Using the chi-square table, we have:
[tex]\chi^{2}_{critical} = 11.070[/tex]
Since, the chi-square test statistic is greater than chi-square critical value. Therefore, we reject the null hypothesis and conclude that the bag of colored candies does not follow the distribution stated.
