The question asks if the distribution of candy colors is significantly different from what's expected. A Chi-square test for goodness of fit is used, comparing observed counts to expected counts. A calculated Chi-square statistic is then compared to a critical value to draw a conclusion.
The student's question addresses whether the observed distribution of candy colors significantly deviates from the manufacturer's expected distribution, given the observed counts for each color. To test this, we can use a Chi-square test for goodness of fit because the manufacturer claims an equal distribution among the colors. Since there are four categories (red, blue, green, yellow), the expected frequency for each color in 200 candies is 200/4 = 50. We then compare the observed frequencies with the expected frequencies using the Chi-square statistical test.
First, calculate the Chi-square statistic using the formula: ² = Σ(O-E)²/E, where O represents the observed count and E represents the expected count. So, we have:
(45-50)²/50 for red
(50-50)²/50 for blue
(35-50)²/50 for green
(70-50)²/50 for yellow
After calculating these values and summing them up, we get the Chi-square statistic. Then we compare this statistic to the critical value from the Chi-square distribution table with 3 degrees of freedom (n-1, where n is the number of categories) at the 5% significance level (0.05).
If the calculated Chi-square statistic is greater than the critical value, we reject the null hypothesis and conclude that the observed distribution significantly deviates from the expected. Otherwise, we do not have sufficient evidence to reject the null hypothesis, and we conclude that there is no significant deviation.