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needhelp83
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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
[tex]\hat{p}=\frac{2048}{4040}=.507[/tex]
Test stat: [tex]\frac{\hat{p}-p_0}{\sqrt{p_0(1-p_0}/n}=\frac{.507-.5}{\sqrt{.5(1-.5}/4040}=\frac{.007}{.008}=.875[/tex]
P-value = [tex]2[1-\Phi(|.875|)]=2(1-.8106)=.3788[/tex]
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
[tex]\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)[/tex]
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!
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
[tex]\hat{p}=\frac{2048}{4040}=.507[/tex]
Test stat: [tex]\frac{\hat{p}-p_0}{\sqrt{p_0(1-p_0}/n}=\frac{.507-.5}{\sqrt{.5(1-.5}/4040}=\frac{.007}{.008}=.875[/tex]
P-value = [tex]2[1-\Phi(|.875|)]=2(1-.8106)=.3788[/tex]
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
[tex]\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)[/tex]
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!