Question

12. suppose an experiment was done to gather data on the number of unpopped kernels of microwave popcorn. half the students in your class took home a bag, put it in the freezer overnight and popped it the next day. the other half took the bag, left it on their kitchen counter and popped it the next day. each group counted the number of unpopped kernels. the class constructs a 95% confidence interval for p(freezer)-p(counter) and finds it to be (-117,-32). what can be concluded?

Answer 1

The 95% confidence interval indicates that popcorn stored in the freezer has significantly fewer unpopped kernels than popcorn left on the counter, with statistical significance since the interval does not include zero.

Step-by-step explanation:

The question pertains to confidence intervals in statistics, specifically to the comparison of two proportions related to an experiment involving microwave popcorn and variables based on storage conditions: freezer and counter. In this case, a 95% confidence interval for the difference in the proportions of unpopped kernels, designated as p(freezer)-p(counter), was calculated to be (-117, -32). This means that we can be 95% confident that the true difference in the proportions of unpopped kernels between the two groups lies within this interval.

Since the entire interval is negative, it indicates that, on average, bags of popcorn stored in the freezer had fewer unpopped kernels than those left on the counter. Moreover, because the interval does not contain zero, we can conclude that there is a statistically significant difference between the two storage methods. This inference is based on the concept that if zero was included in the interval, the difference could potentially be non-existent or could go in either direction.

Overall, the students can conclude that storing popcorn in the freezer before popping is likely to result in a statistically significant reduction in the number of unpopped kernels compared to leaving the popcorn on the counter.