2. A statistics professor has asked his students to flip coins over the years. He has kept track of how many flips land heads and how many land tails. Combining the results of his students over many years, he has formed a 95% confidence interval for the long-run population proportion of heads to be (.497, .513).
a. Why is this interval so narrow?
b. Suppose he were to conduct a hypothesis test of whether the long-run population proportion of heads differs from one-half. Based on this interval (do not conduct the test), would he reject the null hypothesis at the .05 significance level? Explain briefly (no more than one sentence).
c. Does the interval provide strong evidence that the long-run population proportion of heads is much different from one-half? Explain briefly.
It seems like you do not have a normal distribution, since the confidence interval goes from .003 below to .013 above .50. The confidence interval is determined by the standard error of the mean (SEm).
SEm = SD/√n
From your data, you would have an extreme large n.
However, without a normal distribution and more data, I cannot answer your last two questions. What was the sample mean? Do you have a typo?
a. The interval is narrow because the professor has collected a large amount of data over many years, which helps to reduce the variability and increase the precision of the estimated population proportion.
b. Based on this interval, the professor would not reject the null hypothesis at the 0.05 significance level because the interval includes the value of one-half.
c. The interval does not provide strong evidence that the long-run population proportion of heads is much different from one-half because the interval includes the value of one-half, indicating that the difference is not statistically significant at the 0.05 significance level.