Question 1193649
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p = population proportion of the number of times '2' shows up


If we assume each of the six sides are equally likely, then p should be p = 1/6 since one side is labeled '2' out of 6 sides total.
The one-sample proportion test aims to check that claim. 


Hypotheses:
H0: {{{p = 1/6}}}
H1: {{{p > 1/6}}}
The claim is in the alternative hypothesis (H1) because Matt thinks he would roll a '2' more often than expected.
The inequality sign in the alternative hypothesis determines that we have a right-tailed test.


x = number of times a '2' shows up = 11
n = number of rolls = 50


phat = sample proportion that estimates p
phat = x/n
phat = 11/50
phat = 0.22


SE = standard error
SE = sqrt(phat*(1-phat)/n)
SE = sqrt(0.22*(1-0.22)/50)
SE = 0.058583 which is approximate


z = test statistic
z = (phat - p)/SE
z = (0.22 - 1/6)/0.058583
z = 0.91038924830298
z = 0.910389 which is also approximate


Now use a calculator such as this one
<a href = "https://onlinestatbook.com/2/calculators/normal_dist.html">https://onlinestatbook.com/2/calculators/normal_dist.html</a>
to find that P(Z > 0.910389) = 0.1813 approximately
Recall that we're doing a right-tailed test, so this determines which portion we're shading under the curve. Specifically, we're shading to the right of z = 0.910389 and that area is roughly 0.1813


At the common alpha values of things like alpha = 0.05 and alpha = 0.10, the p-value of 0.1813 means we would fail to reject the null. We would conclude that p = 1/6 is indeed the case unless stronger evidence comes along to have us overturn the null. You only reject the null if the p-value is smaller than alpha.


<font color=red>Answer: The p-value is approximately 0.1813</font>
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