SOLUTION: Matt thinks that he has a special relationship with the number 2. In particular, Matt thinks that he would roll a 2 with a fair 6-sided die more often than you'd expect by chance a

Question 1193649: Matt thinks that he has a special relationship with the number 2. In particular, Matt thinks that he would roll a 2 with a fair 6-sided die more often than you'd expect by chance alone. Suppose p is the true proportion of the time Matt will roll a 2.
Now suppose Matt makes n = 50 rolls, and a 2 comes up 11 times out of the 50 rolls. Determine the P-value of the test:
Answer by math_tutor2020(3835) (Show Source): You can put this solution on YOUR website!

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:
H1:
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
https://onlinestatbook.com/2/calculators/normal_dist.html
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.

Answer: The p-value is approximately 0.1813

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