# Are more than 80% of Americans right-handed? To find out, 100 Americans were surveyed using simple random sampling. In that sample of 100 Americans, 87 were right-handed. Is there sufficient evidence at the 5% significance level to show that more than 80% of Americans are, in fact, right-handed? Check for normality to make sure we can proceed to conduct the required hypothesis test.

There is sufficient evidence to support the claim, with 95% confidence, that more than 80% of Americans are right-handed.

We are told that a sample of n=100 Americans were tested and that X=87 of them were right-handed. We are asked to determine, with 95% confidence, if it can be claimed that more than 80% of Americans are right-handed.

So, we have n=100, `p=.8, hat(p)=.87, hat(q)=.13` .

`np=80, nq=20`, and both are greater than 5, so we can perform the hypothesis test.

I. `H_0: p=.8`
`H_1:p>.8`, which is our claim.

II. The critical value is the z-score such that 95% of the population lies below it. From a standard normal table, we find that `z~~1.65` . A test value of 1.65 or greater will lie in the critical region.

III. The test value `z=(hat(p)-p)/sqrt((pq)/n)`, so `z=(.87-.8)/sqrt((.8*.2)/100)=.07/.04=1.75`.

IV. Since z=1.75 is greater than the critical value of 1.65, the test value lies in the critical region. Thus, we would reject the null-hypothesis. Alternatively, we can use a standard normal table to find the probability that a sample such as this or more extreme will occur. We find z=1.75 in a standard normal table and find that the probability that z>1.75 is about .0400. (My calculator gives .0400591135.) Thus, `p~~.0400<.05=alpha`, so we reject the null-hypothesis.

V. There is sufficient evidence to support the claim that at the 5% confidence level, more than 80% of Americans are right-handed.