M & M Project - Speech or Presentation Example

Objective: "You used the wrong numbers. You should use the following numbers: x1 = 990, n1 = 7356, x2 = 999, n2 = 7356."  Above is the Revision comments for the below orderYou can use StatCrunch or the TI to help with this test. Needed information for both tools include: x1 = number of red n1 = total number of candies x2 = number of brown n2 = total number of candies For the TI, you will want 2-PropZTest. Then select the appropriate alternative (not equal), and Calculate then enter. The output will have the test statistic (z), p-value (p), sample p values, weighted p (), then repeat of sample sizes.

"You used the wrong numbers. You should use the following numbers: x1 = 990, n1 = 7356, x2 = 999, n2 = 7356."  Above is the Revision comments for the below orderYou can use StatCrunch or the TI to help with this test. Needed information for both tools include: x1 = number of red n1 = total number of candies x2 = number of brown n2 = total number of candies For the TI, you will want 2-PropZTest. Then select the appropriate alternative (not equal), and Calculate then enter. The output will have the test statistic (z), p-value (p), sample p values, weighted p (), then repeat of sample sizes.

To compare the proportion of red and brown candies in the bag. Let x1 be the no. of red candies in the bag out of n1 candies and x2 be the no. of brown candies out of n2 candies. Here x1 is 990 and x2 is 999. For combined proportion, n=7356 candies. Now the combined proportion is p=(x1+x2)/(n1+n2)=(990+999)/7356=1809/7356=0.2704; q=1-p=1-0.2704=0.7296.Now since the samples are not independent we can use combined proportion in the computation of SE of the estimateNull Hypothesis H0: There is no significant difference between the proportion of red candies and proportion of brown candies. P1=P2.Alternative Hypothesis H1: There is a significant difference between the proportion of red candies and proportion of brown candies. P1#P2.Level of significance: 5% level or α=0.05.Test statistic:Z0=(p1-p2)-(P1-P2)/SE(p1-p2) which follows standard normal distribution under null hypothesis.

=(p1-p2)/sqrt(pq/n) since P1=P2 and p=(x1+x2)/n; q=1-p; n=n1+n2.p1=990/7356=0.1346; p2=999/7356=0.1358 p1-p2=0.1346-0.1358=-0.012;sqrt(pq/n)=sqrt(0.2704*0.7296/7356)=0.005179;therefore the test statistic Z0=-0.012/0.005179=-2.317; Conclusion: Since the test statistic is lesser than -1.96, we have sufficient evidence to REJECT null hypothesis H0. We accept the alternative hypothesis and conclude that there is significant difference between the proportion of red candies and brown candies.