A manufacturer of colored candies states that 13% of the candies in a bag should be brown, 14% yellow, 13% red, 24% blue, 20% orange, and 16% green. A student randomly selected a bag of colored candies He counted the number of candies of each color and obtained the results shown in the table. Test whether the bag of colored candies follows the distribution stated above at the a = 0.05 level of significance Color Brown Yellow Red Blue Orange Green Frequency 59 63 53 59 79 63 Claimed Proportion 0.13 0.14 0.13 0.24 0.20 0.16 What is the test statistic ?

Step 1 ：The problem is asking us to test whether the bag of colored candies follows the distribution stated by the manufacturer at the a = 0.05 level of significance. We are given the observed frequencies of each color and the claimed proportions.

Step 2 ：We can use the chi-square goodness-of-fit test to solve this problem. The test statistic for a chi-square goodness-of-fit test is calculated using the formula: \[X^2 = \sum \frac{(O_i - E_i)^2}{E_i}\] where \(O_i\) is the observed frequency, \(E_i\) is the expected frequency. The expected frequency can be calculated by multiplying the total number of observations by the claimed proportion for each category.

Step 3 ：The observed frequencies for each color are: Brown: 59, Yellow: 63, Red: 53, Blue: 59, Orange: 79, Green: 63.

Step 4 ：The claimed proportions for each color are: Brown: 0.13, Yellow: 0.14, Red: 0.13, Blue: 0.24, Orange: 0.20, Green: 0.16.

Step 5 ：The total number of candies in the bag is 376.

Step 6 ：We can now calculate the expected frequencies for each color: Brown: \(376 \times 0.13 = 48.88\), Yellow: \(376 \times 0.14 = 52.64\), Red: \(376 \times 0.13 = 48.88\), Blue: \(376 \times 0.24 = 90.24\), Orange: \(376 \times 0.20 = 75.2\), Green: \(376 \times 0.16 = 60.16\).

Step 7 ：Next, we substitute the observed and expected frequencies into the formula for the test statistic and calculate the test statistic for each color. Then, we sum them up to get the final test statistic.

Step 8 ：The final test statistic for the chi-square goodness-of-fit test is \(\boxed{15.622}\).