Question 1159397
You can use a chi-square test of goodness of fit to solve this. The formula for finding the test statistic (x^2) is: *[illustration Chi_square_goodness_of_fit_formula]. O is the observed value (for example, 269 peanuts), and E is the expected value (ex: 250 peanuts). Though you can manually calculate chi-square, it would be much faster to use a chi-square goodness of fit calculator, which can be found at vassarstats.net under Frequency Data. After running a chi-square test, the test statistic for these data is 10.14, with 3 degrees of freedom. To find if this is statistically significant, you need to compare the test statistic (10.14) to the appropriate critical value for 3 degrees of freedom. You can find the chi-square critical values chart online. The critical value is 7.815. Because 10.14 is bigger than 7.815, your result is statistically significant, and you can reject your null hypothesis (meaning that this nut ratio inside the can does not fit the expected ratio of 5:2:2:1).