Yes, your conclusions for 1 and 2 are correct.

[needhelp83]

A coin that is balanced should come up heads half the time in the long run. The population for coin tossing contains the results of tossing the coin forever. The parameter p is the probability of a head, which is the proportion of all tosses that give a head. The tosses we actually make are a random sample from this population. Count Buffon tosses a coin 4040 times.He got 2048 heads.

1. Test the null hypothesis that Buffon flipped a balanced coin against the t sided alternative. State the null and alt hypothesis in terms of p, calc an appropriate test stat, p-value, and interpret the p-value as it applies to this particular problem.

Ho: p=.5
Ha: p != .5

n = 4040
\hat{p}=\frac{2048}{4040}=.507

Test stat: \frac{\hat{p}-p_0}{\sqrt{p_0(1-p_0}/n}=\frac{.507-.5}{\sqrt{.5(1-.5}/4040}=\frac{.007}{.008}=.875

P-value = 2[1-\Phi(|.875|)]=2(1-.8106)=.3788

The pvalue is very high thus we cannot reject Ho in favor of Ha. Buffon's experiment doesn't show the coin is unbalanced.

2. Argue whether or not an 80% CI for p would contain 0.5
\widehat{p}\pm Z_{\alpha/2}\sqrt{\frac{\widehat{p}(1-\widehat{p}}{n}}=.507\pm 1.282\sqrt{\frac{.507(1-.507)}{4040}}=(.497,.517)

Yes, we are 80% confident that the probability of a head would range between .497 and .517. 0.5 is contained in this interval.

My questions are if I have properly concluded 1 and 2. Thanks!

 

[needhelp83]

Assuming this is correct, how do I perform Continuity Correction on this problem. I am not exactly sure how this works.

From what I am aware, you have to add .5 to the discrete value. Is this right?